Question 12 Marks
Solve the following quadratic equation:$\sqrt7\text{x}^2-\text{6x}-13\sqrt7=0$
Answer$\sqrt7\text{x}^2-\text{6x}-13\sqrt7=0$$\Rightarrow\sqrt7\text{x}^2-13\text{x}+7\text{x}-13\sqrt7=0$
$\Rightarrow\text{x}\big(\sqrt7\text{x}-13\big)+\sqrt7\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\big(\text{x}+\sqrt7\big)\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\text{x}+\sqrt7=0$ or $\sqrt7\text{x}-13=0$
$\Rightarrow\text{x}=-\sqrt7$ or $\text{x}=\frac{13\sqrt7}{7}$
Hence, $-\sqrt7$ and $\frac{13\sqrt7}{7}$ are the roots of the given equation.
View full question & answer→Question 22 Marks
Find the value of $\alpha$ for which the equation $(\alpha+12)\text{x}^2+2(\alpha-12)\text{x}+2=0$ has equal roots.
AnswerGiven:$(\alpha+12)\text{x}^2+2(\alpha-12)\text{x}+2=0$
Here, $\text{a}=(\alpha-12),\ \text{b}=2(\alpha-12)$ and $\text{c}=2$ It is given that the roots of the given equation are equal; therefore, we have: $\text{D}=0$ $\Rightarrow(\text{b}^2-\text{4ac})=0$ $\Rightarrow\big\{2(\alpha-12)\big\}^2-4\times(\alpha-12)\times2=0$ $\Rightarrow4(\alpha^2-24\alpha+144)-8(\alpha-12)=0$ $\Rightarrow4\alpha^2-96\alpha+576-8\alpha+96=0$ $\Rightarrow4\alpha^2-104\alpha+672=0$ $\Rightarrow\alpha^2-26\alpha+168=0$ $\Rightarrow\alpha^2-14\alpha-12\alpha+168=0$ $\Rightarrow\alpha(\alpha-14)-12(\alpha-14)=0$ $\Rightarrow(\alpha-14)(\alpha-12)=0$ $\therefore\ \alpha=14$ or $\alpha=12$ If the value of $\alpha$ is 12, the given equation becomes non-quadratic.Therefore, the value of $\alpha$ will be 14 for the equation to have equal roots.
View full question & answer→Question 32 Marks
Find the discriminant of the following equation:
$(x - 1)(2x - 1) = 0$
Answer$(x - 1)(2x - 1) = 0$
$\Rightarrow 2x^2 - 3x + 1 = 0$
Comparing it with $ax^2 + bx + c = 0$, we get
$a = 2,$
$b = -3,$
$c = 1$
$\therefore$ Discriminant, $D = b^2 - 4ac$
$= (-3)^2- 4 \times 2 \times 1$
$= 9 - 8$
$= 1$
View full question & answer→Question 42 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$25x^2 + 30x + 7 = 0$
AnswerGiven,
$25x^2 + 30x + 7 = 0$
On comparing it with $ax^2 + bx + c = 0$, we get:
$a = 25, b = 30$ and $c = 7$
Discriminant $D$ is given by:
$D = (b^2 - 4ac)$
$= 30^2 - 4 \times 25 \times 7$
$= 900 - 700$
$= 200$
$= 200 > 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{-30+\sqrt{200}}{2\times25}$
$=\frac{-30+10\sqrt2}{50}$
$=\frac{10\big(-3+\sqrt2\big)}{50}$
$=\frac{\big(-3+\sqrt2\big)}{5}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-30-\sqrt{200}}{2\times25}$
$=\frac{-30-10\sqrt2}{50}$
$=\frac{10\big(-3-\sqrt2\big)}{50}$
$=\frac{\big(-3-\sqrt5\big)}{5}$
Thus, the roots of the equation are $\frac{-3+\sqrt2}{5}$ and $\frac{-3-\sqrt2}{5}$
View full question & answer→Question 52 Marks
The following are quadratic equations in x?$(2x + 3)(3x + 2) = 6(x - 1)(x - 2)$
Answer$(2x + 3)(3x + 2) = 6(x - 1)(x - 2)$
$\Rightarrow 6x^2+ 4x + 9x + 6 = 6(x^2- 2x - x + 2)$
$\Rightarrow 6x^2+ 13x + 6 = 6x^2- 18x + 2$
$\Rightarrow 13x + 18x + 6 - 2 = 0$
$\Rightarrow 31x + 4 = 0$
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 62 Marks
Find the values of p for which the quadratic equation $(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$ has real and equal roots.
AnswerThe given equation is $(2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0$
This is of the form $ax^2 + bx + c = 0$, where $a = 2p + 1, b = -(7p + 2)$ and $c = 7p - 3.$
$\therefore D = b^2 - 4ac$
$D =[-(7p + 2)^2 - 4 \times (2p + 1) \times (7p - 3)]$
$D = (49p^2 + 28p + 4) - 4(14p^2 + p - 3)$
$D = 49p^2 + 28p + 4 - 56p^2 - 4p + 12$
$D = -7p^2 + 24p + 16$
The given equation will have real and equal roots if D = 0.
