Question 15 Marks
Solve the following quadratic equation:
$\frac{4}{\text{x}}-3=\frac{5}{\text{2x}+3},$ $\text{x}\neq=0,\ \frac{-3}{2}$
Answer$\frac{4}{\text{x}}-3=\frac{5}{\text{2x}+3}$ $\Rightarrow\frac{4-\text{3x}}{\text{x}}=\frac{5}{\text{2x}+3}$
$\Rightarrow (4 - 3x)(2x + 3) = 5x$
$\Rightarrow 8x + 12 - 6x^2 - 9x = 5x$
$\Rightarrow 12 - 6x^2 - x = 5x$
$\Rightarrow 12 - 6x^2 - 6x = 0$
$\Rightarrow 6x^2 + 6x - 12 = 0$
$\Rightarrow x^2 + x - 2 = 0$
$\Rightarrow x^2 + 2x - x - 2 = 0$
$\Rightarrow x(x + 2) - 1(x + 2) = 0$
$\Rightarrow (x + 2)(x - 1) = 0$
$\Rightarrow x + 2 = 0$ or $x - 1 = 0$
$\Rightarrow x = -2 $ or $x = 1$
View full question & answer→Question 25 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:$12abx^2 - (9a^2 - 8b^2)x - 6ab = 0$,
where $ \text{a}\neq0$ and $\text{b}\neq0$
AnswerGiven,$12abx^2 - (9a^2 - 8b^2)x - 6ab = 0$
On comparing it with $Ax^2 + Bx + C = 0$, we get:
$A = 12ab, B = -(9b^2 - 8b^2)$ and $C = -6ab$
Discriminant D is given by:
$D = B^2 - 4AC$
$= [-(9a^2 - 8b^2)]^2 - 4 \times 12ab \times (-6ab)$
$= 81a^4 - 144a^2b^2 + 64b^4 + 288a^2b^2$
$= 81a^4 + 144a^2b^2 + 64b^4$
$= (9a^2 + 8b^2) > 0$
Hence, the roots of the equation are equal.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(9\text{a}^2-8\text{b}^2)]+\sqrt{(9\text{a}^2+8\text{b}^2)^2}}{2\times12\text{a}\text{b}}$
$=\frac{9\text{a}^2-8\text{b}^2+9\text{a}^2+8\text{b}^2}{24\text{ab}}$
$=\frac{18\text{a}^2}{24\text{ab}}$
$=\frac{3\text{a}}{4\text{b}}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(9\text{a}^2-8\text{b}^2)]-\sqrt{(9\text{a}^2+8\text{b}^2)^2}}{2\times12\text{ab}}$
$=\frac{9\text{a}^2-8\text{b}^2-9\text{a}^2-8\text{b}^2}{24\text{ab}}$
$=\frac{-16\text{b}^2}{24\text{ab}}$
$=\frac{-2\text{b}}{3\text{a}}$
Thus, the roots of the equation are $\frac{3\text{a}}{4\text{b}}$ and $\frac{-2\text{b}}{3\text{a}}.$
View full question & answer→Question 35 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$a^2b^2x^2 - (4b^4 - 3a^4)x - 12a^2b^2 = 0$, $ \text{a}\neq0$ and $\text{b}\neq0$
AnswerThe Given equation is $a^2b^2x^2 - (4b^4 - 3a^4)x - 12a^2b^2 = 0.$
Comparing it with $Ax^2 + Bx + C = 0$, we get
$A = a^2b^2, B = -(4b^4 - 3a)$ and $C = -12a2b^2$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= [-(4b^4 - 3a^4)]^2 - 4 \times a^2b^2 \times (-12a^2b^2)$
$= 16b^8 - 24a^4b^4 + 9a^8 + 48a^4b^4$
$=16b^8 + 24a^4b^4 + 9a^8$
$= (4b^4 + 3a^4) > 0$
So, the given equation has real rooots.
Now, $\sqrt{\text{D}}=\sqrt{-(4\text{b}^4-3\text{a}^4)^2}$
$=4\text{b}^4+3\text{a}^4$
$\therefore$ $\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(4\text{b}^4-3\text{a}^4)]+(4\text{b}^4+3\text{a}^4)}{2\times\text{a}^2\text{b}^2}$
$=\frac{8\text{b}^4}{2\text{a}^2\text{b}^2}$
$=\frac{4\text{b}^2}{\text{a}^2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(4\text{b}^4-3\text{a}^4)]-(4\text{b}^4+3\text{a}^4)}{2\times\text{a}^2\text{b}^2}$
$=\frac{-6\text{a}^4}{2\text{a}^2\text{b}^2}$
$=\frac{-3\text{a}^2}{\text{b}^2}$
Hence, $\frac{4\text{b}^2}{\text{a}^2}$ and $-\frac{3\text{a}^2}{\text{b}^2}$ are the roots of the given equation.
View full question & answer→Question 45 Marks
Solve the following quadratic equation:
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$
$\text{x}\neq2,4$
Answer$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$\Rightarrow\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}-1)(\text{x}-4)+(\text{x}-3)(\text{x}-2)}{(\text{x}-2)(\text{x}-4)}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}^2-\text{4x}-\text{x}+4)+(\text{x}^2-\text{2x}-\text{3x}+6)}{\text{x}^2-\text{4x}-\text{2x}+8}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-5\text{x}+4+\text{x}^2-5\text{x}+6}{\text{x}^2-6\text{x}+8}=\frac{10}{3}$
$\Rightarrow\frac{\text{2x}^2-\text{10x}+10}{\text{x}^2-6\text{x}+8}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-\text{5x}+5}{\text{x}^2-6\text{x}+8}=\frac{5}{3}$
$\Rightarrow 3x^2 - 15x + 15 = 5x^2 - 30x + 40$
$\Rightarrow 2x^2 - 15x + 25 = 0$
$\Rightarrow 2x^2 - 10x - 5x + 25 = 0$
$\Rightarrow 2x(x - 5) - 5(x - 5) = 0$
$\Rightarrow (x - 5)(2x - 5) = 0$
$\Rightarrow x - 5 = 0$ or $2x - 15 = 0$
$\Rightarrow x = 5$ or $\text{x}=\frac{5}{2}$
View full question & answer→Question 55 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$4x^2 - 4a^2x + (a^4 - b^4) = 0$
AnswerThe given equation is $4x^2 - 4a^2x + (a^4 - b^4) = 0.$
Comparing it with $Ax^2 + Bx + C = 0$, we get
$A = 4, B = -4a^2$ and $C = a^4 - b^4$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= (-4a^2)^2 - 4 \times 4 \times (a^4 - b^4)$
$= 16a^4 - 16a^4 + 16b^4$
$= 16b^4 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{16\text{b}^4}=4\text{b}^2$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a}^2)+4\text{b}^2}{2\times4}$
$=\frac{4(\text{a}^2+\text{b}^2)}{8}$
$=\frac{\text{a}^2+\text{b}^2}{2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a}^2)-4\text{b}^2}{2\times4}$
$=\frac{4(\text{a}^2-\text{b}^2)}{8}$
$=\frac{\text{a}^2-\text{b}^2}{2}$
Hence, $\frac{1}{2}(\text{a}^2+\text{b}^2)$ and $\frac{1}{2}(\text{a}^2-\text{b}^2)$ are the roots of the given equation.
