Question 13 Marks
Solve the following quadratic equation:$3\text{x}^2-2\sqrt6\text{x}+2=0$
Answer$3\text{x}^2-2\sqrt6\text{x}+2=0$$\Rightarrow3\text{x}^2-\sqrt6\text{x}-\sqrt6\text{x}+2=0$
$\Rightarrow\sqrt3\text{x}\big(\sqrt3\text{x}-\sqrt2\big)-\sqrt2\big(\sqrt3\text{x}-\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-\sqrt2\big)\big(\sqrt3\text{x}-\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-\sqrt2\big)^2=0$
$\Rightarrow\sqrt3\text{x}-\sqrt2=0$
$\Rightarrow\text{x}=\frac{\sqrt2}{\sqrt3}$ (repeated root)
View full question & answer→Question 23 Marks
Solve the following quadratic equation:
$x^2 + 5x - (a^2 + a - 6) = 0$
Answer$x^2 + 5x - (a^2 + a - 6) = 0$
$\Rightarrow x^2 + 5x - (a^2 + 3a - 2a - 6) = 0$
$\Rightarrow x^2 + 5x - [a(a + 3) - 2(a + 3)] = 0$
$\Rightarrow x^2 + 5x - [(a + 3)(a - 2)] = 0$
$\Rightarrow x^2 + (a + 3)x - (a - 2)x - (a + 3)(a - 2) = 0$
$\Rightarrow x[x + (a + 3)] - (a + 2)[x + (a + 3)] = 0$
$\Rightarrow [x + (a + 3)][x - (a - 2)] = 0$
$\Rightarrow x + (a + 3) = 0$ or $x - (a - 2) = 0$
$\Rightarrow x = -(a + 3)$ or $x = a - 2$
View full question & answer→Question 33 Marks
Solve the following quadratic equation: $x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
Answer$x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + (b^2 - 5b + 4b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + [b(b - 5) + 4(b - 5)] = 0$
$\Rightarrow x^2- (2b - 1)x + (b - 5)(b + 4) = 0$
$\Rightarrow x^2 - (2b - 1)x - (b + 4)x + (b - 5)(b + 4) = 0$
$\Rightarrow x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0$
$\Rightarrow [x - (b - 5)][x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0$ or $x - (b + 4) = 0$
$\Rightarrow x = b - 5$ or $x = b + 4$
View full question & answer→Question 43 Marks
Solve the following quadratic equation: $2x^2 + ax - a^2 = 0$
Answer$2x^2 + ax - a^2 = 0$
$\Rightarrow 2x^2 + 2ax - ax - a^2 + 2 = 0$
$\Rightarrow 2x(x + a) - a(x + a) = 0$
$\Rightarrow (x + a)(2x - a) = 0$
$\Rightarrow x + a = 0$ or $2x - a = 0$
$\Rightarrow x = -a$, $\text{x}=\frac{\text{a}}{2}$
View full question & answer→Question 53 Marks
If $-4$ is a root of the quadratic equation $x^2 + 2x + 4p = 0$, find the value of $k$ for which the quadratic equation $x^2 + px(1 + 3k) + 7(3 + 2k) = 0$ has equal roots.
AnswerIt is given that $-4$ is a root of the quadratic equation $x^2+2 x+4 p=0$.
$\therefore(-4)^2+2 \times(-4)+4 p=0$
$\Rightarrow 16-8+4 p=0$
$\Rightarrow 4 p+8=0$
$\Rightarrow p=-2$
The equation $x ^2+ px (1+3 k )+7(3+2 k )=0$ has equal roots.
$\therefore D=0$
$\Rightarrow[p(1+3 k)]^2-4 \times 1 \times 7(3+2 k)=0$
$\Rightarrow-[2(1+3 k)]^2-28(3+2 k)=0$
$\Rightarrow 4\left(1+6 k+9 k^2\right)-28(3+2 k)=0$
$\Rightarrow 4\left(1+6 k+9 k^2-21-14 k\right)=0$
$\Rightarrow 9 k^2-8 k-20=0$
$\Rightarrow 9 k^2-18 k+10 k-20=0$
$\Rightarrow 9 k(k-2)+10(k-2)=0$
$\Rightarrow(k-2)(9 k+10)=0$
$\Rightarrow k-2=0 \text { or } 9 k+10=0$
$\Rightarrow k=2 \text { or } k=-\frac{10}{9}$
Hence, the required value of $k$ is $2$ or $-\frac{10}{9}$.
View full question & answer→Question 63 Marks
Solve the following quadratic equation:$\text{10x}-\frac{1}{\text{x}}=3$
Answer$\text{10x}-\frac{1}{\text{x}}=3$
$\Rightarrow 10x^2 - 1 = 3x$
$\Rightarrow 10x^2 - 5x + 2x + 1 = 0$
$\Rightarrow 5x(2x - 1) + 1(2x - 1) = 0$
$\Rightarrow (5x + 1)(2x - 1) = 0$
$\Rightarrow (5x + 1)(2x - 1) = 0$
$\Rightarrow 5x + 1 = 0$ or $2x - 1 = 0$
$\Rightarrow\text{x}=\frac{-1}{5}$ or $\text{x}=\frac{1}{2}$
Hence, $\frac{-1}{5}$ and $\frac{1}{2}$ are the roots of the given equation.