$\therefore$ -$7p^2 + 24p + 16 = 0$
$\Rightarrow 7p^2 - 24p - 16 = 0$
$\Rightarrow 7p^2 - 28p + 4p - 16 = 0$
$\Rightarrow 7p(p - 4) +4(p - 4) = 0$
$\Rightarrow (p - 4)(7p + 4) = 0$
$\Rightarrow p - 4 = 0 or 7p + 4 = 0$
$\Rightarrow p = 4$ or $\text{p}=-\frac{4}{7}$
Hence, 4 and $-\frac{4}{7}$ are the required values of p.
View full question & answer→Question 72 Marks
Find the nature of the roots of the following quadratic equations:$\text{12x}^2-4\sqrt{15}\text{x}+5=0$
AnswerThe given equation is $\text{12x}^2-4\sqrt{15}\text{x}+5=0$This is of the form $ax^2 + bx + c = 0$, where $\text{a}=12,\ \text{b}=-4\sqrt{15}$ and $c = 5$
$\therefore$ Discriminant, $D = b^2- 4ac$
$\text{D}=\big(-4\sqrt{15}\big)^2-4\times12\times5$
$\text{D}=240-240=0$
Hence, the given equation has real and equal roots.
View full question & answer→Question 82 Marks
Solve: $2x^2 + ax - a^2 = 0$
Answer$2x^2 + ax - a^2 = 0$
$\Rightarrow 2x^2+ 2ax - ax - a^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$ or $\text{x}=\frac{\text{a}}2{}$
View full question & answer→Question 92 Marks
Show that the roots of the equation $x^2 + px - q^2 = 0$ are real for all real value of $p$ and $q.$
AnswerGiven:
$x^2 + px - q^2 = 0$
Here,
$a = 1, b = p$ and$ c = -q^2$
Discriminant D is given by:
$D = (b^2 - 4ac)$
$D = p^2 - 4 \times 1 \times (-q^2)$
$D = (p^2 + 4q^2) > 0$
$D > 0$ for all real values of p and q.
Thus, the roots of the equation are real.
View full question & answer→Question 102 Marks
Find the nature of the roots of the following quadratic equations:$\text{3x}^2-2\sqrt6\text{x}+2=0$
AnswerThe given equation is $\text{3x}^2-2\sqrt6\text{x}+2=0$This is of the form $ax^2 + bx + c = 0$, where $\text{a}=3,\ \text{b}=-2\sqrt6$ and c = 2
$\therefore$ Discriminant, $D = b^2- 4ac$
$\text{D}=\big(2\sqrt6\big)^2-4\times3\times2$
$\text{D}=24-24=0$
Hence, the given equation has real and equal roots.
View full question & answer→Question 112 Marks
Find the nature of the roots of the following quadratic equations:$2x^2 - 8x + 5 = 0$
AnswerThe given equation is $2x^2 - 8x + 5 = 0$ This is of the form $ax^2 + bx + c = 0$, where $a = 2, b = -8$ and $c = 5$
$\therefore$ Discriminant, $D = b^2- 4ac$
$D = (-8)^2 - 4 \times 2 \times 5$
$D = 64 - 40$
$D = 24 > 0$
Hence, the given equation has real and unequal roots.
View full question & answer→Question 122 Marks
Find the discriminant of the following equation:
$\sqrt3\text{x}^2+2\sqrt2\text{x}-2\sqrt3=0$
Answer$\sqrt3\text{x}^2+2\sqrt2\text{x}-2\sqrt3=0$
Here,
$\text{a}=\sqrt3,$
$\text{b}=2\sqrt2,$
$\text{c}=-2\sqrt3$
Discriminant D is given by:
$\text{D}=\text{b}^2-4\text{ac}$
$=\big(2\sqrt2\big)^2-4\times\sqrt3\times\big(-2\sqrt3\big)$
$=(4\times2)+(8\times3)$
$=8+24$
$=32$
View full question & answer→Question 132 Marks
Find the nature of the roots of the following quadratic equations:$5x^2 - 4x + 1 = 0$
AnswerThe given equation is $5 x ^2-4 x +1=0$ This is of the form $a x ^2+ bx + c =0$, where $a =5, b=-4$ and $c =1$
$\therefore$ Discriminant, $D = b^2- 4ac$
$D = (-4)^2 - 4 \times 5 \times 1$
$D = 16 - 20$
$D = -4 > 0$
Hence, the given equation has no real roots.
