MCQ 11 Mark
The ratio of the sum and product of the roots of the equation $7 x^2-12 x+18=0$ is:
- A
$7 : 12$
- B
$7 : 18$
- C
$3 : 2$
- ✓
$2 : 3$
AnswerCorrect option: D. $2 : 3$
$7 x^2-12 x+18=0$
Comparing with $a x^2+b x+c=0$, we have
$a = 7, b = -12, c = 18$
Sum of the roots $=-\frac{\text{b}}{\text{a}}=-\frac{(-12)}{7}=\frac{12}{7}$
Product of the roots $=\frac{\text{c}}{\text{a}}=\frac{18}{7}$
Now,
$\frac{\text{Sum of the roots}}{\text{Product of the roots}}=\frac{\frac{12}{7}}{\frac{18}{7}}=\frac{12}{18}$
$=\frac{2}{3}=2:3$
View full question & answer→MCQ 21 Mark
If the equation $x^2- kx + 1 =0 $ has no real roots, then:
- A
$k < -2$
- B
$k > 2$
- ✓
$-2 < k < 2$
- D
AnswerCorrect option: C. $-2 < k < 2$
Since the equation $x^2+ 5kx + 16 = 0$ has no real roots,
$\Rightarrow D < 0$
$\Rightarrow b^2- 4ac > 0$
$\Rightarrow (-k)^2- 4 \times 1 \times 1 < 0$
$\Rightarrow k^2- 4 < 0$
$\Rightarrow k^2 < 4$
$\Rightarrow\text{k}<\sqrt{4}$ or $\text{k}>-\sqrt{4}$
$\Rightarrow k < 2$ or $k > -2$
$\Rightarrow -2 < k < 2$
View full question & answer→MCQ 31 Mark
If $x = 3$ is a solution of the equation $3x^2+ (k - 1)x + 9 = 0$ then k = ?
- A
$11$
- ✓
$-11$
- C
$13$
- D
$-13$
AnswerSince $x = 3$ is a solution of the equation $3x^2+ (k - 1)x + 9 = 0$, we have
$3(3)^2+ (k - 1)3 + 9 = 0$
$\Rightarrow 27 + 3k - 3 + 9 = 0$
$\Rightarrow 3k + 33 = 0$
$\Rightarrow 3k = -33$
$\Rightarrow k = -11$
View full question & answer→MCQ 41 Mark
If the product of the roots of the equation $x^2- 3x + k = 10$ is $-2$ then the value of $k$ is:
Answer$x^2- 3x + k = 10$
$\Rightarrow x^2- 3x + (k - 10) = 0$
Comparing with $ax^2+ bx + c = 0,$ we have
$a = 1, b = -3, c = k - 10$
Product of the roots $= -2$
$ \Rightarrow\frac{\text{c}}{\text{a}}=-2$
$\Rightarrow\text{k}-10=-2$
$\Rightarrow\text{k}=8$
View full question & answer→MCQ 51 Mark
If the equation $x^2+ 5kx + 16 = 0$ has no real roots then:
AnswerCorrect option: C. $\frac{-8}{5}<\text{k}<\frac{8}{5}$
Since the equation $x^2+5 k x+16=0$ has no real roots,
$ \Rightarrow D<0 $
$ \Rightarrow b^2-4 a c>0 $
$ \Rightarrow(5 k)^2-4 \times 16<0 $
$ \Rightarrow 25 k^2-64<0 $
$ \Rightarrow 25 k^2<64$
$\Rightarrow\text{k}^2<\frac{64}{25}$
$\Rightarrow\text{k}<\sqrt{\frac{64}{25}}$ or $\text{k}>-\sqrt{\frac{64}{25}}$
$\Rightarrow\text{k}<\frac{8}{5}$ or $\text{k}>-\frac{8}{5}$
$\Rightarrow-\frac{8}{5}<\text{k}<\frac{8}{5}$
View full question & answer→MCQ 61 Mark
If the equation $9x^2+ 6kx + 4 = 0$ has equal roots then $k = ?$
- A
$2$ or $0$
- B
$-2$ or $0$
- ✓
$2$ or $-2$
- D
$0$ only
AnswerCorrect option: C. $2$ or $-2$
Since the roots of the equation $9 x^2+6 k x+4=0$ are equal,
$ D=0$
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow(6 k)^2-4 \times 9 \times 4=0 $
$\Rightarrow 36 k^2-144=0 $
$\Rightarrow 36 k^2=144$
$\Rightarrow k^2=4$
$\Rightarrow\text{k}=\pm2$
View full question & answer→MCQ 71 Mark
If one root of $5x^2+ 13x + k = 0$ be the reciprocal of the other root, then the value of $k$ is:
AnswerLet one root of the given equation be $\alpha.$
Then, its root will be $\frac{1}{\alpha}.$
Given equation is $5x^2+ 13x + k = 0$
Comparing with $ax^2+ bx + c = 0,$ we have
$a = 5, b = 13, c = k$
Now,
Product of the roots $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\times\frac{1}{\alpha}=\frac{\text{k}}5{}$
$\Rightarrow\text{k}=5$
View full question & answer→MCQ 81 Mark
Which of the following is a quadratic equation?