View full question & answer→Question 73 Marks
Solve the following quadratic equation:$3x^2 - 2x - 1 = 0$
Answer$3x^2 - 2x - 1 = 0$
$\Rightarrow 3x^2 - 3x + 1x - 1 = 0$
$\Rightarrow 3x(x - 1) + 1(x - 1) = 0$
$\Rightarrow (x - 1)(3x + 1) = 0$
$\Rightarrow x - 1 = 0$ or $3x + 1 = 0$
$\Rightarrow x = 1$ or $\text{x}=\frac{-1}{3}$
Hence, $1$ and $\frac{-1}{3}$ are the roots of the equation $3x^2 - 2x - 1 = 0$
View full question & answer→Question 83 Marks
Solve the following equations by using the method of completing the square:
$x^2 - 4x + 1 = 0$
Answer$x^2 - 4x + 1 = 0$
$\Rightarrow x^2 - 4x = -1$
$\Rightarrow x^2 - 2 \times x \times 2 + 2^2 = -1 + 2^2$ (Adding $2^2$ on both sides)
$\Rightarrow (x - 2)^2= -1 + 4 = 3$
$\Rightarrow\text{x}-2=\pm\sqrt3$ (Taking square root on both sides)
$\Rightarrow\text{x}-2=\sqrt3$ or $\text{x}-2=-\sqrt3$
$\Rightarrow\text{x}=2+\sqrt3$ or $\text{x}=2-\sqrt3$
Hence, $2+\sqrt3$ and $2-\sqrt3$ are the roots of the given equation.
View full question & answer→Question 93 Marks
Determine the values of p for which the quadratic equation $2x^2 + px + 8 = 0$ has real roots.
AnswerGiven:
$2x^2 + px + 8 = 0$
Here,
$a = 2, b = p$ and $c = 8$
Discriminant $D$ is given by:
$D = (b^2 - 4ac)$
$D = p^2- 4 \times 2 \times 8$
$D = (p^2 - 64)$
If $\text{D}\ge0,$ the roots of the equation will be real.
$\Rightarrow(\text{p}^2-64)\ge0$
$\Rightarrow(\text{p}+8)(\text{p}-8)\ge0$
$\Rightarrow\text{p}\ge8$ and $\text{p}\le-8$
Thus, the roots of the equation are real for $\text{p}\ge8$ and $\text{p}\le-8.$
View full question & answer→Question 103 Marks
Solve the following equations by using the method of completing the square:
$8x^2 - 14x - 15 = 0$
Answer$8x^2 - 14x - 15 = 0$
$\Rightarrow 16x^2 - 28x - 30 = 0$ (Multiplying both sides by $2$)
$\Rightarrow 16x^2 - 28x = 30$
$\Rightarrow(\text{4x})^2-2\times\text{4x}\times\frac{7}{2}+\Big(\frac{7}{2}\Big)^2\\=30+\Big(\frac{7}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{7}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{4x}-\frac{7}{2}\Big)^2$
$=30+\frac{49}{4}$
$=\frac{169}{4}=\Big(\frac{13}{2}\Big)^2$
$\Rightarrow\text{4x}-\frac{7}{2}=\pm\frac{13}{2}$ (Taking square root on both sides)
$\Rightarrow\text{4x}-\frac{7}{2}=\frac{13}{2}$ or $\text{4x}-\frac{7}{2}=-\frac{13}{2}$
$\Rightarrow\text{4x}=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10$ or $\text{4x}=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3$
$\Rightarrow\text{x}=\frac{5}{2}$ or $\text{x}=-\frac{3}4{}$
Hence, $\frac{5}{2}$ and $-\frac{3}{4}$ are the roots of the given equation.
View full question & answer→Question 113 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$kx^2 + 6x + 1 = 0$
AnswerThe given equation is $kx^2 + 6x + 1 = 0$
$\therefore$ $D = 6^2 - 4 \times k \times 1$
$D = 36 - 4k$
The given equation has real and distinct roots if $D > 0$.
$\therefore$ $36 - 4k > 0$
$\Rightarrow 4k < 36$
$\Rightarrow k < 9$
View full question & answer→Question 123 Marks
Solve the following quadratic equation:$\text{2x}^2-\text{x}+\frac{1}{8}=0$
Answer$\text{2x}^2-\text{x}+\frac{1}{8}=0$
$\Rightarrow 16x^2 - 8x + 1 = 0$
$\Rightarrow 16x^2 - 4x - 4x + 1 = 0$
$\Rightarrow 4x(4x - 1) - 1(4x - 1) = 0$
$\Rightarrow (4x - 1)(4x - 1) = 0$
$\Rightarrow (4x - 1)^2 = 0$
$\Rightarrow 4x - 1 = 0$
$\Rightarrow\text{x}=\frac{1}{4}$ (repeated root)
View full question & answer→Question 133 Marks
If the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ are real and equal, show that either $a = 0 or (a^3 + b^3 + c^3) = 3abc$.