View full question & answer→Question 142 Marks
Find the discriminant of the following equation:
$3x^2 - 2x + 8 = 0$
Answer$3x^2 - 2x + 8 = 0$
Here,
$a = 3,$
$b = -2,$
$c = 8$
Discriminant D is given by:
$D = b^2 - 4ac$
$D = (-2)^2 - 4 \times 3 \times 8$
$D = 4 - 96$
$D = -92$
View full question & answer→Question 152 Marks
Solve the following quadratic equation:$x^2 + 12x + 35 = 0$
Answer$x^2 + 12x + 35 = 0$
$\Rightarrow x^2 + 7x + 5x + 35 = 0$
$\Rightarrow x(x + 7) + 5(x + 7) = 0$
$\Rightarrow (x + 7)(x + 5) = 0$
$\Rightarrow x + 7 = 0 or x + 5 = 0$
$\Rightarrow x = -7 or x = -5$
View full question & answer→Question 162 Marks
Solve the following quadratic equation:$3x^2 - 243 = 0$
Answer$3x^2 - 243 = 0$
$\Rightarrow 3(x^2 - 81) = 0$
$\Rightarrow x^2 = 81$
$\Rightarrow\text{x}=\pm\sqrt{81}=\pm9$
$\Rightarrow x = 9, -9$
Hence 9 and -9 are the roots of the equation $3x^2 - 243 = 0$
View full question & answer→Question 172 Marks
Solve the following quadratic equation:$4x^2 + 5x = 0$
Answer$4x^2 + 5x = 0$
$\Rightarrow x(4x + 5) = 0$
$\Rightarrow x = 0 or 4x + 5 = 0$
$\Rightarrow x = 0$ or $\text{x}=-\frac{5}{4}$
View full question & answer→Question 182 Marks
Find the values of p for which the quadratic equation $(p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0, p \neq -1$ has equal roots. Hence, find the roots of the equation.
AnswerThe given equation is $(p+1) x^2-6(p+1) x+3(p+9)=0$
This is of the form $a x^2+b x+c=0$, where $a=p+1, b=-6(p+1)$ and $c=3(p+9)$.
$\therefore$ $D = b^2 - 4ac$
$D =[-6(p + 1)]^2 - 4 \times (p + 1) \times 3(p + 9)]$
$D = 12(p + 1)[3(p + 1) - (p + 9)]$
$D = 12(p + 1)(2p - 6)$
The given equation will have real and equal roots if $D = 0.$
$\therefore$ $12(p + 1)(2p - 6) = 0$
$\Rightarrow p + 1 = 0 or 2p - 6 = 0$
$\Rightarrow p = -1 or p = 3$
But, $p \neq -1$ (Given)
Thus, the value of p is $3$
Putting $p = 3$, the given equation becomes $4x^2 - 24x + 36 = 0.$
$4x^2 - 24x + 36 = 0$
$\Rightarrow 4(x^2 - 6x + 9) = 0$
$\Rightarrow (x - 3)^2 = 0$
$\Rightarrow x - 3 = 0$
$\Rightarrow x = 3$
Hence, 3 is the repeated root of this equation.
View full question & answer→Question 192 Marks
The following are quadratic equations in x?
$\text{x}^2-3\text{x}-\sqrt{\text{x}}+4=0$
Answer$\text{x}^2-3\text{x}-\sqrt{\text{x}}+4$ is not a quadratic polynomial since it contains $\sqrt{\text{x}}$ or $\text{x}^{\frac{1}{2}}$ in which power $\frac{1}{2}$ of x is not an integer.
$\therefore\ \text{x}^2-3\text{x}-\sqrt{\text{x}}+4=0$ is a quadratic equation.
View full question & answer→Question 202 Marks
Solve: $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)[2x + (b - a)] = 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 + (b - 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 212 Marks
If a and b are distinct real numbers, show that the quadratic equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ has no real roots.
AnswerThe given equation is $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$$\therefore$ $D = [2(a + b)]^2 - 4 \times 2(a^2 + b^2) \times 1$
$= 4(a^2 + 2ab + b^2) - 8(a^2 + b^2)$
$= 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2$
$= -4a^2 + 8ab - 4b^2$
$= -4(a^2 - 2ab + b^2)$
$= -4(a - b)^2 < 0$
Hence, the given equation has no real roots.