View full question & answer→Question 65 Marks
Solve the following quadratic equation:
$\Big(\frac{\text{4x}-3}{\text{2x}+1}\Big)-10\Big(\frac{\text{2x}+1}{\text{4x}-3}\Big)=3,$ $\text{x}\neq\frac{-1}{2},\ \frac{3}{4}$
Answer$\Big(\frac{\text{4x}-3}{\text{2x}+1}\Big)-10\Big(\frac{\text{2x}+1}{\text{4x}-3}\Big)=3$
Taking $\frac{\text{4x}-3}{\text{2x}+1}=\text{y},$
we have $\text{y}-\frac{10}{\text{y}}=3$
$\Rightarrow\frac{\text{y}^2-10}{\text{y}}=3$
$\Rightarrow y^2 - 10 = 3y $
$\Rightarrow y^2 - 3y - 10 = 0 $
$\Rightarrow y^2 - 5y + 2y - 10 = 0 $
$\Rightarrow y(y - 5) + 2(y - 5) = 0 $
$\Rightarrow (y - 5)(y + 2) = 0 $
$\Rightarrow y - 5 = 0$ or $y + 2 = 0$
$\Rightarrow y = 5$ or $y = -2$
$\Rightarrow\frac{\text{4x}-3}{\text{2x}+1}=5$ or $\frac{\text{4x}-3}{\text{2x}+1}=-2$
$\Rightarrow 4x - 3 = 10x + 5$ or $4x - 3 = -4x - 2$
$\Rightarrow 6x = -8$ or $8x = 1$
$\Rightarrow\text{x}=\frac{-4}{3}$ or $\text{x}=\frac{1}{8}$
View full question & answer→Question 75 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$x^2 - 4ax - b^2 + 4a^2 = 0$
AnswerThe given equation is $x^2 - 4ax - b^2 + 4a^2 = 0.$
Comparing it with $Ax^2 + Bx + C = 0$, we get $A = 1, B = -4a$ and $C = -b^2+ 4a^2$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= (-4a)^2 - 4 \times 1 \times (-b^2 + 4a^2)$
$ = 16a^2 + 4b^2 - 16a^2 $
$= 4b^2 > 0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{4\text{b}^2}=2\text{b}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a})+2\text{b}}{2\times1}$
$=\frac{4\text{a}+2\text{b}}{2}$
$=2\text{a}+\text{b}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a})-2\text{b}}{2\times1}$
$=\frac{4\text{a}-2\text{b}}{2}$
$=2\text{a}-\text{b}$
Hence, $(2a + b)$ and $(2a - b)$ are the roots of the given equation.
View full question & answer→Question 85 Marks
Solve the following quadratic equation:
$\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=3\frac{1}{3},$ $\text{x}\neq5,7$
Answer$\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=3\frac{1}{3}$
$\Rightarrow\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}-4)(\text{x}-7)+(\text{x}-6)(\text{x}-5)}{(\text{x}-5)(\text{x}-7)}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}^2-\text{7x}-\text{4x}+28)+(\text{x}^2-\text{5x}-\text{6x}+30)}{\text{x}^2-\text{7x}-\text{5x}+35}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-11\text{x}+28+\text{x}^2-11\text{x}+30}{\text{x}^2-12\text{x}+35}=\frac{10}{3}$
$\Rightarrow\frac{\text{2x}^2-\text{22x}+58}{\text{x}^2-12\text{x}+35}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-\text{11x}+29}{\text{x}^2-12\text{x}+35}=\frac{5}{3}$
$\Rightarrow 3x^2 - 33x + 87 = 5x^2 - 60x + 175$
$\Rightarrow 2x^2 - 27x + 88 = 0$
$\Rightarrow 2x^2 - 16x - 11x + 88 = 0$
$\Rightarrow 2x(x - 8) - 11(x - 8) = 0$
$\Rightarrow (x - 8)(2x - 11) = 0$
$\Rightarrow x - 8 = 0$ or $2x - 11 = 0$
$\Rightarrow x = 8$ or $\text{x}=\frac{11}{2}$
$\Rightarrow x = 8$ or $\text{x}=5\frac{1}{2}$
View full question & answer→Question 95 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 + 6x - (a^2 + 2a -8) = 0$
AnswerThe given equation is $x^2 + 6x - (a^2 + 2a - 8) =0$.
Comparing it with $Ax^2 + Bx + C = 0$, we get $A = 1, B = 6$ and $C = -(a^2 + 2a - 8)$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= 6^2- 4 \times 1 \times [-(a^2 + 2a - 8)] $
$= 36 + 4a^2 + 8a - 32 $
$= 4a^2 + 8a + 4 $
$= 4(a^2 + 2a +1) $
$= 4(a + 1)^2 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{4(\text{a}+1)^2}=2(\text{a}+1)$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-6+2(\text{a}+1)}{2\times1}$
$=\frac{2\text{a}-4}{2}$
$=\text{a}-2$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-6-2(\text{a}+1)}{2\times1}$
$=\frac{-2\text{a}-8}{2}$
$=-\text{a}-4$
$=-(\text{a}+4)$
Hence, $(a - 2)$ and $-(a + 4)$ are the roots of the given equation.