AnswerGiven:
$(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$
Here,
$a = (c^2 - ab), b = -2(a^2 - bc), c = (b^2 - ac)$
It is given that the roots of the equation are real and equal; therefore, we have:
$D = 0$
$\Rightarrow (b^2 - 4ac) = 0$
$\Rightarrow {-2(a^2 - bc)}^2 - 4 \times (c^2 - ab) \times (b^2 - ac) = 0$
$\Rightarrow 4(a^4 - 2a^2bc + b^2c^2) - 4(b^2c^2 - ac^3 - ab^3 + a^2bc) = 0$
$\Rightarrow a^4 - 2a^2bc + b^2c^2 - b^2c^2 + ac^3 + ab^3 - a^2bc = 0$
$\Rightarrow a^4 - 3a^2bc + ac^3 + ab^3 = 0$
$\Rightarrow a(a^3 - 3abc + c^3 + b^3) = 0$
Now,
$a = 0$ or $a^3 - 3abc + c^3 + b^3 = 0$
$a = 0$ or $a^3 + b^3 + c^3 = 3abc$
View full question & answer→Question 143 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$36x^2 - 12ax + (a^2 - b^2) = 0$
AnswerThe given equation is $36x^2 - 12ax + (a^2 - b^2) = 0.$
Comparing it with $Ax^2 + Bx + C = 0,$
we get $A = 36, B = -12a$ and $C = a^2 - b^2$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= (-12a)^2 - 4 \times 36 \times (a^2 - b^2) = 144a^2 - 144a^2 + 144b^2 = 144b^2 > 0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{144\text{b}^2}=12\text{b}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-12\text{a}+12\text{b})}{2\times36}$
$=\frac{12(\text{a}+\text{b})}{72}$
$=\frac{\text{a}+\text{b}}{6}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-12\text{a})-12\text{b}}{2\times36}$
$=\frac{12(\text{a}-\text{b})}{72}$
$=\frac{\text{a}-\text{b}}{6}$
Hence, $\frac{\text{a}+\text{b}}{6}$ and $\frac{\text{a}-\text{b}}{6}$ are the roots of the given equation.
View full question & answer→Question 153 Marks
The following are the roots of $3x^2 + 2x - 1 = 0$?
$-\frac{1}{2}$
AnswerThe given equation is $3x^2 + 2x - 1 = 0$
On substituting $\text{x}=-\frac{1}{2}$ in the equation, we get
$\text{LHS}=3\times\Big(\frac{-1}{2}\Big)^2+2\times\Big(\frac{-1}{2}\Big)-1=0$
$=\frac{3}{4}-1+1\neq0$
$\therefore\ \text{LHS}\neq\text{RHS}$
$\therefore\ \text{x}=\frac{-1}{2}$ is not a solution of $3x^2 + 2x - 1 = 0$
View full question & answer→Question 163 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2x^2 + x - 4 = 0$
AnswerGiven, $2x^2 + x - 4 = 0$ On comparing it with $ax^2 + bx + c = 0$, we get
$a = 2, b = 1$ and $c = -4$ Discriminant $D$ is given by:
$D = (b^2 - 4ac) = (1)^2 - 4 \times 2 \times (-4) = 1 + 32 = 33 > 0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{33}$
$\therefore\ \alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-1+\sqrt{33}}{2\times2}$
$=\frac{-1+\sqrt{33}}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-1-\sqrt{33}}{2\times2}$ $=\frac{-1-\sqrt{33}}{4}$
Hence $\frac{-1+\sqrt{33}}{4}$ and $\frac{-1-\sqrt{33}}{4}$ are the roots of the given equation.
View full question & answer→Question 173 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$4\sqrt3\text{x}^2+5\text{x}-2\sqrt3=0$
AnswerThe given equation is:$4\sqrt3{\text{x}}^2+5\text{x}-2\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=4\sqrt3,\ \text{b}=5$ and $\text{c}=-2\sqrt3$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=5^2-4\times4\sqrt3\times\big(-2\sqrt3\big)$
$=25+96$
$=121>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{121}=11$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5+11}{2\times4\sqrt3}$
$=\frac{6}{8\sqrt3}$
$=\frac{\sqrt3}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5-11}{2\times4\sqrt3}$
$=\frac{-16}{8\sqrt3}$
$=-\frac{2\sqrt3}{3}$
Hence, $\frac{\sqrt3}{4}$ and $-\frac{2\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 183 Marks
Solve the following equations by using the method of completing the square:
$5x^2 - 6x - 2 = 0$
Answer$5x^2 - 6x - 2 = 0 $
$\Rightarrow 25x^2 - 30x - 10 = 0$ (Multiplying both sides by $5$)
$\Rightarrow 25x^2 - 30x = 10 $
$\Rightarrow (5x)^2 - 2 \times 5x \times 3 + 3^2 = 10 + 3^2$ [Adding $3^2$ on both sides]
$\Rightarrow (5x - 3)^2 = 10 + 9 = 19$
$\Rightarrow\text{5x}-3=\pm19$ (Taking square root on both sides)
$\Rightarrow\text{5x}-3=\sqrt{19}$ or $\text{5x}- 3=-\sqrt{19}$
$\Rightarrow\text{5x}=3+\sqrt{19}$ or $\text{5x}=3-\sqrt{19}$
$\Rightarrow\text{x}=\frac{3+\sqrt{19}}{5}$ or $\text{x}=\frac{3-\sqrt{19}}{5}$
Hence, $\frac{3+\sqrt{19}}{5}$ and $\frac{3-\sqrt{19}}{5}$ are the roots of the given equation.
View full question & answer→Question 193 Marks
Find the non-zero value of k for which the roots of the quadratic equation $9x^2 - 3kx + k = 0$ are real and equal.
AnswerThe given equation is $9 x^2-3 k x+k=0$.
This is of the form $a x^2+b x+c=0$, where $a=9, b=-3 k$ and $c=k$.
$\therefore D=b^2-4 a c$
$D=(-3 k)^2-4 \times 9 \times k$
$D=9 k^2-36 k$
The given equation will have real and equal roots if $D=0$.
$\therefore 9 k^2-36 k=0$
$\Rightarrow 9 k(k-4)=0$
$\Rightarrow k=0 \text { or } k-4=0$
$\Rightarrow k=0 \text { or } k=4$
But, $k \neq 0$ (Given)
Hence, the required value of $k$ is $4$ .