View full question & answer→Question 222 Marks
The following are the roots of $3x^2 + 2x - 1 = 0?$
$\frac{1}3{}$
AnswerThe given equation is $3x^2 + 2x - 1 = 0$
On substituting $\text{x}=\frac{1}{3}$ in the equation, we get
$\text{LHS}=3\times\Big(\frac{1}{3}\Big)^2+2\times\Big(\frac{1}{3}\Big)-1=0$
$=\Big(\frac{1}{3}+\frac{2}{3}-1\Big)=0=\text{RHS}$
$\therefore\text{x}=\frac{1}{3}$ is a solution of $3x^2 + 2x - 1 = 0$
View full question & answer→Question 232 Marks
Solve the following quadratic equation:
$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1},$ $\text{x}\neq0,\ 0,\ 1$
Answer$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1}$ $\Rightarrow\frac{16-\text{x}}{\text{x}}=\frac{15}{\text{x}+1}$
$\Rightarrow (16 - x)(x + 1) = 15x$
$\Rightarrow 16x + 16 - x^2 - x = 15x$
$\Rightarrow 15x + 16 - x^2 = 15x$
$\Rightarrow x^2 - 16 = 0$
$\Rightarrow (x - 4)(x + 4) = 0$
$\Rightarrow x - 4 = 0 or x + 4 = 0$
$\Rightarrow x = 4 or x = -4$
View full question & answer→Question 242 Marks
Solve the following quadratic equation:$\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
Answer$\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$$\Rightarrow\text{x}^2-\sqrt3\text{x}-\text{x}+\sqrt3=0$
$\Rightarrow\text{x}\big(\text{x}-\sqrt3\big)-1\big(\text{x}-\sqrt3\big)=0$
$\Rightarrow\big(\text{x}-\sqrt3\big)\big(\text{x}-1\big)=0$
$\Rightarrow\text{x}-\sqrt3=0$ or $\text{x}-1=0$
$\Rightarrow\text{x}=\sqrt{3}$ or $\text{x}=1$
View full question & answer→Question 252 Marks
Solve the following quadratic equation:$x^2 + 6x + 5 = 0$
Answer$x^2 + 6x + 5 = 0$
$\Rightarrow x^2 + 5x + x + 5 = 0$
$\Rightarrow x(x + 5) + 1(x + 5) = 0$
$\Rightarrow (x + 5)(x + 1) = 0$
$\Rightarrow x + 5 = 0 or x + 1 = 0$
$\Rightarrow x = -5 or x = -1$
View full question & answer→Question 262 Marks
Show that $\text{x}=-\frac{\text{bc}}{\text{ad}}$ is a solution of the quadratic equation $\text{ad}^2\Big(\frac{\text{ax}}{\text{b}}+\frac{\text{2c}}{\text{d}}\Big)\text{x}+\text{bc}^2=0$
Answer$\text{ad}^2\Big(\frac{\text{ax}}{\text{b}}+\frac{\text{2c}}{\text{d}}\Big)\text{x}+\text{bc}^2=0$By multiplying $ad^2x$ by $\frac{\text{ax}}{\text{b}}$ we get
$\text{ad}^2\text{x}\Big(\frac{\text{ax}}{\text{b}}\Big)$
$=\frac{\text{a}^2\text{d}^2\text{x}^2}{ \text{b}}$
By multiplying $ad^2x$ by $\frac{\text{2c}}{\text{d}}$ we get
$\text{ad}^2\text{x}\Big(\frac{\text{2c}}{\text{d}}\Big)$
$=\frac{\text{2ca}\text{d}^2\text{x}}{\text{d}}$
By following the equation
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\frac{\text{2ca}\text{d}^2\text{x}}{\text{d}}+\text{bc}^2=0$
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\text{2cad}^{2-1}\text{x}+\text{bc}^2=0$
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\text{2cad}\text{x}+\text{bc}^2=0$
$a^2d^2x + 2abcdx + b^2c^2 = 0$ (by multiplying the denominator 'b' to 2cadx & $bc^2$)
$\Rightarrow (adx + bc) = 0 (Squaring)$
$\Rightarrow adx + bc = 0$
$\Rightarrow adx = -bc$
$\Rightarrow\text{x}=\frac{-\text{bc}}{\text{ad}}$
View full question & answer→Question 272 Marks
Find the discriminant of the following equation:
$2x^2 - 7x + 6 = 0$
Answer$2x^2 - 7x + 6 = 0$
Here,
$a = 2,$
$b = -7,$
$c = 6$
Discriminant D is given by:
$D = b^2 - 4ac$
$D = (-7)^2 - 4 \times 2 \times 6$
$D = 49 - 48$
$D = 1$
View full question & answer→Question 282 Marks
Solve the following quadratic equation:$3\sqrt7\text{x}^2+\text{4x}-\sqrt7=0$
Answer$3\sqrt7\text{x}^2+\text{4x}-\sqrt7=0$$\Rightarrow3\sqrt7\text{x}^2+7\text{x}-3\text{x}-\sqrt7=0$
$\Rightarrow\sqrt7\text{x}\big(3\text{x}+\sqrt7\big)-1\big(3\text{x}+\sqrt7\big)=0$
$\Rightarrow\big(\text{3x}+\sqrt{7}\big)\big(\sqrt7\text{x}-1\big)=0$
$\Rightarrow\text{3x}+\sqrt7=0$ or $\sqrt7\text{x}-1=0$
$\Rightarrow\text{3x}=-\sqrt7$ or $\text{x}=\frac{1}{\sqrt7}$
$\Rightarrow\text{x}=\frac{-\sqrt7}{3}$ or $\text{x}=\frac{1\times\sqrt7}{\sqrt7\times\sqrt7}=\frac{\sqrt7}{7}$
Hence, $\frac{-\sqrt7}{3}$ and $\frac{\sqrt7}{7}$ are the roots of the given equation.