View full question & answer→Question 105 Marks
Solve the following quadratic equation:
$9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0$
Answer$9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2)$
$\Rightarrow 9x^2 - 9(a + b)x + (2a^2 + 4ab + ab + 2b^2) = 0$
$\Rightarrow 9x^2 - 9(a + b)x + [2a(a + 2b) + b(a + 2b)] = 0$
$\Rightarrow 9x^2 - 9(a + b)x + (a + 2b)(2a + b) = 0$
$\Rightarrow 9x^2 - 3(a + 2b)x - 3(2a + b)x + (a + 2b)(2a + b) = 0$
$\Rightarrow 3x[3x - (a + 2b)] - (2a + b)[3x - (a + 2b)] = 0$
$\Rightarrow [3x - (a + 2b)][3x - (2a + b)] = 0$
$\Rightarrow 3x - (a + 2b) = 0$ or $3x - (2a + b) = 0$
$\Rightarrow\text{x}=\frac{\text{a}+\text{2b}}{3}$ or $\text{x}=\frac{\text{2a}+\text{b}}{3}$
View full question & answer→Question 115 Marks
Solve the following quadratic equation:
$\frac{1}{\text{2x}-3}+\frac{1}{\text{x}-5}=1\frac{1}{9},$ $\text{x}\neq\frac{3}{2},\ 5$
Answer$\frac{1}{\text{2x}-3}+\frac{1}{\text{x}-5}=1\frac{1}{9}$
$\Rightarrow\frac{1}{\text{2x}-3}+\frac{1}{\text{x}-5}=\frac{10}{9}$
$\Rightarrow\frac{\text{x}-5+\text{2x}-3}{(\text{2x}-3)(\text{x}-5)}=\frac{10}{9}$
$\Rightarrow\frac{\text{3x}-8}{(\text{2x}-3)(\text{x}-5)}=\frac{10}{9}$
$\Rightarrow 27x - 72 = 10[(2x - 3)(x - 5)]$
$\Rightarrow 27x - 72 = 10[2x^2 - 10x - 3x + 15]$
$\Rightarrow 27x - 72 = 10[2x^2 - 13x + 15]$
$\Rightarrow 27x - 72 = 20x^2 - 130x + 150$
$\Rightarrow 20x^2 - 157x + 222 = 0$
So, $a = 20, b = -157, c = 222$
So, $\text{x}=\frac{157\pm\sqrt{(-157)^2-4(20)(222)}}{40}$
$\Rightarrow\text{x}=\frac{157\pm\sqrt{24649-17760}}{40}$
$\Rightarrow\text{x}=\frac{157\pm\sqrt{6889}}{40}$
$\Rightarrow\text{x}=\frac{157\pm83}{40}$
$\Rightarrow\text{x}=\frac{157+83}{40}$ or $\text{x}=\frac{157-83}{40}$
$\Rightarrow\text{x}=6$ or $\text{x}=1.85$
View full question & answer→Question 125 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$4x^2 + 4bx - (a^2 - b^2) = 0$
AnswerThe given equation is $4x^2 + 4bx - (a^2 - b^2) = 0.$
Comparing it with $Ax^2 + Bx + C = 0$, we get
$A = 4, B = 4b$ and $C = -(a^2 - b^2)$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= (4b)^2 - 4 \times 4 \times [-(a^2 - b^2)]$
$= 16b^2 + 16a^2 -16b^2$
$= 16a^2 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{16\text{a}^2}=4\text{a}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-4\text{b}+4\text{a}}{2\times4}$
$=\frac{4(\text{a}-\text{b})}{8}$
$=\frac{\text{a}-\text{b}}{2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-4\text{b}-4\text{a}}{2\times4}$
$=\frac{-4(\text{a}+\text{b})}{8}$
$=-\frac{\text{a}+\text{b}}{2}$
Hence, $\frac{1}{2}(\text{a}-\text{b})$ and $-\frac{1}{2}(\text{a}+\text{b})$ are the roots of the given equation.
View full question & answer→Question 135 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\text{x}+\frac{1}{\text{x}}=3,\ \text{x}\neq0$
AnswerThe given equation is:$\text{x}+\frac{1}{\text{x}}=3,\ \text{x}\neq0$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=3$
$\Rightarrow\text{x}^2+1=3\text{x}$
$\Rightarrow\text{x}^2-3\text{x}+1=0$
The equation is of the form $ax^2 + bx + c = 0$, where $a = 1, b = -3$ and $c = 1$.
$\therefore$ Discriminant,$ D = b^2- 4ac=(-3)^2 - 4 \times 1 \times 1$
$= 9 - 4 = 5 > 0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt5$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-3)+\sqrt5}{2\times1}$
$=\frac{3+\sqrt5}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-3)-\sqrt5}{2\times1}$
$=\frac{3-\sqrt5}{2}$
Hence, $\frac{3+\sqrt5}{2}$ and $\frac{3-\sqrt5}{2}$ are the roots of the given equation.
View full question & answer→Question 145 Marks
Solve the following quadratic equation:
$\frac{\text{3x}-4}{\text{7}}+\frac{\text{7}}{\text{3x}-4}=\frac{5}{2},$ $\text{x}\neq\frac{4}{3}$
Answer$\frac{\text{3x}-4}{\text{7}}+\frac{\text{7}}{\text{3x}-4}=\frac{5}{2}$
Taking $\frac{\text{3x}-4}{7}=\text{y},$ we have
$\text{y}+\frac{1}{\text{y}}=\frac{5}{2}$
$\Rightarrow\frac{\text{y}^2+1}{\text{y}}=\frac{5}{2}$
$\Rightarrow 2x^2 + 2 = 5y = 0$
$\Rightarrow 2y^2 - 5y + 2 = 0$
$\Rightarrow 2y^2 - 4y - y + 2 = 0$
$\Rightarrow 2y(y - 2) - 1(y - 2) = 0$
$\Rightarrow (y - 2)(2y - 1) = 0$
$\Rightarrow y - 2 = 0$ or $2y - 1 = 0$
$\Rightarrow y = 2$ or $\text{y}=\frac{\text{1}}{2}$
$\Rightarrow\frac{\text{3x}-4}{7}=2$ or $\frac{\text{3x}-4}{7}=\frac{1}{2}$
$\Rightarrow 3x - 4 = 14$ or $6x - 8 = 7$
$\Rightarrow 3x = 18$ or $6x = 15$
$\Rightarrow x = 6$ or $\text{x}=\frac{5}{2}$
View full question & answer→Question 155 Marks
Solve the following quadratic equation:
$3\Big(\frac{\text{3x}-1}{\text{2x}+3}\Big)-2\Big(\frac{\text{2x}+3}{\text{3x}-1}\Big)=5$ $\text{x}\neq\frac{1}{3},\ \frac{-3}{2}$
Answer$3\Big(\frac{\text{3x}-1}{\text{2x}+3}\Big)-2\Big(\frac{\text{2x}+3}{\text{3x}-1}\Big)=5$
Taking $\frac{\text{3x}-1}{\text{2x}+3}=\text{y},$ we have
$\text{3y}-\frac{2}{\text{y}}=5$
$\Rightarrow\frac{\text{3y}^2-2}{\text{y}}=5$
$\Rightarrow 3y^2 - 2 = 5y$
$\Rightarrow 3y^2 - 5y - 2 = 0$
$\Rightarrow 3y^2 - 6y + y - 2 = 0$