AnswerCorrect option: C. $ (\sqrt{2} x+3)^2=2 x^2+6 $
- $\text { a. } 3 x-x^2=x^2+5 $
$ \Rightarrow 2 x^2-3 x+5=0$
which is a quadratic equation.
- $(x+2)^2=2\left(x^2-5\right)$
$\Rightarrow x^2+4 x+4=2 x^2-10 $
$ \Rightarrow x^2-4 x-14=0,$
which is a quadratic equation.
- $\big(\sqrt2\text{x}+3\big)^2=\text{2x}^2+6$
$\Rightarrow\text{2x}^2+6\sqrt2\text{x}+9=\text{2x}^2+6$
$\Rightarrow6\sqrt2\text{x}+3=0$
This is not an quadratic equation, since it contains a term involving $\sqrt{\text{x}},$ i.e., $\text{x}^{\frac{1}{2}},$ where $\frac{1}{2}$ is not an integer.
- $(x-1)^2=3 x^2+x-2$
$\Rightarrow x^2-2 x+1=3 x^2+x-2$
$ \Rightarrow 2 x^2+3 x-3=0,$
which is a quadratic equation. View full question & answer→MCQ 91 Mark
If the sum of the roots of the equation $kx^2+ 2x + 3x = 0$ is equal to their product, then the value of $k$ is:
- A
$\frac{1}{3}$
- B
$\frac{-1}{3}$
- C
$\frac{2}{3}$
- ✓
$\frac{-2}{3}$
AnswerCorrect option: D. $\frac{-2}{3}$
Given equation is $kx^2+ 2x + 3x = 0$
Comparing with $ax^2+ bx + c = 0,$ we have
$a = k, b = 2, c = 3k$
Now,
Sum of the roots = product of the roots
$\Rightarrow\frac{-2}{\text{k}}=\frac{\text{3k}}{\text{k}}$
$\Rightarrow\text{3k}=-2$
$\Rightarrow\text{k}=\frac{-2}3{}$
View full question & answer→MCQ 101 Mark
The roots of the equation $2 x^2-6 x+3=0$ are:
- A
Real, unequal and rational.
- ✓
Real, unequal and irrational.
- C
- D
AnswerCorrect option: B. Real, unequal and irrational.
Given equation $2 x^2-6 x+3=0$
Here, $a=2, b=-6, c=3$
Discriminant, $D=b^2-4 a c$
$ =(-6)^2-4 \times 2 \times 3$
$=36-24 $
$ =12<0$
Also, 12 is not a perfect square.
Hence, the roots of the given equation are real, unequal and irrational.