View full question & answer→Question 292 Marks
$x^2 + 6x - (a^2 + 2a - 8) = 0$
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 302 Marks
Find the value of k for which the equation $x^2 + k(2x + k - 1) + 2 = 0$ has real and equal roots.
AnswerThe given equation is $x^2+ k(2x + k - 1) + 2 = 0.$
$\Rightarrow x^2+ 2kx + k(k - 1) + 2 = 0$
So, $a = 1, b = 2k, c = k(k - 1) + 2$
We know $D = b^2- 4ac$
$\Rightarrow D = (2k)^2- 4 \times 1 \times [k(k - 1) + 2]$
$\Rightarrow D = 4k^2- 4[k^2- k + 2]$
$\Rightarrow D = 4k^2- 4k^2 + 4k - 8$
$\Rightarrow D = 4k - 8 = 4(k - 2)$
For equal roots, $D = 0$
Thus, $4(k - 2) = 0$
So, $k = 2.$
View full question & answer→Question 312 Marks
Solve: $4\sqrt3\text{x}^2+5\text{x}-2\sqrt3=0$
Answer$4\sqrt3\text{x}^2+5\text{x}-2\sqrt3=0$$\Rightarrow4\sqrt3\text{x}^2+\text{8x}-\text{3x}-2\sqrt3=0$
$\Rightarrow\text{4x}\big(\sqrt3\text{x}+2\big)-\sqrt3\big(\sqrt3\text{x}+2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}+2\big)\big(4\text{x}-\sqrt3\big)=0$
$\Rightarrow\sqrt3\text{x}+2=0$ or $4\text{x}-\sqrt3=0$
$\Rightarrow\text{x}=-\frac{2}{\sqrt3}=-\frac{2\sqrt3}{3}$ or $\text{x}=\frac{\sqrt3}{4}$
View full question & answer→Question 322 Marks
For what values of p are the roots of the equation $4x^2 + px + 3 = 0$ real and equal?
AnswerThe given equation is $4x^2 + px + 3 = 0.$
This is of the form $ax^2 + bx + c = 0$, where $a = 4, b = p$ and $c = 3.$
$\therefore$ $D = b^2 - 4ac$
$D = p^2 - 4 \times 4 \times 3$
$D = p^2 - 48$
The given equation will have real and equal roots if $D = 0.$
$\therefore$ $p^2 - 48 = 0$
$\Rightarrow p^2 = 48$
$\Rightarrow\text{p}=\pm48=\pm4\sqrt3$
Hence, $4\sqrt3$ and $-4\sqrt3$ are the required values of p.
View full question & answer→Question 332 Marks
Find the nature of the roots of the following quadratic equations:$x^2 - x + 2 = 0$
AnswerThe given equation is $x^2-x+2=0$ This is of the form $a x^2+b x+c=0$, where $a=1, b=-1$ and $c=2$
$\therefore$ Discriminant, $D = b^2- 4ac$
$D = (-1)^2 - 4 \times 1 \times 2$
$D = 1 - 8$
$D = -7 < 0$
Hence, the given equation has no real roots.
View full question & answer→Question 342 Marks
Solve: $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 352 Marks
Solve the following quadratic equation:$15x^2 - 28 = x$
Answer$15x^2 - 28 = x$
$\Rightarrow 15x^2 - x - 28 = 0$
$\Rightarrow 15x^2 - 21x + 20x - 28 = 0$
$\Rightarrow 3x(5x - 7) + 4(5x - 7) = 0$
$\Rightarrow (5x - 7)(3x + 4) = 0$
$\Rightarrow 5x - 7 = 0 or 3x + 4 = 0$
$\Rightarrow\text{x}=\frac{7}{5}$ or $\text{x}=\frac{-4}{3}$
Hence, $\frac{7}{5}$ and $\frac{-4}{3}$ are the roots of the equation $15x^2 - 28 = x$
View full question & answer→Question 362 Marks
Find the discriminant of the following equation:
$\text{2x}^2-5\sqrt{2\text{x}}+4=0$
Answer$\text{2x}^2-5\sqrt{2\text{x}}+4=0$
Here,
$\text{a} = 2,$
$\text{b}=-5\sqrt2,$
$\text{c}=4$
Discriminant D is given by:
$\text{D} = \text{b}^2 - \text{4ac}$
$\text{D}=\big(-5\sqrt2\big)^2-4\times2\times4$
$\text{D} = 49 - 48$
$\text{D} = 1$
View full question & answer→Question 372 Marks
Solve the following quadratic equation:$6x^2 + x - 12 = 0$
Answer$6x^2 + x - 12 = 0$
$\Rightarrow 6x^2 + 9x - 8x - 12 = 0$
$\Rightarrow 3x(2x + 3) - 4(2x + 3) = 0$
$\Rightarrow (2x + 3)(3x - 4) = 0$
$\Rightarrow 2x + 3 = 0 or 3x - 4 = 0$
$\Rightarrow\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{4}{3}$
Hence, $\frac{-3}{2}$ and $\frac{4}{3}$ are the roots of $6x^2 + x - 12 = 0$
View full question & answer→Question 382 Marks
Solve the following quadratic equation:$2x^2 + x - 6 = 0$
Answer$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 392 Marks
$x^2 - 4ax + 4a^2 - b^2 = 0$
Answer$x^2 - 4ax + 4a^2 - b^2 = 0$
$\Rightarrow x^2 - 4ax + (4a^2 - b^2) = 0$