$\Rightarrow 3y(y - 2) + 1(y - 2) = 0$
$\Rightarrow (y - 2)(3y + 1) = 0$
$\Rightarrow y - 2 = 0$ or $3y + 1 = 0$
$\Rightarrow y = 2$ or $\text{y}=\frac{-1}{3}$
$\Rightarrow\frac{\text{3x}-1}{\text{2x}+3}=2$ or $\frac{\text{3x}-1}{\text{2x}+3}=\frac{-1}{3}$
$\Rightarrow 3x - 1 = 4x + 6$ or $9x - 3 = -2x - 3$
$\Rightarrow x = -7$ or $11x = 0$
$\Rightarrow x = -7$ or $x = 0$
View full question & answer→Question 165 Marks
Solve the following quadratic equation:
$\frac{\text{1}}{\text{x}+1}+\frac{\text{3}}{\text{5x}+1}=\frac{5}{\text{x}+4},$ $\text{x}\neq-1,-\frac{1}{5},-4$
Answer$\frac{\text{1}}{\text{x}+1}+\frac{\text{3}}{\text{5x}+1}=\frac{5}{\text{x}+4}\ \dots(1)$
Equation (1):
$\frac{\text{1}}{\text{x}+1}+\frac{\text{3}}{\text{5x}+1}=\frac{5}{\text{x}+4}$
Taking LCM
$\frac{(\text{5x}+1)+3(\text{x}+1)}{(\text{x}+1)(\text{5x}+1)}=\frac{5}{\text{x}+4}$
$\frac{\text{5x}+1+3\text{x}+3}{(\text{5x}^2+\text{6x}+1)}=\frac{5}{\text{x}+4}$
$\frac{\text{8x}+4}{(\text{5x}^2+\text{6x}+1)}=\frac{5}{\text{x}+4}$
Cross multiply
$\Rightarrow (8x + 4)(x + 4) = (5x^2 + 6x + 1)5$
$\Rightarrow 8x^2 - 25x^2 + 36x - 30x + 16 - 5 = 0$
$\Rightarrow -17x^2 + 6x + 11 = 0$
$\Rightarrow 17x^2 - 6x - 11 = 0$
$\Rightarrow 17x^2 - 17x + 11x - 11 = 0$ (Middle term split)
$\Rightarrow (x - 1)(17x + 11) = 0$
$\Rightarrow x = 1$ or $\text{x}=\frac{-11}{17}$
View full question & answer→Question 175 Marks
Solve the following quadratic equation:
$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7},$ $\text{x}\neq1,\ -5$
Answer$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7}$
$\Rightarrow\frac{\text{x}+5-\text{x}+1}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{6}{\text{x}^2+\text{5x}-\text{x}-5}=\frac{6}{7}$
$\Rightarrow\frac{1}{\text{x}^2+\text{4x}-5}=\frac{1}{7}$
$\Rightarrow 7 = x^2 + 4x - 5$
$\Rightarrow x^2 + 4x - 12 = 0$
$\Rightarrow x^2 + 6x - 2x - 12 = 0$
$\Rightarrow x(x + 6) - 2(x + 6) = 0$
$\Rightarrow (x + 6)(x - 2) = 0$
$\Rightarrow x + 6 = 0$ or $x - 2 = 0$
$\Rightarrow x = -6$ or $x = 2$
View full question & answer→Question 185 Marks
Solve the following quadratic equation:
$\frac{\text{x}}{\text{x}+1}+\frac{\text{x}+1}{\text{x}}=2\frac{4}{15},$ $\text{x}\neq0,-1$
Answer$\frac{\text{x}}{\text{x}+1}+\frac{\text{x}+1}{\text{x}}=2\frac{4}{15}$
$\Rightarrow\frac{\text{x}}{\text{x}+1}+\frac{\text{x}+1}{\text{x}}=\frac{34}{15}$
Taking $\frac{\text{x}}{\text{x}+1}=\text{y},$ we have
$\text{y}+\frac{1}{\text{y}}=\frac{34}{15}$
$\Rightarrow\frac{\text{y}^2+1}{\text{y}}=\frac{34}{15}$
$\Rightarrow 15y^2 + 15 = 34y = 0$
$\Rightarrow 15y^2 - 34y + 15 = 0$
$\Rightarrow 15y^2 - 25y - 9y + 15 = 0$
$\Rightarrow 5y(3y - 5) - 3(3y - 5) = 0$
$\Rightarrow (3y - 5)(5y - 3) = 0$
$\Rightarrow 3y - 5 = 0$ or $5y - 3 = 0$
$\Rightarrow\text{y}=\frac{5}{3}$ or $\text{y}=\frac{\text{3}}{5}$
$\Rightarrow\frac{\text{x}}{\text{x}+1}=\frac{5}{3}$ or $\frac{\text{x}}{\text{x}+1}=\frac{3}{5}$
$\Rightarrow 3x = 5x + 5$ or $5x = 3x + 3$
$\Rightarrow 2x = -5$ or $2x = 3$
$\Rightarrow\text{x}=\frac{-5}{2}$ or $\text{x}=\frac{3}{2}$
View full question & answer→Question 195 Marks
Solve the following quadratic equation:
$4^{(x+1)} + 4^{(1-x)} = 10$
Answer$4^{(x+1)} + 4^{(1-x)} = 10$
$4^x.4^1 + 4^1.4^{-x} = 10$
$\Rightarrow\text{4y}+\frac{4}{\text{y}}=10$ where $4^x = y$
$\Rightarrow 4y^2 - 10y + 4 = 0$
$\Rightarrow 4y^2 - 8y - 2y + 4 = 0$
$\Rightarrow 4y(y - 2) - 2(y - 2) = 0$
$\Rightarrow (y - 2)(4y - 2) = 0$
$\Rightarrow y - 2 = 0$ or $4y - 2 = 0$
$\Rightarrow y = 2$ or $\text{y}=\frac{2}{4}=\frac{1}2{}$
$\Rightarrow y = 2$ or $\text{y}=\frac{1}{2}$
In case I:
$\Rightarrow 4^x = 2$
$\Rightarrow (2)^{2x} = (2)^1$
$\Rightarrow 2x = 1$
$\Rightarrow\text{x}=\frac{1}{2}$
In case II:
$\Rightarrow4^\text{x}=\frac{1}{2}$
$\Rightarrow(2)^\text{2x}=\Big(\frac{1}{2}\Big)^1$
$\Rightarrow(2)^{\text{2x}}=(2)^{-1}$
$\Rightarrow\text{x}=-\frac{1}{2}$
Hence, $\frac{1}{2},-\frac{1}2{}$ are the roots of given equation.
View full question & answer→Question 205 Marks
Solve the following quadratic equation:
$\frac{\text{a}}{(\text{x}-\text{b})}+\frac{\text{b}}{(\text{x}-\text{a})}=2,$ $\text{x}\neq-\text{b},\ \text{a}$
AnswerThe given equation:
$\Big(\frac{\text{a}}{\text{x}-\text{b}}-1\Big)+\Big(\frac{\text{b}}{\text{x}-\text{a}}-1\Big)=0$
$\Rightarrow\frac{(\text{a}-\text{x}+\text{b})}{(\text{x}-\text{b})}+\frac{(\text{b}-\text{x}+\text{a})}{(\text{x}-\text{a})}=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})\Big[\frac{1}{(\text{x}-\text{b})}+\frac{1}{(\text{x}-\text{a})}\Big]=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})\Big[\frac{\text{2x}-(\text{a}+\text{b})}{(\text{x}-\text{a})(\text{x}-\text{b})}\Big]=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})[\text{2x}-(\text{a}+\text{b})]=0$
$\Rightarrow\text{x}=(\text{a}+\text{b})$ or $\text{x}=\frac{(\text{a}+\text{b})}{2}$
Hence, (a + b) and $\frac{(\text{a}+\text{b})}{2}$ is the root of the given equation.
