MCQ 11 Mark
The ratio of the sum and product of the roots of the equation $7x^2- 12x + 18 = 0$ is:
- A
$7 : 12$
- B
$7 : 18$
- C
$3 : 2$
- ✓
$2 : 3$
AnswerCorrect option: D. $2 : 3$
$7x^2- 12x + 18 = 0$
Comparing with $ax^2+ bx + c = 0$, 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 $k =$ ?
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 $x^2 + 5kx + 16 = 0$ has no real roots,
$\Rightarrow D < 0$
$\Rightarrow b^2 - 4ac > 0$
$\Rightarrow (5k)^2 - 4 \times 16 < 0$
$\Rightarrow 25k^2 - 64 < 0$
$\Rightarrow 25k^2 < 64$
$\Rightarrow\text{k}^2<\frac{64}{25}$
$\Rightarrow\text{k}<\sqrt{\frac{64}{25}}$ 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 $9x^2+ 6kx + 4 = 0$ are equal,
$D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow (6k)^2 - 4 \times 9 \times 4 = 0$
$\Rightarrow 36k^2 - 144 = 0$
$\Rightarrow 36k^2 = 144$
$\Rightarrow k^2 = 4$
$\Rightarrow\text{k}=\pm2$
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. $\big(\sqrt2\text{x}+3\big)^2=\text{2x}^2+6$
$a. 3x - x^2 = x^2 + 5$ $\Rightarrow 2x^2 - 3x + 5 = 0,$
which is a quadratic equation.
$b. (x + 2)^2 = 2(x^2 - 5)$
$\Rightarrow x^2 + 4x + 4 = 2x^2 - 10$
$\Rightarrow x^2 - 4x - 14 = 0$,
which is a quadratic equation.
$c. \big(\sqrt2\text{x}+3\big)^2=\text{2x}^2+6$
$\Rightarrow\text{2x}^2+6\sqrt2\text{x}+9=\text{2x}^2+6$
$\Rightarrow6\sqrt2\text{x}+3=0$
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.
$d. (x - 1)^2 = 3x^2 + x - 2$
$\Rightarrow x^2- 2x + 1 = 3x^2 + x - 2$
$\Rightarrow 2x^2 + 3x - 3 = 0,$
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 $2x^2 - 6x + 3 = 0$ are:
- A
Real, unequal and rational.
- ✓
Real, unequal and irrational.
- C
- D
AnswerCorrect option: B. Real, unequal and irrational.
Given equation $2x^2 - 6x + 3 = 0$
Here, $a = 2, b = -6, c = 3$
Discriminant, $D = b^2 - 4ac$
$= (-6)^2 - 4 \times 2 \times 3$
$= 36 - 24$
$= 12 < 0$
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$
Now Area $= 240\ m^2$
$\Rightarrow$ Length $\times$ Breadth $= 240$
$\Rightarrow x(x + 8) = 240$
$\Rightarrow x^2 + 8x = 240$
$\Rightarrow x^2 + 8x - 240 = 0$
$\Rightarrow x^2 + 20x - 12x - 240 = 0$
$\Rightarrow x(x + 20) - 12(x + 20) = 0$
$\Rightarrow (x + 20)(x - 12) = 0$
$\Rightarrow x + 20 = 0$ 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. $x^3 - x^2 = (x - 1)^3$
$1. (x^2 + 1) = (2 - x)^2 + 3$
$2. \Rightarrow x^2 + 1 = 4 - 4x + x^2$
$\Rightarrow 4x - 3,$
This is not an equation of degree $2$.
$1. x^3 - x^2= (x - 1)^2$
$\Rightarrow x^3 - x^2 = x^3 - 3x^2 + 3x - 1$
$\Rightarrow 2x^2 - 2x + 1 = 0$
This is a quadratic equation.
$2. 2x^2 + 3 = (5 + x)(2x - 3)$
$\Rightarrow 2x^3 + 3 = 10x - 15 + 2x^2 - 3x$
$\Rightarrow 2x^3 - 2x^2 - 7x + 18 = 0$
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:
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, $\text{x}+\frac{1}{\text{x}}=\frac{41}{20}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{41}{20}$
$\Rightarrow 20x^2 + 20 = 41x$
$\Rightarrow 20x^2 - 41x + 20 = 0$
$\Rightarrow 20x^2 - 25x - 16x + 20 = 0$
$\Rightarrow 5x(4x - 5) - 4(4x - 5) = 0$
$\Rightarrow (4x - 5)(5x - 4) = 0$
$\Rightarrow 4x - 5 = 0$ 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. $\pm\frac{4}{3}$
Since the roots of the equation $4x^2- 3kx + 1 = 0$ are equal,
$D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow (-3k)^2 - 4 \times 4 \times 1 = 0$
$\Rightarrow 9k^2 - 16 = 0$
$\Rightarrow 9k^2 = 16$
$\Rightarrow\text{k}^2=\frac{16}{9}$
$\Rightarrow\text{k}=\pm\frac{4}{3}$
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 $5x^2 - kx + 1 = 0$ are real and distinct, then:
AnswerCorrect option: D. either $\text{k}>2\sqrt5$ or $\text{k}<-2\sqrt5$
Given, the roots of $5x^2 - kx + 1 = 0$ are real and distinct.