View full question & answer→MCQ 111 Mark
If $\alpha$ and $\beta$ are the roots of the equation $3x^2+ 8x + 2 = 0$ then $\Big(\frac{1}{\alpha}+\frac{1}{\beta}\Big)=?$
- A
$\frac{-3}{8}$
- B
$\frac{2}{3}$
- ✓
$-4$
- D
$4$
AnswerThe given equation is $3x^2+ 8x + 2 = 0$
Here, $a = 3, b = 8, c = 2$
$\alpha+\beta=-\frac{8}{3}$ and $\alpha\times\beta=\frac{2}3{}$
Now, $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}$
$=\frac{\frac{-8}{3}}{\frac{2}{3}}=-4$
View full question & answer→MCQ 121 Mark
The sum of the roots of the equation $x^2- 6x + 2 = 0$ is:
Answer$x^2- 6x + 2 = 0$
Comparing with $ax^2+ bx + c = 0, $ we have
$a = 1, b = -6, c = 2$
Sum of the roots $=-\frac{\text{b}}{\text{a}}=-\frac{(-6)}{1}=6$
View full question & answer→MCQ 131 Mark
For what values of $k,$ the equation $kx^2- 6x - 2 = 0$ has real roots?
AnswerCorrect option: B. $\text{k}\ge\frac{-9}{2}$
Given, the roots of $kx^2- 6x - 2 = 0$ are real
$\Rightarrow\text{D}\ge0$
$\Rightarrow\text{b}^2-\text{4ac}\ge0$
$\Rightarrow(-6)^2-4\times\text{k}\times(-2)\ge0$
$\Rightarrow36+\text{8k}\ge0$
$\Rightarrow\text{8k}\ge-36$
$\Rightarrow\text{k}\ge-\frac{9}{2}$
View full question & answer→MCQ 141 Mark
If one root of the equation $3x^2- 10x + 3 = 0$ is $\frac{1}{3}$ then the other root is:
- A
$\frac{-1}{3}$
- B
$\frac{1}{3}$
- C
$-3$
- ✓
$3$
AnswerLet the other root be $\alpha.$
Given equation is $3x^2- 10x + 3 = 0$
Comparing with $ax^2+ bx + c = 0,$ we have
$a = 3, b = -10, c = 3$
Product of the roots $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\times\frac{1}{3}=\frac{3}{3}$
$\Rightarrow\alpha=3$
View full question & answer→MCQ 151 Mark
The length of a rectangular field exceeds its breadth by $8\ m$ and the area of the field is $240\ m^2$. The breadth of the field is:
- A
$20\ m$
- B
$30\ m$
- ✓
$12\ m$
- D
$16\ m$
AnswerCorrect option: C. $12\ m$
Let the breadth of the rectangle be $x$ m.
Then, length of the rectangle $= (x + 8)m$
$ \text { Now Area }=240 m^2 $
$ \Rightarrow \text { Length } \times \text { Breadth }=240 $
$ \Rightarrow x(x+8)=240 $
$ \Rightarrow x^2+8 x=240 $
$ \Rightarrow x^2+8 x-240=0 $
$ \Rightarrow x^2+20 x-12 x-240=0 $
$\Rightarrow x(x + 20) - 12(x + 20) = 0$
$\Rightarrow (x + 20)(x - 12) = 0$
$\Rightarrow x + 20 = 0$ or $x - 12 = 0$
$\Rightarrow x = -20$ or $x = 12$
$\Rightarrow x = 12$ (Breadth cannot be negative).
View full question & answer→MCQ 161 Mark
Which of the following is a quadratic equation?
AnswerCorrect option: B. $ x^3-x^2=(x-1)^3 $
- $\left(x^2+1\right)=(2-x)^2+3$
$\Rightarrow x^2+1=4-4 x+x^2 $
$\Rightarrow 4 x-3,$
This is not an equation of degree $2$.
- $x^3-x^2=(x-1)^2$
$\Rightarrow x^3-x^2=x^3-3 x^2+3 x-1$
$\Rightarrow 2 x^2-2 x+1=0$
This is a quadratic equation.