View full question & answer→Question 215 Marks
Solve the following quadratic equation:
$\frac{\text{x}}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}=4\frac{1}{4},$ $\text{x}\neq0,1$
Answer$\frac{\text{x}}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}=4\frac{1}{4}$
$\Rightarrow\frac{\text{x}}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}=\frac{17}{4}$
Taking $\frac{\text{x}}{\text{x}-1}=\text{y},$ we have
$\text{y}+\frac{1}{\text{y}}=\frac{17}{4}$
$\Rightarrow\frac{\text{y}^2+1}{\text{y}}=\frac{17}{4}$
$\Rightarrow 4y^2 + 4 = 17y = 0$
$\Rightarrow 4y^2 - 17y + 4 = 0$
$\Rightarrow 4y^2 - 16y - y + 4 = 0$
$\Rightarrow 4y(y - 4) - 1(y - 4) = 0$
$\Rightarrow (y - 4)(4y - 1) = 0$
$\Rightarrow y - 4 = 0$ or $4y - 1 = 0$
$\Rightarrow y = 4$ or $\text{y}=\frac{\text{1}}{4}$
$\Rightarrow\frac{\text{x}}{\text{x}-1}=4$ or $\frac{\text{x}}{\text{x}-1}=\frac{1}{4}$
$\Rightarrow x = 4x - 4$ or $4x = x - 1$
$\Rightarrow 3x = 4$ or $3x = -1$
$\Rightarrow\text{x}=\frac{4}{3}$ or $\text{x}=\frac{-1}{3}$
View full question & answer→Question 225 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$x^2 + 5x - (a^2 + a - 6) = 0$
AnswerThe given equation is $x^2 + 5x - (a^2 + a - 6) = 0$. Comparing it with $Ax^2 + Bx + C = 0,$
we get $A = 1, B = 5 $and $C = -(a^2 + a - 6)$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= 5^2 - 4 \times 1 \times [-(a^2 + a - 6)] = 25 + 4a^2 + 4a - 24 = 4a^2 + 4a + 1 = (2a + 1)^2 > 0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{(2\text{a}+1)^2}=2\text{a}+1$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-5+2\text{a}+1}{2\times1}$
$=\frac{2\text{a}-4}{2}$
$=\text{a}-2$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-5-(2\text{a}+1)}{2\times1}$
$=\frac{-2\text{a}-6}{2}$
$=-\text{a}-3$
$=-(\text{a}+3)$
Hence, (a - 2) and -(a + 3) are the roots of the given equation.
View full question & answer→Question 235 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 2ax - (4b^2 - a^2) = 0$
AnswerThe given equation is $x^2 - 2ax - (4b^2 - a^2) = 0$. Comparing it with $Ax^2 + Bx + C = 0,$
we get $A = 1, B = -2a$ and $C = -(4b^2 - a^2)$
$\therefore$ Discrimiant,$ D = B^2 - 4AC$
$= (-2a)^2 - 4 \times 1 \times [-(4b^2 - a^2)] = 4a^2 + 16b^2 - 4a^2 = 16b^2 > 0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{16\text{b}^2}=4\text{b}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-2\text{a})+4\text{b}}{2\times1}$
$=\frac{2(\text{a}+2\text{b})}{2}$
$=\text{a}+2\text{b}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-2\text{a}-4\text{b})}{2\times1}$
$=\frac{2(\text{a}-2\text{b})}{2}$
$=\text{a}-2\text{b}$
Hence, a + 2b and a - 2b are the roots of the given equation.
View full question & answer→Question 245 Marks
Solve the following quadratic equation:
$\Big(\frac{\text{x}}{\text{x}}+1\Big)^2-5\Big(\frac{\text{x}}{\text{x}+1}\Big)+6=0,$ $\text{x}\neq-1$
AnswerPutting $\Big(\frac{\text{x}}{\text{x}+1}\Big)=\text{y},$ the given equation become.
$\Rightarrow y^2 - 5y + 6 = 0 $
$\Rightarrow y^2 - 3y - 2y + 6 = 0 $
$\Rightarrow y(y - 3) - 2(y - 3) = 0 $
$\Rightarrow (y - 3)(y - 2) = 0 $
$\Rightarrow y - 3 = 0$ or $y - 2 = 0$
$\Rightarrow y = 3$ or $y = 2$
Case I:
$y = 3$
$\Rightarrow\frac{\text{x}}{\text{x}+1}=3 $
$\Rightarrow 3x + 3 = x $
$\Rightarrow 3x - x = 3 $
$\Rightarrow 2x = 3$
$\Rightarrow\text{x}=\frac{-3}{2}$
Case II:
$y = 2$
$\Rightarrow\frac{\text{x}}{\text{x}+1}=2 $
$\Rightarrow 2x + 2 = x $
$\Rightarrow 2x - x = -2$
$\Rightarrow x = -2$
Hence, $\frac{-3}{2},\ -2$ are the roots of the given equation.
View full question & answer→Question 255 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula: $x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
AnswerThe given equation is $x^2 - (2b - 1)x + (b^2 - b - 20) = 0.$
Comparing it with $Ax^2 + Bx + C = 0$, we get
$A = 1, B = -(2b - 1)$ and $C = b^2 - b - 20$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= [-(2b - 1)]^2 - 4 \times 1 \times (b^2 - b - 20)$
$= 4b^2 - 4b + 1 - 4b^2 + 4b + 80$
$= 81 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{81}=9$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(2\text{b}-1)]+9}{2\times1}$
$=\frac{2\text{b}+8}{2}$
$=\text{b}+4$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(2\text{b}-1)]-9}{2\times1}$
$=\frac{2\text{b}-10}{2}$
$=\text{b}-5$
Hence, $(b + 4)$ and $(b - 5)$ are the roots of the given equation.