View full question & answer→Question 203 Marks
Solve the following quadratic equation:$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$
Answer$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$$\Rightarrow\sqrt2\text{x}^2+5\text{x}+2\text{x}+50\sqrt2=0$
$\Rightarrow\text{x}\big(\sqrt2\text{x}+5\big)-\sqrt2\big(\sqrt2\text{x}+5\big)=0$
$\Rightarrow\big(\sqrt2\text{x}+5\big)\big(\text{x}+\sqrt2\big)=0$
$\Rightarrow\sqrt2\text{x}+5=0$ or $\text{x}+\sqrt2=0$
$\Rightarrow\text{x}=\frac{-5}{\sqrt2}$ or $\text{x}=-\sqrt2$
View full question & answer→Question 213 Marks
Solve the following quadratic equation:$\sqrt3\text{x}^2+\text{10x}+7\sqrt3=0$
Answer$\sqrt3\text{x}^2+\text{10x}+7\sqrt3=0$$\Rightarrow\sqrt3\text{x}^2+12\text{x}-2\text{x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+4\sqrt3\big)-2\big(\text{x}+4\sqrt{3}\big)=0$
$\Rightarrow\big(\sqrt{3}\text{x}-2\big)\big(\text{x}+4\sqrt3\big)=0$
$\Rightarrow\sqrt3\text{x}-2=0$ or $\text{x}+4\sqrt3=0$
$\Rightarrow\text{x}=\frac{2}{\sqrt3}$ or $\text{x}=-4\sqrt3$
View full question & answer→Question 223 Marks
Solve the following quadratic equation:$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$
Answer$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$$\Rightarrow4\sqrt6\text{x}^2-16\text{x}+3\text{x}-2\sqrt6=0$
$\Rightarrow4\sqrt2\text{x}\big(\sqrt3\text{x}-2\sqrt2\big)+\sqrt3\big(\sqrt3\text{x}-2\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-2\sqrt2\big)\big(4\sqrt2\text{x}+\sqrt3\big)=0$
$\Rightarrow\Big(\frac{2\sqrt2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\Big)=\frac{2\sqrt6}{3}$ or $\Big(\frac{-\sqrt3}{4\sqrt2}\times\frac{\sqrt2}{\sqrt2}\Big)=\frac{-\sqrt6}{8}$
Hence, $\frac{2\sqrt6}{3}$ and $\frac{-\sqrt6}{8}$ are the roots of the given equation.
View full question & answer→Question 233 Marks
If $-5$ is a root of the quadratic equation $2 x^2+p x-15=0$ and the quadratic equation $p\left(x^2+x\right)+k=0$ has equal roots, find the value of $k$.
AnswerIt is given that -5 is a root of the quadratic equation $2 x ^2+ px -15=0$.
$\therefore 2(-5)^2+p \times(-5)-15=0$
$\Rightarrow-5 p+35=0$
$\Rightarrow p=7$
The roots of the equation $px ^2+ px + k =0$ are equal.
$\therefore D=0$
$\Rightarrow p^2-4 p k=0$
$\Rightarrow(7)^2-4 \times 7 \times k=0$
$\Rightarrow 49-28 k=0$
$\Rightarrow k=\frac{49}{28}=\frac{7}{4}$
Thus, the value of $k$ is $\frac{7}{4}$.
View full question & answer→Question 243 Marks
Solve the following quadratic equation:$\text{x}^2-3\sqrt5\text{x}+10=0$
Answer$\text{x}^2-3\sqrt5\text{x}+10=0$$\Rightarrow\text{x}^2-\sqrt5\text{x}-2\sqrt{5}\text{x}+10=0$
$\Rightarrow\text{x}\big(\text{x}-\sqrt5\big)-2\sqrt5\big(\text{x}-\sqrt5\big)=0$
$\Rightarrow\big(\text{x}-\sqrt5\big)\big(\text{x}-2\sqrt5\big)=0$
$\Rightarrow\text{x}-\sqrt5=0$ or $\text{x}-2\sqrt5=0$
$\Rightarrow\text{x}=\sqrt{5}$ or $\text{x}=2\sqrt5$
View full question & answer→Question 253 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
AnswerGiven: $\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
On comparing it with $ax^2 + bx + c = 0$, we get:
$\text{x}=\sqrt3,\ \text{b}=10$ and $\text{c}=-8\sqrt3$
Discriminant $D$ is given by:
$D = (b^2 - 4ac)$
$=(10)^2-4\times\sqrt3\times\big(8\sqrt3\big)$
$=100+96$
$=196>0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-10+\sqrt{196}}{2\sqrt3}$
$=\frac{-10+14}{2\sqrt3}$
$=\frac{4}{2\sqrt3}$
$=\frac{2}{\sqrt3}$
$=\frac{2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\frac{2\sqrt3}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-10-\sqrt{196}}{2\sqrt3}$
$=\frac{-10-14}{2\sqrt3}$
$=\frac{-24}{2\sqrt3}$
$=\frac{-12}{\sqrt3}$
$=\frac{-12}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\frac{-12\sqrt3}{3}$
$=-4\sqrt3$
Thus, the roots of the equation are $\frac{2\sqrt3}{3}$ and $-4\sqrt3.$
View full question & answer→Question 263 Marks
Solve the following quadratic equation:$5x^2 + 13x + 8 = 0$
Answer$5x^2 + 13x + 8 = 0$
$\Rightarrow 5x^2 + 5x + 8x + 8 = 0$
$\Rightarrow 5x(x + 1) + 8(x + 1) = 0$
$\Rightarrow (x + 1)(5x + 8) = 0$
$\Rightarrow x + 1 = 0$ or $5x + 8 = 0$
$\Rightarrow x = -1$ or $\text{x}=\frac{-8}{5}$
View full question & answer→Question 273 Marks
If the equation $\left(1+m^2\right) x^2+2 m c x+\left(c^2-a^2\right)=0$ has equal roots, prove that $c^2=a^2\left(1+m^2\right)$.