- $2 x^2+3=(5+x)(2 x-3)$
$\Rightarrow 2 x^3+3=10 x-15+2 x^2-3 x$
$ \Rightarrow 2 x^3-2 x^2-7 x+18=0$
This is an equation of degree $3$. View full question & answer→MCQ 171 Mark
The roots of the equation $ax^2+ bx + c = 0$ will be reciprocal of each other if:
- A
$a = b$
- B
$b= c$
- ✓
$c = a$
- D
AnswerCorrect option: C. $c = a$
Product of the roots $=\frac{\text{c}}{\text{a}}$
Also, ${\alpha}\times\frac{1}{\alpha}=1$
$\Rightarrow\frac{\text{c}}{\text{a}}=1$
$\Rightarrow\text{c}=\text{a}$
View full question & answer→MCQ 181 Mark
The sum of a number and its reciprocal is $2\frac{1}{20}.$ The number is:
- ✓
$\frac{5}{4}$ or $\frac{4}{5}$
- B
$\frac{4}{3}$ or $\frac{3}{4}$
- C
$\frac{5}{6}$ or $\frac{6}{5}$
- D
$\frac{1}{6}$ or 6
AnswerCorrect option: A. $\frac{5}{4}$ or $\frac{4}{5}$
Let the number be $x$.
Then, $\text{x}+\frac{1}{\text{x}}=\frac{41}{20}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{41}{20}$
$ \Rightarrow 20 x^2+20=41 x $
$ \Rightarrow 20 x^2-41 x+20=0 $
$ \Rightarrow 20 x^2-25 x-16 x+20=0 $
$\Rightarrow 5x(4x - 5) - 4(4x - 5) = 0$
$\Rightarrow (4x - 5)(5x - 4) = 0$
$\Rightarrow 4x - 5 = 0$ or $5x - 4 = 0$
$\Rightarrow 4x = 5$ or $5x = 4$
$\Rightarrow\text{x}=\frac{5}4{}$ or $\text{x}=\frac{4}{5}$
View full question & answer→MCQ 191 Mark
If the equation $4x^2- 3kx + 1 = 0$ has equal roots then $k = ?$
- A
$\pm\frac{2}{3}$
- B
$\pm\frac{1}{3}$
- C
$\pm\frac{3}{4}$
- ✓
$\pm\frac{4}{3}$
AnswerCorrect option: D. $\pm\frac{4}{3}$
$\text { Since the roots of the equation } 4 x^2-3 k x+1=0 \text { are equal, }$
$D=0$
$ \Rightarrow b^2-4 a c=0$
$ \Rightarrow(-3 k)^2-4 \times 4 \times 1=0 $
$ \Rightarrow 9 k^2-16=0$
$ \Rightarrow 9 k^2=16$
$\Rightarrow\text{k}^2=\frac{16}{9}$
$\Rightarrow\text{k}=\pm\frac{4}{3}$
View full question & answer→MCQ 201 Mark
The roots of $ax^2+ bx + c = 0,$ $\text{a}\neq0$ are real and unequal, if $(b^2- 4ac)$ is:
AnswerSince the roots of the equation $ax^2+ bx + c = 0,$ $\text{a}\neq0$ are real and unequal,
we must have $D > 0$
$\Rightarrow b^2- 4ac > 0$
View full question & answer→MCQ 211 Mark
If the roots of $5 x^2-k x+1=0$ are real and distinct, then:
AnswerCorrect option: D. either $\text{k}>2\sqrt5$ or $\text{k}<-2\sqrt5$
$\text { Given, the roots of } 5 x^2-k x+1=0 \text { are real and distinct. }$
$ \Rightarrow D>0 $
$ \Rightarrow b^2-4 a c>0 $
$ \Rightarrow(-k)^2-4 \times 5 \times 1>0 $
$ \Rightarrow k^2-20>0 $
$ \Rightarrow k^2>20$
$\Rightarrow\text{k}>\sqrt{20}$ or $\text{k}<-\sqrt{20}$
$\Rightarrow\text{k}>2\sqrt5$ or $\text{k}<-2\sqrt5$
View full question & answer→MCQ 221 Mark
In the equation $a x^2+b x+c=0$, it is given that $D=\left(b^2-4 a c\right)>0$. Then, the roots of the equation are:
AnswerFor equation $a x^2+b x+c=0$, it is given that $D=\left(b^2-4 a c\right)>0$.