$\Rightarrow x^2 - 4ax + (2a + b)(2a - b) = 0$
$\Rightarrow x^2 - (2a + b)x - (2a - b)x + (2a + b)(2a - b) = 0$
$\Rightarrow x[x - (2a + b)] - (2a - b)[x + (2a + b)] = 0$
$\Rightarrow [x - (2a + b)][x - (2a - b)] = 0$
$\Rightarrow x - (2a + b) = 0 or x - (2a - b) = 0$
$\Rightarrow x = 2a + b or x = 2a - b$
View full question & answer→Question 402 Marks
Solve: $\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
Answer$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$$\Rightarrow\sqrt3\text{x}^2-3\sqrt2\text{x}+\sqrt2\text{x}-2\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}-\sqrt6\big)+\sqrt2\big(\text{x}-\sqrt6\big)=0$
$\Rightarrow\big(\sqrt3\text{x}+\sqrt2\big)\big(\text{x}-\sqrt6\big)=0$
$\Rightarrow\sqrt3\text{x}+\sqrt2$ or $\text{x}-\sqrt6$
$\Rightarrow\text{x}=\frac{-\sqrt2}{\sqrt3}$ or $\text{x}=\sqrt6$
View full question & answer→Question 412 Marks
The following are quadratic equations in x?
$\text{x}-\frac{6}{\text{x}}=3$
Answer$\text{x}-\frac{6}{\text{x}}=3$
$\Rightarrow x^2- 6 = 3x$
$\Rightarrow x^2 - 3x - 6 = 0$
And $(x^2- 3x - 6)$ Being a polynomial of degree 2, it is a quadratic polynomial.
Hence $\text{x}-\frac{6}{\text{x}}=3$ is a quadratic equation.
View full question & answer→Question 422 Marks
For what value of k are the roots of the quadratic equation $\text{kx}\big(\text{x}-2\sqrt5\big)+10=0$ real and equal?
AnswerThe given equation is:
$\text{kx}\big(\text{x}-2\sqrt5\big)+10=0$
$\Rightarrow\text{kx}^2-2\sqrt5\text{kx}+10=0$
This is of the form ax2 + bx + c = 0, where $\text{a}=\text{k},\ \text{b}=-2\sqrt5\text{k}$ and c = 10
$\therefore\text{D}=\text{b}^2-\text{4ac}$
$\text{D}=\big(-2\sqrt5\text{k}\big)^2-4\times\text{k}\times10$
$\text{D}=\text{20k}^2-\text{40k}$
The given equation will have real and equal roots if D = 0.
$\therefore 20k^2- 40k = 0$
$\Rightarrow 20k(k - 2) = 0$
$\Rightarrow k = 0 or k - 2 = 0$
$\Rightarrow k = 0 or k = 2$
But, for k = 0, we get 10 = 0, which is not true.
Hence, 2 is the required value of k.
View full question & answer→Question 432 Marks
Solve the following quadratic equation:$4 - 11x = 3x^2$
Answer$4 - 11x = 3x^{2}$
$\Rightarrow 3x^2 + 11x - 4 = 0$
$\Rightarrow 3x^2 + 12x - x - 4 = 0$
$\Rightarrow 3x(x + 4) - 1(x + 4) = 0$
$\Rightarrow (x + 4)(3x - 1) = 0$
$\Rightarrow x + 4 = 0$ or $3x - 1 = 0$
$\Rightarrow x = -4$ or $\text{x}=\frac{1}{3}$
Hence, $-4$ and $\frac{1}{3}$ are the roots of the equation $4 - 11x = 3x^2$
View full question & answer→Question 442 Marks
Solve the following quadratic equation:(2x - 3)(3x + 1) = 0
Answer(2x - 3)(3x + 1) = 0⇒ 2x - 3 = 0 or 3x + 1 = 0
⇒ 2x = 3 or 3x = -1
$\Rightarrow\text{x}=\frac{3}{2}$ or $\text{x}=-\frac{1}{3}$
View full question & answer→Question 452 Marks
The following are the roots of $3x^2 + 2x - 1 = 0?$
$-1$
AnswerThe given equation is $3x^2 + 2x - 1 = 0$
On substituting x = -1 in the equation, we get
$LHS = 3 \times (-1)^2 + 2 \times (-1) - 1$
$= 3 - 2 - 1 = 0 = RHS$
$\therefore x = -1$ is a solution of $3x^2 + 2x - 1 = 0$
View full question & answer→Question 462 Marks
Solve the following quadratic equation:$\text{x}^2-3\sqrt3-30=0$
Answer$\text{x}^2-3\sqrt3-30=0$$\Rightarrow\text{x}^2+5\sqrt3\text{x}-2\sqrt3\text{x}-30=0$
$\Rightarrow\text{x}\big(\text{x}+5\sqrt3\big)-2\sqrt3\big(\text{x}+5\sqrt3\big)=0$
$\Rightarrow\big(\text{x}+5\sqrt3\big)\big(\text{x}-2\sqrt3\big)=0$
$\Rightarrow\text{x}+5\sqrt3=0$ or $\text{x}-2\sqrt3=0$
$\Rightarrow\text{x}=-5\sqrt{3}$ or $\text{x}=2\sqrt3$
View full question & answer→Question 472 Marks
Solve the following quadratic equation:$100x^2 - 20x + 1 = 0$
Answer$100x^2 - 20x + 1 = 0$
$\Rightarrow 100x^2 - 10x - 10x + 1 = 0$
$\Rightarrow 10x(10x - 1) - 1(10x - 1) = 0$