View full question & answer→Question 265 Marks
Solve the following quadratic equation: $12abx^2 - (9a^2 - 8b^2)x - 6ab = 0$
Answer$12abx^2 - (9a^2 - 8b^2)x - 6ab = 0$
$\Rightarrow 12abx^2 - 9a^2x + 8b^2x - 6ab = 0$
$\Rightarrow 3ax(4bx - 3a) + 2b(4bx - 3a) = 0$
$\Rightarrow (4bx - 3a)(3ax + 2b) = 0$
$\Rightarrow (4bx - 3a) = 0$ or $(3ax + 2b) = 0$
$\Rightarrow 4bx = 3a$ or $3ax = -2b$
$\Rightarrow\text{x}=\frac{\text{3a}}{\text{4b}}$ or $\text{x}=\frac{-\text{2b}^2}{\text{3a}}$
Hence, $\frac{\text{3a}}{\text{4b}}$ and $\frac{-\text{2b}}{\text{3a}}$ are the roots of the given equation.
View full question & answer→Question 275 Marks
Solve the following quadratic equation:
$\frac{\text{x}+3}{\text{x}-2}-\frac{1-\text{x}}{\text{x}}=4\frac{1}{4},$ $\text{x}\neq2,\ 0$
Answer$\frac{\text{x}+3}{\text{x}-2}-\frac{1-\text{x}}{\text{x}}=4\frac{1}{4}$
$\Rightarrow\frac{\text{x}(\text{x}+3)-(1-\text{x})(\text{x}-2)}{\text{x}(\text{x}-2)}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+\text{3x}-(\text{x}-2-\text{x}^2+\text{2x})}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+\text{3x}-(\text{3x}-2-\text{x}^2)}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+\text{3x}-\text{3x}+2+\text{x}^2}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{2x}^2+2}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow 8x^2 + 8 = 17x^2 - 34x = 0$
$\Rightarrow 9x^2 - 34x - 8 = 0$
$\Rightarrow 9x^2 - 36x + 2x - 8 = 0$
$\Rightarrow 9x(x - 4) + 2(x - 4) = 0$
$\Rightarrow (x - 4)(9x + 2) = 0$
$\Rightarrow x - 4 = 0$ or $9x + 2 = 0$
$\Rightarrow x = 4$ or $\text{x}=\frac{-\text{2}}{9}$
View full question & answer→Question 285 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$9x^2 + 3kx + 4 = 0$
AnswerThe given equation is $9x^2 + 3kx + 4 = 0$
$\therefore$ $D = (3k)^2 - 4 \times 9 \times 4$
$D = 9k^2 - 144$
The given equation has real and distinct roots if $D > 0$.
$\therefore$ $9k^2 - 144 > 0$
$\Rightarrow 9(k^2 - 16) > 0$
$\Rightarrow (k - 4)(k + 4) > 0$
$\Rightarrow k < -4 or k > 4$
View full question & answer→Question 295 Marks
Solve the following quadratic equation:
$\frac{1}{\text{2a}+\text{b}+\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}+\frac{1}{\text{2x}}$
Answer$\frac{1}{\text{2a}+\text{b}+\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}+\frac{1}{\text{2x}}$
$\Rightarrow\frac{1}{\text{2a}+\text{b}+\text{2x}}-\frac{1}{\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}$
$\Rightarrow\frac{\text{2x}-(\text{2a}+\text{b}+\text{2x})}{\text{2x}(\text{2a}+\text{b}+\text{2x})}=\frac{\text{b}+\text{2a}}{\text{2ab}{}}$
$\Rightarrow\frac{\text{2x}-\text{2a}-\text{b}-\text{2x}}{\text{4ax}+\text{2bx}+\text{4x}^2}=\frac{\text{b}+\text{2a}}{\text{2ab}{}}$
$\Rightarrow\frac{-(\text{2a}+\text{b})}{\text{2x}(\text{2a}+\text{b}+\text{2x})}=\frac{\text{b}+\text{2a}}{\text{2ab}}$
$\Rightarrow\frac{-1}{\text{x}(\text{2a}+\text{b}+\text{2x})}=\frac{1}{\text{ab}}$
$\Rightarrow -ab = 2ax + bx + 2x^2$
$\Rightarrow 2x^2 + bx + 2ax + ab = 0$
$\Rightarrow 2x^2 + 2ax + bx + ab = 0$
$\Rightarrow 2x(x + a) + b(x + a) = 0$
$\Rightarrow (x + a)(2x + b) = 0$
$\Rightarrow x + a = 0$ or $2x + b = 0$
$\Rightarrow x = -a$ or $\text{x}=\frac{-\text{b}}{2}$
View full question & answer→Question 305 Marks
Solve the following equations by using the method of completing the square:
$7x^2 + 3x - 4 = 0$
Answer$7x^2 + 3x - 4 = 0$
$\Rightarrow 49x^2 + 21x - 28 = 0$ (Multiplying both sides by $7$)
$\Rightarrow 49x^2 + 21x = 28$
$\Rightarrow(\text{7x})^2+2\times\text{7x}\times\frac{3}{2}+\Big(\frac{3}{2}\Big)^2\\=28+\Big(\frac{3}{2}\Big)^2$
$\Big[$Adding $\Big(\frac{3}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{7x}+\frac{3}{2}\Big)^2$
$=29+\frac{9}{4}$
$=\frac{121}{4}=\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{7x}-\frac{3}{2}=\pm\frac{11}{2}$ (Taking square root on both sides)
$\Rightarrow\text{7x}+\frac{3}{2}=\frac{11}{2}$ or $\text{7x}+\frac{3}{2}=-\frac{11}{2}$
$\Rightarrow\text{7x}=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4$ or $\text{7x}=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$
$\Rightarrow\text{x}=\frac{4}{7}$ or $x = -1$
Hence, $\frac{4}{7}$ and $-1$ are the roots of the given equation.