This means that the roots of the equation are real and unequal.
View full question & answer→MCQ 231 Mark
If the roots of the equation $ax^2+ bx + c = 0$ are equal, then then $c = ?$
- A
$\frac{-\text{b}}{\text{2a}}$
- B
$\frac{\text{b}}{\text{2a}}$
- C
$\frac{-\text{b}^2}{\text{4a}}$
- ✓
$\frac{\text{b}^2}{\text{4a}}$
AnswerCorrect option: D. $\frac{\text{b}^2}{\text{4a}}$
Since roots of the equation $ax^2+ bx + c = 0$ are equal,
$D = 0$
$\Rightarrow b^2- 4ac = 0$
$\Rightarrow b^2= 4ac$
$\Rightarrow\text{c}=\frac{\text{b}^2}{\text{4a}}$
View full question & answer→MCQ 241 Mark
The perimeter of a rectangle is $82m$ and its area is $400m^2$. The breadth of the rectangle is :
AnswerPerimeter of a rectangle $= 82m$
Let the breadth of the rectangle be $x m$.
Then, length of the rectangle $=\frac{\text{Perimeter}}{2}-\text{Breadth}$
$=\frac{82}{2}-\text{x}=(41-\text{x})\text{m}$
$ \text { Now Area }=400 m^2 $
$ \Rightarrow \text { Length } \times \text { Breadth }=400 $
$ \Rightarrow x(41-x)=400 $
$ \Rightarrow 41 x-x^2=400 $
$ \Rightarrow x^2-41 x+400=0 $
$ \Rightarrow x^2-25 x-16 x+400=0 $
$ \Rightarrow x(x-25)-16(x-25)=0 $
$ \Rightarrow(x-25)(x-16)=0 $
$ \Rightarrow x-25=0 $ or $ x-16=0 $
$ \Rightarrow x=25 $ or $ x=16 $
Hence, the length is $25m$ and the breadth is $16m$.
View full question & answer→MCQ 251 Mark
If the equation $x^2+ 6(k + 2)x + 9k = 0$ has equal roots then $k = ?$
- ✓
$1$ or $4$
- B
$-1$ or $4$
- C
$1$ or $-4$
- D
$-1$ or $-4$
AnswerCorrect option: A. $1$ or $4$
Since the roots of the equation $x^2+6(k+2) x+9 k=0$ are equal,
$ D=0 $
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow[2(k+2)]^2-4 \times 1 \times 9 k=0 $
$ \Rightarrow 4\left(k^2+4 k+4\right)-36 k=0 $
$ \Rightarrow 4 k^2+16 k+16-36 k=0 $
$ \Rightarrow 4 k^2-20 k+16=0 $
$ \Rightarrow k^2-5 k+4=0 $
$ \Rightarrow k^2-4 k-k+4=0 $
$ \Rightarrow k(k-4)-1(k-4)=0 $
$ \Rightarrow(k-4)(k-1)=0 $
$ \Rightarrow k-4=0 \text { or } k-1=0 $
$ \Rightarrow k=4 \text { or } k=1$
View full question & answer→MCQ 261 Mark
If the sum of the roots of a quadratic equation is $6$ and their product is $6,$ the equation is:
- ✓
$ x^2-6 x+6=0 $
- B
$ x^2+6 x-6=0 $
- C
$ x^2-6 x-6=0 $
- D
$ x^2+6 x+6=0 $
AnswerCorrect option: A. $ x^2-6 x+6=0 $
Sum of the roots $= 6$
Product of the roots $= 6 × 6 = 36$
Required equation $= x^2- ($Sum of roots$)x + $Product of roots$ = 0$
$ ⇒x^2-6 x+6=0 $
View full question & answer→MCQ 271 Mark
If one root of the equation $2 x^2+a x+6=0$ is $2$ then $a = ?$
- A
$7$
- ✓
$-7$
- C
$\frac{7}{2}$
- D
$\frac{-7}{2}$
AnswerSince $x=3$ is a solution of the equation $2 x^2+a x+6=0$, we have
$ 2(2)^2+a(2)+6=0$
$ \Rightarrow 8+2 a+6=0 $
$ \Rightarrow 2 a=-14$
$ \Rightarrow a=-7$
View full question & answer→MCQ 281 Mark
The roots of the equation $2 x^2-6 x+7=0$ are:
- A
Real, unequal and rational.