$\Rightarrow (10x - 1)(10x - 1) = 0$
$\Rightarrow (10x - 1)^2 = 0$
$\Rightarrow 10x - 1 = 0$
$\Rightarrow\text{x}=\frac{1}{10}$ (repeated root)
View full question & answer→Question 482 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$2\sqrt3\text{x}^2-5\text{x}+\sqrt3=0$
AnswerThe given equation is:$2\sqrt3\text{x}^2-5\text{x}+\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=2\sqrt3,\ \text{b}=-5$ and $\text{c}=\sqrt3$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=(-5)^2-4\times2\sqrt3\times\sqrt3$
$=25-24$
$=1>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt1=1$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-5)+1}{2\times2\sqrt3}$
$=\frac{6}{4\sqrt3}$
$=\frac{\sqrt3}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-5)-1}{2\times2\sqrt3}$
$=\frac{4}{4\sqrt3}$
$=\frac{\sqrt3}{3}$
Hence, $\frac{\sqrt3}{2}$ and $\frac{\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 492 Marks
The following are quadratic equations in x?
$(x + 2)^3 = x^3 - 8$
Answer$(x + 2)^3 = x^3 - 8$
$\Rightarrow x^3+ 8 + 6x(x + 2) = x^3- 8$
$\Rightarrow x^3+ 8 + 6x^2 + 12x = x^3- 8$
$\Rightarrow 6x^2 + 12x + 16 = 0$
This is of form $ax^2 + bx + c$, where
$a = 6, b = 12$ and $c = 16$
Hence, the given equation is a quadratic equation.
View full question & answer→Question 502 Marks
The following are quadratic equations in x?
$\text{x}+\frac{2}{\text{x}}=\text{x}^2$
Answer$\text{x}+\frac{2}{\text{x}}=\text{x}^2$
$\Rightarrow x^2 + 2 = x^3$
$\Rightarrow x^3 - x^2 - 2 = 0$
And $(x^3- x^2 - 2)$ Being a polynomial of degree 3, it is a quadratic polynomial.
Hence $\text{x}+\frac{2}{\text{x}}=\text{x}^2$ is a quadratic equation.
View full question & answer→Question 512 Marks
Solve: $\text{3x}^2+5\sqrt5\text{x}-10=0$
Answer$\text{3x}^2+5\sqrt5\text{x}-10=0$$\Rightarrow\text{3x}^2+6\sqrt5\text{x}-\sqrt5\text{x}-10=0$
$\Rightarrow\text{3x}\big(\text{x}+2\sqrt5\big)-\sqrt5\big(\text{x}+2\sqrt5\big)=0$
$\Rightarrow\big(\text{x}+2\sqrt5\big)\big(\text{3x}-\sqrt5\big)=0$
$\Rightarrow\text{x}+2\sqrt5=0$ or $\text{3x}-\sqrt5=0$
$\Rightarrow\text{x}=-2\sqrt5$ or $\text{x}=\frac{\sqrt5}{3}$
View full question & answer→Question 522 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 4x - 1 = 0$
AnswerGiven, $x^2 - 4x - 1 = 0$ On comparing it with $ax^2 + bx + c = 0,$
we get : $a = 1, b = -4$ and $c = -1$ Discriminant $D$ is
given by : $D = (b^2 - 4ac) = (-4)^2 - 4 \times 1 \times (-1) = 16 + 4 = 20 = 20 > 0$
Hence, the roots of the equation are real. Roots $\alpha$ and $\beta$ are given by: $\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-4)+\sqrt{20}}{2\times1}$
$=\frac{4+2\sqrt5}{2}$
$=\frac{2\big(2+\sqrt5\big)}{2}$
$=\big(2+\sqrt5\big)$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-4)-\sqrt{20}}{2}$
$=\frac{4-2\sqrt5}{2}$
$=\frac{2\big(2-\sqrt5\big)}{2} $
$=\big(2-\sqrt5\big)$
Thus, the roots of the equation are $\big(2+\sqrt5\big)$ and $\big(2-\sqrt5\big)$
View full question & answer→Question 532 Marks
Solve: $\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
AnswerThe given equation is $\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=1,\ \text{b}=-\big(\sqrt3+1\big),\ \text{c}=\sqrt3$
$\therefore\text{D}=\text{b}^2-\text{4ac}$
$=\big[-\big(\sqrt3+1\big)\big]^2-4\times1\times\sqrt3$
$=\big(\sqrt3-1\big)^2>0$
So, the given equation has real roots, given by
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]+\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1+\sqrt3-1}{2}$
$=\frac{2\sqrt3}{2}$
$=\sqrt3$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]-\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1-\sqrt3-1}{2}$
$=\frac{2}{2}$
$=1$
Hence, $\sqrt3$ and 1 are the roots of the given equation.