View full question & answer→Question 315 Marks
Solve the following quadratic equation:
$\frac{1}{\text{x}-2}+\frac{2}{\text{x}-1}=\frac{6}{\text{x}}$ $\text{x}\neq0,\ 1,\ 2$
Answer$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x} \Rightarrow \frac{x-1+2 x-4}{(x-2)(x-1)}=\frac{6}{x} \Rightarrow \frac{3 x-5}{x^2-x-2 x+2}=\frac{6}{x} \Rightarrow \frac{3 x-5}{x^2-3 x+2}=\frac{6}{x} \Rightarrow 3 x^2-5 x=6 x^2-18 x+12$
$\Rightarrow 3 x^2-13+12=0$
$\Rightarrow 3 x^2-9 x-4 x+12=0$
$\Rightarrow 3 x(x-3)-4(x-3)=0$
$\Rightarrow(x-3)(3 x-4)=0$
$\Rightarrow x-3=0 \text { or } 3 x-4=0$
$\Rightarrow x=3 \text { or } x=\frac{4}{3}$
View full question & answer→Question 325 Marks
Solve the following quadratic equation:
$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{\text{3x}-1},$ $\text{x}\neq-1,\ \frac{1}{3}$
Answer$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{\text{3x}-1}$
$\Rightarrow\frac{3}{\text{x}+1}-\frac{2}{\text{3x}-1}=\frac{1}{2}$
$\Rightarrow\frac{3(\text{3x}-1)-2(\text{x}+1)}{(\text{x}+1)(\text{3x}-1)}=\frac{1}{2}$
$\Rightarrow\frac{\text{9x}-3-\text{2x}-2}{\text{3x}^2-\text{x}+\text{3x}-1}=\frac{1}{2}$
$\Rightarrow\frac{\text{7x}-5}{\text{3x}^2+\text{2x}-1}=\frac{1}{2}$
$\Rightarrow 14x - 10 = 3x^2 + 2x - 1$
$\Rightarrow 3x^2 + 2x - 1 - 14x + 10 = 0$
$\Rightarrow 3x^2 - 12x + 9 = 0$
$\Rightarrow x^2 - 4x + 3 = 0$
$\Rightarrow x^2 - 3x - x + 3 = 0$
$\Rightarrow x(x - 3) - 1(x - 3) = 0$
$\Rightarrow (x - 3)(x - 1) = 0$
$\Rightarrow x - 3 = 0$ or $x - 1 = 0$
$\Rightarrow x = 3$ or $x = 1$
View full question & answer→Question 335 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$2\text{x}^2+5\sqrt3\text{x}+6=0$
AnswerThe given equation is $2\text{x}^2+5\sqrt3\text{x}+6=0$Comparing it with $ax^2 + bx + c = 0$, we get
$\text{a}=2,\ \text{b}=5\sqrt3$ and $\text{c}=6$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=\big(5\sqrt3\big)^2-4\times2\times6$
$=75-48$
$=27>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{27}=3\sqrt3$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5\sqrt3+3\sqrt3}{2\times2}$
$=\frac{-2\sqrt3}{4}$
$=-\frac{\sqrt3}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5\sqrt3-3\sqrt3}{2\times2}$
$=\frac{-8\sqrt3}{4}$
$=-2\sqrt3$
Hence, $-\frac{\sqrt3}{2}$ and $-2\sqrt3$ are the roots of the given equation.
View full question & answer→Question 345 Marks
Solve the following quadratic equation:
$\frac{\text{a}}{(\text{ax}-\text{1})}+\frac{\text{b}}{(\text{bx}-\text{1})}=(\text{a}+\text{b}),$ $\text{x}\neq\frac{1}{\text{a}},\ \frac{1}{\text{b}}$
Answer$\frac{\text{a}}{(\text{ax}-\text{1})}+\frac{\text{b}}{(\text{bx}-\text{1})}=(\text{a}+\text{b})$
$\Rightarrow\Big[\frac{\text{a}}{\text{ax}-1}-\text{b}\Big]+\Big[\frac{\text{b}}{\text{bx}-1}-\text{a}\Big]=0$
$\Rightarrow\Big[\frac{\text{a}-\text{abx}+\text{b}}{\text{ax}-1}\Big]+\Big[\frac{\text{b}-\text{abx}+\text{a}}{\text{bx}-1}\Big]=0$
$\Rightarrow(\text{a}-\text{abx}+\text{b})\Big[\frac{1}{\text{ax}-1}+\frac{1}{\text{bx}-1}\Big]=0$
$\Rightarrow(\text{a}-\text{abx}+\text{b})$ or $\frac{1}{\text{ax}-1}+\frac{1}{\text{bx}-1}=0$
$\Rightarrow\text{abx}=\text{a}+\text{b}$ or $\frac{1}{\text{ax}-1}=-\frac{1}{\text{bx}-1}$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{bx} - 1 = -\text{ax} + 1$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{bx} + \text{ax} = 2$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{x}(\text{b} + \text{a}) = 2$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{x}=\frac{2}{\text{a}+\text{b}}$
View full question & answer→Question 355 Marks
Solve the following quadratic equation:
$\frac{\text{1}}{\text{x}+1}+\frac{\text{2}}{\text{x}+2}=\frac{5}{\text{x}+4},$ $\text{x}\neq-1,-2,-4$
Answer$\frac{\text{1}}{\text{x}+1}+\frac{\text{2}}{\text{x}+2}=\frac{5}{\text{x}+4}$
$\Rightarrow\frac{\text{x}+2+\text{2x}+2}{(\text{x}+1)(\text{x}+2)}=\frac{5}{\text{x}+4}$
$\Rightarrow\frac{\text{3x}+4}{\text{x}^2+\text{3x}+2}=\frac{5}{\text{x}+4}$
$\Rightarrow (3x + 4)(x + 4) = 5(x^2 + 3x + 2)$
$\Rightarrow 3x^2 + 16x + 16 = 5x^2 + 15x + 10$
$\Rightarrow 2x^2 - x - 6 = 0$
$\Rightarrow 2x^2 - 4x + 3x - 6 = 0$
$\Rightarrow 2x(x - 2) + 3(x - 2) = 0$
$\Rightarrow (x - 2)(2x + 3) = 0$
$\Rightarrow x - 2 = 0$ or $2x + 3 = 0$
$\Rightarrow x = 2$ or $\text{x}=\frac{-3}{2}$
View full question & answer→Question 365 Marks
Solve the following quadratic equation:
$3^{(x+2)} + 3^{-x} = 10$
Answer$3^{(x+2)} + 3^{-x} = 10$
$3^x.3^2 + 3^{-x} = 10$
$\Rightarrow\text{9y}+\frac{1}{\text{y}}=10$ where $3^x = y$
$\Rightarrow 9y^2 - 10y + 1 = 0$
$\Rightarrow 9y^2 - 9y - y + 1 = 0$
$\Rightarrow 9y(y - 1) - 1(y - 1) = 0$
$\Rightarrow (9y - 1)(y - 1) = 0$
$\Rightarrow 9y - 1 = 0$ or $y - 1 = 0$
$\Rightarrow\text{y}=\frac{1}{9}$ or $y = 1$
If $3^\text{x}=\frac{1}{9}$
$\Rightarrow 3x = (3)^{-2}$
$\Rightarrow x = -2$
If $3x = 1 = 30$
$\Rightarrow x = 0$
Hence, $-2, 0$ are the roots of given equation.
View full question & answer→Question 375 Marks
Solve the following equations by using the method of completing the square:
$\sqrt2\text{x}^2-3\text{x}-2\sqrt2=0$
Answer$\sqrt2\text{x}^2-3\text{x}-2\sqrt2=0$
$\Rightarrow2\text{x}^2-3\sqrt2\text{x}-4=0$ $\big($Multiplying both sides by $\sqrt2\big)$
$\Rightarrow\text{2x}^2-3\sqrt2\text{x}=4$
$\Rightarrow\big(\sqrt2\text{x}\big)^2-2\times\sqrt2\text{x}\times\frac{3}2{}+\Big(\frac{3}{2}\Big)^2\\=4+\Big(\frac{3}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{3}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}-\frac{3}{2}\Big)^2=4+\frac{9}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\frac{5}{2}$ or $\sqrt2\text{x}-\frac{3}{2}=-\frac{5}{2}$
$\Rightarrow\sqrt2\text{x}=\frac{5}2{}+\frac{3}{2}=\frac{8}{2}=4$ or $\sqrt2\text{x}=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1$
$\Rightarrow\text{x}=\frac{4}{\sqrt2}=2\sqrt2$ or $\text{x}=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}$
Hence, $2\sqrt2$ and $-\frac{\sqrt2}{2}$ are the roots of the given equation.