$\Rightarrow D > 0$
$\Rightarrow b^2 - 4ac > 0$
$\Rightarrow (-k)^2 - 4 \times 5 \times 1 > 0$
$\Rightarrow k^2 - 20 > 0$
$\Rightarrow k^2 > 20$
$\Rightarrow\text{k}>\sqrt{20}$ 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 $ax^2+ bx + c = 0$, it is given that $D = (b^2- 4ac) > 0$. Then, the roots of the equation are:
AnswerFor equation $ax^2+ bx + c = 0$, it is given that $D = (b^2- 4ac) > 0.$ 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 $82\ m$ and its area is $400\ m^2$. The breadth of the rectangle is:
- A
$25\ m$
- B
$20\ m$
- ✓
$16\ m$
- D
$9\ m$
AnswerCorrect option: C. $16\ m$
Perimeter of a rectangle $= 82m$
Let the breadth of the rectangle be $x m$.
Then, length of the rectangle $=\frac{\text{Perimeter}}{2}-\text{Breadth}$
$=\frac{82}{2}-\text{x}=(41-\text{x})\text{m}$
Now Area$ = 400m^2$
$\Rightarrow $ Length $\times $ Breadth $= 400$
$\Rightarrow x(41 - x) = 400$
$\Rightarrow 41x - x^2 = 400$
$\Rightarrow x^2 - 41x + 400 = 0$
$\Rightarrow x^2 - 25x - 16x + 400 = 0$
$\Rightarrow x(x - 25) - 16(x - 25) = 0$
$\Rightarrow (x - 25)(x - 16) = 0$
$\Rightarrow x - 25 = 0$ or $x - 16 = 0$
$\Rightarrow x = 25$ or $x = 16$
Hence, the length is $25\ m$ and the breadth is $16\ m$.
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 $x^2+ 6(k + 2)x + 9k = 0$ are equal,
$D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow [2(k + 2)]^2 - 4 \times 1 \times 9k = 0$
$\Rightarrow 4(k^2 + 4k + 4) - 36k = 0$
$\Rightarrow 4k^2 + 16k + 16 - 36k = 0$
$\Rightarrow 4k^2 - 20k + 16 = 0$
$\Rightarrow k^2 - 5k + 4 = 0$
$\Rightarrow k^2 - 4k - k + 4 = 0$
$\Rightarrow k(k - 4) -1(k - 4) = 0$
$\Rightarrow (k - 4)(k - 1) = 0$
$\Rightarrow k - 4 = 0$ or $k - 1 = 0$
$\Rightarrow k = 4$ 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:
- ✓
$x^2- 6x + 6 = 0$
- B
$x^2+ 6x - 6 = 0$
- C
$x^2- 6x - 6 = 0$
- D
$x^2+ 6x + 6 = 0$
AnswerCorrect option: A. $x^2- 6x + 6 = 0$
Sum of the roots $= 6$
Product of the roots $= 6 \times 6 = 36$
Required equation $= x^2 - ($Sum of roots$)x +$ Product of roots $= 0$
$\Rightarrow x^2- 6x + 6 = 0$
View full question & answer→MCQ 271 Mark
If one root of the equation $2x^2 + ax + 6 = 0$ is $2$ then $a = ?$
- A
$7$
- ✓
$-7$
- C
$\frac{7}{2}$
- D
$\frac{-7}{2}$
AnswerSince $x = 3$ is a solution of the equation $2x^2 + ax + 6 = 0$, we have
$2(2)^2 + a(2) + 6 = 0$
$\Rightarrow 8 + 2a + 6 = 0$
$\Rightarrow 2a = -14$
$\Rightarrow a = -7$
View full question & answer→MCQ 281 Mark
The roots of the equation $2x^2 - 6x + 7 = 0$ are:
- A
Real, unequal and rational.
- B
Real, unequal and irrational.
- C
- ✓
AnswerGiven equation $2x^2 - 6x + 7 = 0$
Here, $a = 2, b = -6, c = 7$
Discriminant, $D = b^2 - 4ac$
$= (-6)^2 - 4 \times 2 \times 7$
$= 36 - 56$
$= -20 < 0$
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
$x^2- 3x + 10 = 0$
- ✓
$x^2- 3x - 10 = 0$
- C
$x^2+ 3x - 10 = 0$
- D
$x^2+ 3x + 10 = 0$
AnswerCorrect option: B. $x^2- 3x - 10 = 0$
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 $= 0$
$\Rightarrow x^2- 3x - 10 = 0$
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. $2,\ \frac{-3}{2}$
$\Rightarrow 2x^2 - x - 6 = 0$
$\Rightarrow 2x^2 - 4x + 3x - 6 = 0$
$\Rightarrow 2x(x - 2) + 3(x - 2) = 0$
$\Rightarrow (x - 2)(2x + 3) = 0$
$\Rightarrow x - 2 = 0$ or $2x + 3 = 0$
$\Rightarrow x = 2$ or $\text{x}=\frac{-3}{2}$
Thus, the roots of the given equation are $2$ and $\frac{-3}{2}.$
View full question & answer→