AnswerGiven:
$\left(1+m^2\right) x^2+2 m c x+\left(c^2-a^2\right)=0$
Here,
$a=\left(1+m^2\right), b=2 m c \text { and } c=\left(c^2-a^2\right)$
It is given that the roots of the equation are equal; therefore, we have:
$D=0$
$\Rightarrow\left(b^2-4 a c\right)=0$
$\Rightarrow(2 m c)^2-4 \times\left(1+m^2\right) \times\left(c^2-a^2\right)=0$
$\Rightarrow 4 m^2 c^2-4\left(c^2-a^2+m^2 c^2-m^2 a^2\right)=0$
$\Rightarrow 4 m^2 c^2-4 c^2+4 a^2-4 m^2 c^2+4 m^2 a^2=0$
$\Rightarrow-4 c^2+4 a^2+4 m^2 a^2=0$
$\Rightarrow a^2+m^2 a^2=c^2$
$\Rightarrow a^2\left(1+m^2\right)=c^2$
$\Rightarrow c^2=a^2\left(1+m^2\right)$
Hence proved.
View full question & answer→Question 283 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 6x + 4 = 0$
AnswerGiven, $x^2 - 6x + 4 = 0$ On comparing it with $ax^2 + bx + c = 0$, we get
Discriminant $D$ is given by:
$a = 1, b = -6$ and $c = 4 D = (b^2 - 4ac) = (-6)^2 - 4 \times 1 \times 4 = 36 - 16 = 20 > 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{-(-6)+\sqrt{20}}{2\times1}$
$=\frac{6+2\sqrt5}{2}$ $=\frac{2\big(3+\sqrt5\big)}{2}$
$=\big(3+\sqrt5\big)$ $\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)-\sqrt{20}}{2\times1}$ $=\frac{6-2\sqrt5}{2}$ $=\frac{2\big(3-\sqrt5\big)}{2}$
$=\big(3-\sqrt5\big)$
Thus, the roots of the equation are $\big(3+2\sqrt5\big)$ and $\big(3-2\sqrt5\big).$
View full question & answer→Question 293 Marks
Solve the following equations by using the method of completing the square:
$3x^2 - x - 2 = 0$
Answer$3x^2 - x - 2 = 0$
$\Rightarrow 9x^2 - 3x - 6 = 0$ (Multiplying both sides by $3$)
$\Rightarrow 9x^2 - 3x = 6$
$\Rightarrow(\text{3x})^2-2\times\text{3x}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\\=6+\Big(\frac{1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{1}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{3x}-\frac{1}{2}\Big)^2$
$=6+\frac{1}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\text{3x}-\frac{1}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\text{3x}-\frac{1}{2}=\frac{5}{2}$ or $\text{3x}-\frac{1}{2}=-\frac{5}{2}$
$\Rightarrow\text{3x}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$ or $\text{3x}=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2$
$\Rightarrow x = 1$ or $\text{x}=-\frac{2}3{}$
Hence 1 and $-\frac{2}{3}$ are the roots of the given equation.
View full question & answer→Question 303 Marks
For what values of $k$ are the roots of the quadratic equation $3 x^2+2 k x+27=0$ real and equal?
AnswerGiven:
$3 x^2+2 kx+27=0$
Here,
$a=3, b=2 k \text { and } c=27$
It is given that the roots of the equation are real and equal; therefore, we have:
$D=0$
$\Rightarrow(2 k)^2-4 \times 3 \times 27=0$
$\Rightarrow 4 k^2-324=0$
$\Rightarrow 4 k^2=324$
$\Rightarrow k^2=81$
$\Rightarrow k= \pm 9$
$\therefore k=9 \text { or } k=-9$
View full question & answer→Question 313 Marks
Find the value of k for which $x = 1$ is a root of the equation $x^2 + kx + 3 = 0$ Also, find the other root.
AnswerSince $x=1$ is a solution of $x^2+k x+3=0$, it must be satisfy the equation.
$(1)^2+k(1)+3=0$
$\Rightarrow k=-4$
Hence the required value of $k=-4$
Since $x =\frac{3}{4}$ is a root of $a x^2+b x-6=0$, we have $a \times\left(\frac{3}{4}\right)^2+ b \times\left(\frac{3}{4}\right)-6=0$
$\Rightarrow \frac{9 a}{16}+\frac{3 b}{4}-6=0$
$\Rightarrow 9 a+12 b=96$
$\Rightarrow 3 a+4 b=32 \ldots(1)$
Again $x=-2$ being a root of $a x^2+b x-6=0$, we have
$a \times(-2)+b(-2)-6=0$
$\Rightarrow 4 a-2 b=6$
$\Rightarrow 2 a-b=3 \ldots(2)$
View full question & answer→Question 323 Marks
Solve the following equations by using the method of completing the square:
$4x^2 + 4bx - (a^2 - b^2) = 0$
Answer$4x^2 + 4bx - (a^2 - b^2) = 0$
$\Rightarrow 4x^2 + 4bx = a^2 - b^2$
$\Rightarrow (2x)^2 + 2 \times 2x \times b + b^2 = a^2- b^2 + b^2$ [Adding $b^2$ on both sides]
$\Rightarrow (2x + b)^2 = a^2$
$\Rightarrow\text{2x}+\text{b}=\pm\text{a}$ (Taking square root on both sides)
$\Rightarrow 2x + b = a$ or $2x + b = -a$
$\Rightarrow 2x = a - b$ or $2x = -a - b$
$\Rightarrow\text{x}=\frac{\text{a}-\text{b}}{2}$ or $\text{x}=-\frac{\text{a}+\text{b}}{2}$
Hence, $\frac{\text{a}-\text{b}}{2}$ and $-\frac{\text{a}+\text{b}}{2}$ are the roots of the given equation.