View full question & answer→Question 385 Marks
Solve the following quadratic equation: $a^2b^2x^2 + b^2x - a^2x - 1= 0$
Answer$a^2b^2x^2 + b^2x - a^2x - 1= 0$
$\Rightarrow b^2x(a^2x + 1) - 1(a^2x + 1) = 0$
$\Rightarrow (a^2x + 1)(b^2x - 1) = 0$
$\Rightarrow (a^2x + 1) = 0$ or $(b^2x - 1) = 0$
$\Rightarrow\text{x}=\frac{-\text{1}}{\text{a}^2}$ or $\text{x}=\frac{\text{1}}{\text{b}^2}$
Hence, $\frac{-\text{1}}{\text{a}^2}$ and $\frac{\text{1}}{\text{b}^2}$ are the roots of the given equation.
View full question & answer→Question 395 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$3a^2x^2 + 8abx + 4b^2 = 0$, $\text{a}\neq0$
AnswerGiven,
$3a^2x^2 + 8abx + 4b^2 = 0$
On comparing it with $Ax^2 + Bx + C = 0$, we get:
$A = 3a^2, B = 8ab$ and $C = 4b^2$
Discriminant D is given by:
$D = (B^2 - 4AC)$
$= (8ab)^2 - 4 \times 3a^2 \times 4b^2$
$= 16a^2b^2 > 0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-8\text{ab}+\sqrt{16\text{a}^2\text{b}^2}}{2\times3\text{a}^2}$
$=\frac{-8\text{ab}+4\text{ab}}{6\text{a}^2}$
$=\frac{-4\text{ab}}{6\text{a}^2}$
$=\frac{-2\text{b}}{3\text{a}}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-8\text{ab}-\sqrt{16\text{a}^2\text{b}^2}}{2\times3\text{a}^2}$
$=\frac{-8\text{ab}-4\text{ab}}{6\text{a}^2}$
$=\frac{-12\text{ab}}{6\text{a}^2}$
$=\frac{-2\text{b}}{\text{a}}$
Thus, the roots of the equation are $\frac{-2\text{b}}{3\text{a}}$ and $\frac{-2\text{b}}{\text{a}}.$
View full question & answer→Question 405 Marks
Solve the following quadratic equation:
$3\Big(\frac{\text{7x}+1}{\text{5x}-3}\Big)-4\Big(\frac{\text{5x}-3}{\text{7x}+1}\Big)=11$ $\text{x}\neq\frac{3}{5},\ \frac{-1}{7}$
Answer$3\Big(\frac{\text{7x}+1}{\text{5x}-3}\Big)-4\Big(\frac{\text{5x}-3}{\text{7x}+1}\Big)=11$
Taking $\frac{\text{7x}+1}{\text{5x}-3}=\text{y},$ we have
$\text{3y}-\frac{4}{\text{y}}=11$
$\Rightarrow\frac{\text{3y}^2-4}{\text{y}}=11$
$\Rightarrow 3y^2 - 4 = 11y$
$\Rightarrow 3y^2 - 11y - 4 = 0$
$\Rightarrow 3y^2 - 12y + y - 4 = 0$
$\Rightarrow 3y(y - 4) + 1(y - 4) = 0$
$\Rightarrow (y - 4)(3y + 1) = 0$
$\Rightarrow y - 4 = 0$ or $3y + 1 = 0$
$\Rightarrow y = 4$ or $\text{y}=\frac{-1}{3}$
$\Rightarrow\frac{\text{7x}+1}{\text{5x}-3}=4$ or $\frac{\text{7x}+1}{\text{5x}-3}=\frac{-1}{3}$
$\Rightarrow 7x + 1 = 20x - 12$ or $21x + 3 = -5x + 3$
$\Rightarrow 13x = 13$ or $26x = 0$
$\Rightarrow x = 1$ or $x = 0$
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Find the roots of the following equations, if they exist, by applying the quadratic formula:
$3x^2 - 2x + 2 = 0$
AnswerThe given equation is $3x^2 - 2x + 2 = 0$ Comparing it with $ax^2 + bx + c = 0$,
we get $a = 3, b = -2$ and $c = 2$
$\therefore$ Discriminant,$D = b^2 - 4ac$
$= (-2)^2 - 4 \times 3 \times 2 $
$= 4 - 24 = -20 < 0$
Hence, the given equation has no real roots (or real roots does not exist).
View full question & answer→Question 425 Marks
Solve the following quadratic equation:$4x^2 - 2(a^2 + b^2)x + a^2b^2 = 0$
Answer$4x^2 - 2(a^2 + b^2)x + a^2b^2 = 0$
$\Rightarrow 4x^2 - 2a^2x - 2b^2x + a^2b^2 = 0$
$\Rightarrow 2x(2x - a^2) - b^2(2x - a^2) = 0$
$\Rightarrow (2x - a^2)(2x - b^2) = 0$
$\Rightarrow (2x - a^2) = 0$ or $(2x - b^2) = 0$
$\Rightarrow\text{x}=\frac{\text{a}^2}{\text{2}}$ or $\text{x}=\frac{\text{b}^2}{\text{2}}$
Hence, $\frac{\text{a}^2}{\text{2}}$ and $\frac{\text{b}^2}{\text{2}}$ are the roots of the given equation.
View full question & answer→Question 435 Marks
Solve the following quadratic equation:
$\frac{\text{x}-1}{\text{2x}+1}+\frac{\text{2x}+1}{\text{x}-1}=2,$ $\text{x}\neq-\frac{1}{2},\ 1$
Answer$\frac{\text{x}-1}{\text{2x}+1}+\frac{\text{2x}+1}{\text{x}-1}=2$
$\frac{(\text{x}-1)^2+(\text{2x}+1)^2}{(\text{x}-1)(\text{2x}+1)}=2$
$\Rightarrow x^2 - 2x + 1 + 4x^2 + 4x + 1 = 2(x - 1)(2x + 1)$
$\Rightarrow 5x^2 + 2x + 2 = 4x^2 - 2x - 2$
$\Rightarrow x^2 + 4x + 4 = 0$
$\Rightarrow (x + 2)^2 = 0$
$\Rightarrow x = -2, 2$
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