- B
Real, unequal and irrational.
- C
- ✓
AnswerGiven equation $2 x^2-6 x+7=0$
Here, $a=2, b=-6, c=7$
$ \text { Discriminant, } D=b^2-4 a c $
$ =(-6)^2-4 \times 2 \times 7 $
$ =36-56 $
$ =-20<0$
Hence, the roots of the given equation are imaginary.
View full question & answer→MCQ 291 Mark
The root of a quadratic equation are $5$ and $-2$. Then, the equation is:
- A
$ x^2-3 x+10=0 $
- ✓
$ x^2-3 x-10=0 $
- C
$ x^2+3 x-10=0 $
- D
$ x^2+3 x+10=0 $
AnswerCorrect option: B. $ x^2-3 x-10=0 $
Sum of the roots $= 5 + (-2) = 3$
Product of the roots $= 5 × (-2) = -10$
Required equation $= x^2- ($Sum of roots$)x +$ Product of roots $= 0$
$ \Rightarrow x^2-3 x-10=0 $
View full question & answer→MCQ 301 Mark
Which of the following is a quadratic equation?
- A
$\text{x}^2-3\sqrt{\text{x}}+2=0$
- B
$\text{x}+\frac{1}{\text{x}}=\text{x}^2$
- C
$\text{x}^2+\frac{1}{\text{x}^2}=5$
- ✓
$\text{2x}^2-\text{5x}=(\text{x}-1) ^2$
AnswerCorrect option: D. $\text{2x}^2-\text{5x}=(\text{x}-1) ^2$
a.$\text{x}^2-3\sqrt{\text{x}}+2=0$ is not a quadratic equation, since it contains a term involving $\sqrt{\text{x}},$ i.e., $\text{x}^{\frac{1}{2}},$ where $\frac{1}{2}$ is not a integer.
b.$\text{x}+\frac{1}{\text{x}}=\text{x}^2$
$\Rightarrow\text{x}^2+1=\text{x}^4$
$\Rightarrow\text{x}^2-\text{x}^2-1=0,$ which is a polynomial of degree $4$.
c.$\text{x}^2+\frac{1}{\text{x}^2}=\text{5}$
$\Rightarrow\text{x}^4+1=\text{5x}^2$
$\Rightarrow\text{x}^4-\text{5x}^2+1=0,$ which is a polynomial of degree $4$.
d$\text{2x}^2-\text{5x}=(\text{x}-1)^2$
$\Rightarrow\text{2x}^2-\text{5x}=\text{x}^2-\text{2x}+1$
$\Rightarrow\text{x}^2-\text{3x}-1=0$ This is a quadratic equation.
View full question & answer→MCQ 311 Mark
The roots of the quadratic equation $2x^2- x - 6 = 0$ are:
- A
$-2,\ \frac{3}{2}$
- ✓
$2,\ \frac{-3}{2}$
- C
$-2,\ \frac{-3}{2}$
- D
$2,\ \frac{3}{2}$
AnswerCorrect option: B. $2,\ \frac{-3}{2}$
$ \Rightarrow 2 x^2-x-6=0 $
$ \Rightarrow 2 x^2-4 x+3 x-6=0 $
$ \Rightarrow 2 x(x-2)+3(x-2)=0 $
$ \Rightarrow(x-2)(2 x+3)=0 $
$ \Rightarrow x-2=0 \text { or } 2 x+3=0$
$⇒ x = 2$ or $\text{x}=\frac{-3}{2}$
Thus, the roots of the given equation are $2$ and $\frac{-3}{2}.$
View full question & answer→