View full question & answer→Question 333 Marks
Solve the following equations by using the method of completing the square:
$x^2 - 6x + 3 = 0$
Answer$x^2 - 6x + 3 = 0$
$\Rightarrow x^2 - 6x = -3$
$\Rightarrow x^2 - 2 \times x \times 3 + 3^2 = -3 + 3^2$ (Adding $3^2$ on both sides)
$\Rightarrow (x - 3)^2= -3 + 9 = 6$
$\Rightarrow\text{x}-3=\pm\sqrt6$ (Taking square root on both sides)
$\Rightarrow\text{x}-3=\sqrt6$ or $\text{x}-3=-\sqrt6$
$\Rightarrow\text{x}=3+\sqrt6$ or $\text{x}=3-\sqrt6$
Hence, $3+\sqrt6$ and $3-\sqrt6$ are the roots of the given equation.
View full question & answer→Question 343 Marks
Solve the following quadratic equation:$4x^2 + 4bx - (a^2 - b^2) = 0$
Answer$4x^2 + 4bx - (a^2 - b^2) = 0$
$\Rightarrow 4x^2 + 4bx + (b^2 - a^2) = 0$
$\Rightarrow 4x^2 + 2(b + a)x + 2(b - a)x + (b^2 - a^2) = 0$
$\Rightarrow 2x[2x + (b + a)] + (b - a)[2x + (b + a)] = 0$
$\Rightarrow [2x + (b + a)][2x + (b - a)] = 0$
$\Rightarrow 2x + (b - a) = 0 or 2x + 9b - a = 0$
$\Rightarrow\text{x}=\frac{-(\text{b}+\text{a})}{2}$ or $\text{x}=\frac{-(\text{b}-\text{a})}{2}$
$\Rightarrow\text{x}=\frac{-(\text{a}+\text{b})}{2}$ or $\text{x}=\frac{(\text{a}-\text{b})}{2}$
View full question & answer→Question 353 Marks
The following are quadratic equations in x?
$\text{x}^2-\frac{1}{\text{x}^2}=5$
Answer$\text{x}^2-\frac{1}{\text{x}^2}=5$
$\Rightarrow x^4- 1 = 5x^2$
$\Rightarrow x^4 - 5x^2 - 1 = 0$
And $(x^4 - 5x^2 - 1)$ Being a polynomial of degree $4$
$\therefore\ \text{x}^2-\frac{1}{\text{x}^2}=5$ is not a quadratic equation.
View full question & answer→Question 363 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$3\text{x}^2-2\sqrt6\text{x}+2=0$
AnswerThe given equation is:$3\text{x}^2-2\sqrt6\text{x}+2=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=3,\ \text{b}=-2\sqrt6$ and $\text{c}=2$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=\big(-2\sqrt6\big)^2-4\times3\times2$
$=24-24=0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=0$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(2\sqrt6\big)+0}{2\times3}$
$=\frac{2\sqrt6}{6}$
$=\frac{\sqrt6}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(2\sqrt6\big)-0}{2\times3}$
$=\frac{2\sqrt6}{6}$
$=\frac{\sqrt6}{3}$
Hence, $\frac{\sqrt6}{3}$ is the repeated roots of the given equation.
View full question & answer→Question 373 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,\ \text{x}\neq0,\ 2$
AnswerThe given equation is:$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,\ \text{x}\neq0,\ 2$
$\Rightarrow\frac{\text{x}-2-\text{x}}{\text{x}(\text{x}-2)}=3$
$\Rightarrow\frac{-2}{\text{x}^2-2\text{x}}=3$
$\Rightarrow-2=3\text{x}^2-6\text{x}$
$\Rightarrow3\text{x}^2-6\text{x}+2=0$
This equation is of the form $ax^2 + bx + c = 0$, where
$a = 3, b = -6$ and $c = 2.$
$\therefore$ Discriminant, $D = b^2 - 4ac$
$= (-6)^2 - 4 \times 3 \times 2$
$= 36 - 24$
$= 12 > 0$
so, the given equation has real roots.