View full question & answer→Question 542 Marks
Solve the following quadratic equation:$x^2 = 18x - 77$
Answer$x^2 = 18x - 77$
$\Rightarrow x^2 - 18x + 77 = 0$
$\Rightarrow x^2 - 11x - 7x + 77 = 0$
$\Rightarrow x(x - 11) - 7(x - 11) = 0$
$\Rightarrow (x - 11)(x - 7) = 0$
$\Rightarrow x - 11 = 0 or x - 7 = 0$
$\Rightarrow x = 11 or x = 7$
View full question & answer→Question 552 Marks
Solve: $\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
Answer$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$$\Rightarrow\sqrt3\text{x}^2+\text{12x}-\text{2x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+4\sqrt3\big)-2\big(\text{x}+4\sqrt3\big)=0$
$\Rightarrow\big(\text{x}+4\sqrt3\big)\big(\sqrt3\text{x}-2\big)=0$
$\Rightarrow\text{x}+4\sqrt3=0$ or $\sqrt3\text{x}-2=0$
$\Rightarrow\text{x}=-4\sqrt3$ or $\text{x}=\frac{2}{\sqrt3}$
View full question & answer→Question 562 Marks
Find the discriminant of the following equation:
$1 - x = 2x^2$
Answer$1 - x = 2x^2$
$2x^2+ x - 1 = 0$
Here,
$a = 2,$
$b = 1,$
$c = -1$
Discriminant D is given by:
$D = b^2 - 4ac$
$= 1^2 - 4 \times 2(-1)$
$= 1 + 8$
$= 9$
View full question & answer→Question 572 Marks
Solve the following quadratic equation:$9x^2 - 3x - 2 = 0$
Answer$9x^2 - 3x - 2 = 0$
$\Rightarrow 9x^2 - 6x + 3x - 2 = 0$
$\Rightarrow 3x(3x - 2) + 1(3x - 2) = 0$
$\Rightarrow (3x - 2)(3x + 1) = 0$
$\Rightarrow 3x - 2 = 0$ or $3x + 1 = 0$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}3{}$
View full question & answer→Question 582 Marks
Find the nature of the roots of the following quadratic equations:$5x(x - 2) + 6 = 0$
AnswerThe given equation is $5x(x - 2) + 6 = 0$
$\Rightarrow 5x^2 - 10x + 6 = 0$
This is of the form $ax^2 + bx + c = 0$, where $a = 5, b = -10$ and $c = 6$
$\therefore$ Discriminant, $D = b^2- 4ac$
$D = (-10)^2 - 4 \times 5 \times 6$
$D = 100 - 120$
$D = -20 > 0$
Hence, the given equation has no real roots.
View full question & answer→Question 592 Marks
Solve the following quadratic equation:$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$
Answer$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$ Here, $6\sqrt3\times\sqrt3=6\times3=18$ and 9 × 2 = 18 & 9 + 2 = 11 $\Rightarrow\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$$\Rightarrow\sqrt3\text{x}^2+9\text{x}+2\text{x}+6\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+3\sqrt3\big)+2\big(\text{x}+3\sqrt{3}\big)$
$\Rightarrow\big(\text{x}+3\sqrt{3}\big)\big(\sqrt3\text{x}+2\big)=0$
$\Rightarrow\text{x}+3\sqrt3=0$ or $\sqrt3\text{x}+2=0$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2}{\sqrt3}$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2\times\sqrt3}{\sqrt3\times\sqrt3}=\frac{-2\sqrt3}{3}$
Hence, $-3\sqrt3$ and $\frac{-2\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 602 Marks
Solve the following quadratic equation:$6x^2 + 11x + 3 = 0$
Answer$6x^2 + 11x + 3 = 0$
$\Rightarrow 6x^2 + 9x + 2x + 3 = 0$
$\Rightarrow 3x(2x + 3) + 1(2x + 3) = 0$
$\Rightarrow (2x + 3)(3x + 1) = 0$
$\Rightarrow (2x + 3) = 0 or (3x + 1) = 0$
$\Rightarrow\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{-1}{3}$
Hence, $\frac{-3}{2}$ and $\frac{-1}{3}$ are the roots of $6x^2 + 11x + 3 = 0$
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