Now,$\sqrt{\text{D}}=\sqrt{12}=2\sqrt3$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)+2\sqrt3}{2\times3}$
$=\frac{6+2\sqrt3}{6}$
$=\frac{3+\sqrt3}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)-2\sqrt3}{2\times3}$
$=\frac{6-2\sqrt3}{6}$
$=\frac{3-\sqrt3}{3}$
Hence, $\frac{3+\sqrt3}{3}$ and $\frac{3-\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 383 Marks
Solve the following quadratic equation:$x^2 - 2ax - (4b^2 - a^2) = 0$
Answer$x^2 - 2ax - (4b^2 - a^2) = 0$
$\Rightarrow x^2 - 2ax + (a^2 - 4b^2) = 0$
$\Rightarrow x^2 - 2ax + (a - 2b)(a + 2b) = 0$
$\Rightarrow x^2 - (a - 2b)x - (a + 2b)x + (a - 2b)(a + 2b) = 0$
$\Rightarrow x[x + (a - 2b)] - (a + 2b)[x - (a - 2b)] = 0$
$\Rightarrow [x - (a - 2b)][x - (a + 2b)] = 0$
$\Rightarrow x - (a - 2b) = 0$ or $x - (a + 2b) = 0$
$\Rightarrow x = a - 2b$ or $x = a + 2b$
View full question & answer→Question 393 Marks
Solve the following quadratic equation:$4x^2 - 9x = 100$
Answer$4x^2 - 9x = 100\Rightarrow 4x^2 - 9x - 100 = 0$
$\Rightarrow 4x^2 - 25x + 16x - 100 = 0$
$\Rightarrow x(4x - 25) + 4(4x - 25) = 0$
$\Rightarrow (4x - 25)(x + 4) = 0$
$\Rightarrow 4x - 25 = 0$ or $x + 4 = 0$
$\Rightarrow\text{x}=\frac{25}{4}$ or$ x = -4$
Hence, $\frac{25}{4}$ and -4 are the roots of the equation $4x^2 - 9x = 100$
View full question & answer→Question 403 Marks
Find the values k for which of roots of $9x^2 + 8kx + 16 = 0$ are real and equal
AnswerGiven:
$9x^2 + 8kx + 16 = 0$
Here,
$a = 9, b = 8k$ and $c = 16$
It is given that the roots of the equation are real and equal; therefore, we have:
$D = 0$
$\Rightarrow (b^2 - 4ac) = 0$
$\Rightarrow (8k)^2 - 4 \times 9 \times 16 = 0$
$\Rightarrow 64k^2 - 576 = 0$
$\Rightarrow 64k^2 = 576$
$\Rightarrow k^2 = 9$
$\Rightarrow k = ±3$
$\therefore$ $k = 3$ or $k = -3$
View full question & answer→Question 413 Marks
Solve the following quadratic equation:
$9x^2 + 6x + 1 = 0$
Answer$9x^2 + 6x + 1 = 0$
$\Rightarrow 9x^2 + 3x + 3x + 1 = 0$
$\Rightarrow 3x(3x + 1) + 1(3x + 1) = 0$
$\Rightarrow (3x + 1)(3x + 1) = 0$
$\Rightarrow (3x + 1)^2 = 0$
$\Rightarrow 3x + 1 = 0$
$\Rightarrow\text{x}=\frac{-1}{3}$ (repeated root)
View full question & answer→Question 423 Marks
Solve the following quadratic equation:$x^2 + 6x - (a^2 + 2a - 8) = 0$
Answer$x^2 + 6x - (a^2 + 2a - 8) = 0$
$\Rightarrow x^2 + 6x - (a^2 + 4a - 2a - 8) = 0$
$\Rightarrow x^2 + 6x - [a(a + 4) - 2(a + 4)] = 0$
$\Rightarrow x^2+ 6x - (a + 4)(a - 2) = 0$
$\Rightarrow x^2 + (a + 4)x - (a - 2)x - (a + 4)(a - 2) = 0$
$\Rightarrow x[x + (a + 4)] - (a - 2)[x + (a + 4)] = 0$
$\Rightarrow [x + (a + 4)][x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0$ or $x - (a - 2) = 0$
$\Rightarrow x = -(a + 4)$ or $x = (a - 2)$
View full question & answer→Question 433 Marks
The following are quadratic equations in x?
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=2\Big(\text{x}+\frac{1}{\text{x}}\Big)+3$
Answer$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=2\Big(\text{x}+\frac{1}{\text{x}}\Big)+3$
$\Rightarrow\text{x}^2+2+\frac{1}{\text{x}^2}=\text{2x}+\frac{2}{\text{x}}+3$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}-\text{2x}-\frac{2}{\text{x}}-1=0$
$\Rightarrow\text{x}^4+1-\text{2x}^3-\text{2x}-\text{x}^2=0$
$\Rightarrow\text{x}^4-\text{2x}^3-\text{x}^2-\text{2x}+1=0$
Clearly, $x^4 - 2x^3 - x^2- 2x + 1$ is a polynomial of degree $4$
This is not of the form $ax^2 + bx + c = 0$
Hence, the given equation is not a quadratic equation.
View full question & answer→Question 443 Marks
Solve the following quadratic equation:$\text{x}^2-\big(1+\sqrt2\big)\text{x}+\sqrt2=0$
Answer$\text{x}^2-\big(1+\sqrt2\big)\text{x}+\sqrt2=0$
$\Rightarrow\text{x}^2-1.\text{x}-\sqrt2\text{x}+\sqrt2=0$
$\Rightarrow\text{x}(\text{x}-1)-\sqrt2(\text{x}-1)=0$
$\Rightarrow(\text{x}-1)\big(\text{x}-\sqrt2\big)=0$
$\Rightarrow(\text{x}-1)=0$ or $\text{x}-\sqrt2=0$
$\Rightarrow\text{x}=1$ or $\text{x}=\sqrt2$
Hence, 1 and $\sqrt2$ are the roots of the given equation.
View full question & answer→Question 453 Marks
Solve the following quadratic equation:$\text{x}^2+2\sqrt{2}\text{x}-6=0$
Answer$\text{x}^2+2\sqrt{2}\text{x}-6=0$$\Rightarrow\text{x}^2+3\sqrt{2}\text{x}-\sqrt{2}\text{x}-6=0$
$\Rightarrow\text{x}\big(\text{x}+3\sqrt{2}\big)-\sqrt{2}\big(\text{x}+3\sqrt{2}\big)$
$\Rightarrow\big(\text{x}+3\sqrt{2}\big)\big(\text{x}-\sqrt2\big)=0$
$\Rightarrow\text{x}=-3\sqrt2$ or $\text{x}=\sqrt2$
View full question & answer→Question 463 Marks
Solve the following quadratic equation:$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
Answer$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$$\Rightarrow\sqrt{3}\text{x}^2-3\sqrt2\text{x}-\sqrt{2}\text{x}-2\sqrt{3}=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}-\sqrt6\big)-\sqrt2\big(\text{x}-\sqrt6\big)=0$
$\Rightarrow\big(\text{x}-\sqrt6\big)\big(\sqrt3\text{x}+\sqrt2\big)=0$
$\Rightarrow\text{x}-\sqrt6=0$ or $\sqrt{3}\text{x}+\sqrt2=0$
$\Rightarrow\text{x}=\sqrt{6}$ or $\text{x}=\frac{-\sqrt{2}}{\sqrt{3}}$
View full question & answer→Question 473 Marks
Find the values of k for which the quadratic equation $(3k + 1)x^2 + 2(k + 1)x + 1 = 0$ has real and equal roots.
AnswerThe given equation is $(3k + 1)x^2 + 2(k + 1)x + 1 = 0.$
This is of the form $ax^2 + bx + c = 0$, where$ a = 3k +1, b = 2(k + 1)$ and $c = 1$.
$\therefore$ $D = b^2 - 4ac$
$D = [2(k + 1)]^2 - 4 \times (3k + 1) \times 1$
$D = (k^2 + 2k + 1) - 4(3k + 1)$
$D = 4k^2 + 8k + 4 - 12k - 4$
$D = 4k^2- 4k$
The given equation will have real and equal roots if $D = 0.$
$\therefore$ $4k^2 - 4k = 0$
$\Rightarrow 4k(k - 1) = 0$
$\Rightarrow k = 0$ or $k - 1 = 0$
Hence, $0$ and $1$ are the required values of $k$.
View full question & answer→Question 483 Marks
Solve the following quadratic equation:$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
Answer$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$ Multiplying by $x^22 - 5x + 2x^2 = 0$ or $2x^2 - 5x + 2 = 0$
$\Rightarrow 2x^2 - 4x - x + 2 = 0$
$\Rightarrow 2x(x - 2) - 1(x - 2) = 0$
$\Rightarrow (x - 2)(2x - 1) = 0$
$\therefore$ $x - 2 = 0$ or $2x - 1 = 0$
$⇒ x = 2$, $\text{x}=\frac{1}{2}$
Hence, $2$ and $\frac{1}{2}$ are the roots of the given equation.
View full question & answer→Question 493 Marks
If $3$ is a root of the quadratic equation $x^2 - x + k = 0$, find the value of p so that the roots of the equation $x^2+ k(2x + k + 2) + p = 0$ are equal.
AnswerIt is given that 3 is a root of the quadratic equation $x^2 - x + k = 0.$
$\therefore$ $(3)^2 - 3 + k = 0$
$\Rightarrow k + 6 = 0$
$\Rightarrow k = -6$
The roots of the equation $x^2 + 2kx + (k^2 + 2k + p) = 0$ are equal.
$\therefore$ $D = 0$
$\Rightarrow (2k)^2 - 4 \times 1 \times (k^2 + 2k + p) = 0$
$\Rightarrow 4k2 - 4k2 - 8k - 4p = 0$
$\Rightarrow -8k - 4p = 0$
$\Rightarrow\text{p}=\frac{\text{8k}}{-4}=-\text{2k}$
$\Rightarrow p = -2 \times (-6) = 12$
Hence, the value of $p$ is $12$.
View full question & answer→Question 503 Marks
Solve the following equations by using the method of completing the square:
$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
Answer$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
$\Rightarrow\frac{2-\text{5x}+\text{2x}^2}{\text{x}^2}=0$
$\Rightarrow 2x^2 - 5x + 2 = 0$
$\Rightarrow 4x^2 - 10x + 4 = 0$ (Multiplying both sides by $2$)
$\Rightarrow 4x^2 - 10x = -4$
$\Rightarrow(\text{2x})^2-2\times\text{2x}\times\frac{5}{2}+\Big(\frac{5}{2}\Big)^2\\=-4+\Big(\frac{5}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{5}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}-\frac{5}{2}\Big)^2$
$=-4+\frac{25}{4}$
$=\frac{9}{4}=\Big(\frac{3}{2}\Big)^2$
$\Rightarrow\text{2x}-\frac{5}{2}=\pm\frac{3}{2}$ (Taking square root on both sides)
$\Rightarrow\text{2x}-\frac{5}{2}=\frac{3}{2}$ or $\text{2x}-\frac{5}{2}=-\frac{3}{2}$
$\Rightarrow\text{2x}=\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4$ or $\text{2x}=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1$
$\Rightarrow x = 2$ or $\text{x}=\frac{1}{2}$
Hence, $2$ and $\frac{1}{2}$ are the roots of the given equation.
View full question & answer→Question 513 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$16x^2 = 24x + 1$
AnswerGiven,
$16x^2 = 24x + 1$
$16x^2 - 24x - 1 = 0$
On comparing it with $ax^2 + bx + c = 0$
$a = 16, b = -24$ and $c = -1$
Discriminant D is given by:
$D = (b^2 - 4ac)$
$= (-24)^2 - 4 \times 16 \times (-1)$
$= 576 + (64)$
$= 640 > 0$
Hence, the roots of the equation are real,
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-24)+\sqrt{640}}{2\times16}$
$=\frac{24+8\sqrt{10}}{32}$
$=\frac{8\big(3+\sqrt{10}\big)}{32}$
$=\frac{\big(3+\sqrt{10}\big)}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-24)+\sqrt{640}}{2\times16}$
$=\frac{24-8\sqrt{10}}{32}$
$=\frac{8\big(3-\sqrt{10}\big)}{32}$
$=\frac{\big(3-\sqrt{10}\big)}{4}$
Thus, the roots of the equation are $\frac{\big(3+\sqrt{10}\big)}{4}$ and $\frac{\big(3-\sqrt{10}\big)}{4}.$
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