MCQ 11 Mark
If the roots of $x^2− bx + c = 0$ are two consecutive integers, then $b^2 − 4\ c$ is:
AnswerGiven equation: $x^2 − bx + c = 0$
Let $\alpha$ and $\alpha+1$ be the two consecutive roots of the equation.
Sum of the roots $=\alpha+\alpha+1=2\alpha+1$
Product of the roots $=\alpha(\alpha+1)=\alpha^2+\alpha$
So, sum of the roots $=2\alpha+1=\frac{-\text{Coeffecient of x}}{\text{Coeffecient of x}^2}=\frac{\text{b}}{1}=\text{b}$
Product of the roots $=\alpha^2+\alpha=\frac{\text{Constant term}}{\text{Coeffecient of x}^2}=\frac{\text{c}}{1}=\text{c}$
Now,$\text{b}^2-4\text{c}=(2\alpha+1)^2-4(\alpha^2+\alpha)$
$=4\alpha^2+4\alpha+1-4\alpha^2-4\alpha$
$=1$
View full question & answer→MCQ 21 Mark
The number of real roots of the equation $(x^2 + 2x)^2− (x + 1)^2 − 55 = 0:$
Answer$(x^2 + 2x)^2 - (x + 1)^2 - 55 = 0$
$\Rightarrow (x^2+ 2x + 1 - 1)^2 - (x + 1)^2 - 55 = 0$
$\Rightarrow\Big\{(\text{x}+1)^2-1\Big\}^2-(\text{x}+1)^2-55=0$
$\Rightarrow\Big\{(\text{x}+1)^2\Big\}^2+1-3(\text{x}+1)^2-55=0$
$\Rightarrow\Big\{(\text{x}+1)^2\Big\}^2-3(\text{x}+1)^2-54=0$
Let $p = (x + 1)^2$
$\Rightarrow p^2 - 3p - 54 = 0$
$\Rightarrow p^2 - 9p + 6p - 54 = 0$
$\Rightarrow (p + 6) (p - 9) = 0$
$\Rightarrow p = 9$ or $ p = -6$
Rejecting $p = −6$
$\Rightarrow (x + 1)^2 = 9$
$\Rightarrow x^2 + 2x - 8 = 0$
$\Rightarrow (x + 4) (x - 2) = 0$
$\Rightarrow x = 2, x = -4$
View full question & answer→MCQ 31 Mark
If $\alpha$ and $\beta$ are the roots of $4x^2+ 3x + 7 = 0,$ then the value of $\frac{1}{\alpha}+\frac{1}{\beta}$ is:
- A
$\frac{4}{7}$
- ✓
$-\frac{3}{7}$
- C
$\frac{3}{7}$
- D
$-\frac{3}{4}$
AnswerCorrect option: B. $-\frac{3}{7}$
Given equation: $4x^2 + 3x + 7 = 0$
Also, $\alpha$ and $\beta$ are the roots of the equation.
Then, sum of the roots $=\alpha + \beta =\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\frac{3}{4}$
Product of the roots $=\alpha\beta=\frac{\text{constant term}}{\text{Coefficient of x}^2}=\frac{7}{4}$
$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-\frac{3}{4}}{\frac{7}{4}}=-\frac{3}{7}$
View full question & answer→MCQ 41 Mark
If one root of the equation $x^2+ px + 12 = 0,$ is $4$, while the equation $x^2+ px + q = 0$ has equal roots, the value of $q$ is:
- ✓
$\frac{49}{4}$
- B
$\frac{4}{49}$
- C
$4$
- D
AnswerCorrect option: A. $\frac{49}{4}$
It is given that, $4$ is the root of the equation $x^2+ px + 12 = 0.$
$\therefore 16 + 4p + 12 = 0$
$\Rightarrow p = -7$
It is also given that, the equation $x^2 + px + q = 0$ has equal roots.
So, the discriminant of:
$x^2 + px + q = 0$ will be zero.
$\therefore p^2 - 4q = 0$
$\Rightarrow 4q = (-7)^2 = 49$
$\Rightarrow\text{q}=\frac{49}{4}$
View full question & answer→MCQ 51 Mark
The values of $x$ satisfying $\log_3 (x^2+ 4x + 12) = 2$ are:
- A
$2, −4$
- B
$1, −3$
- C
$−1, 3$
- ✓
$−1, −3$
AnswerCorrect option: D. $−1, −3$
The given equation is $\log_3(x^2 + 4x + 12) = 2$
$\Rightarrow x^2 + 4x + 12 = 3^2= 9$
$\Rightarrow x^2 + 4x + 3 = 0$
$\Rightarrow (x + 1) (x + 3) = 0$
$\Rightarrow x = -1, -3$
View full question & answer→MCQ 61 Mark
The complete set of values of $k,$ for which the quadratic equation $x^2− kx + k + 2 = 0$ has equal roots, consists of:
- A
$2+\sqrt{2}$
- ✓
$2\pm\sqrt{12}$
- C
$2-\sqrt{12}$
- D
$-2-\sqrt{12}$
AnswerCorrect option: B. $2\pm\sqrt{12}$
Since the equation has real roots.
$\Rightarrow D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow K^2 - 4 (1)(K + 2) = 0$
$\Rightarrow K^2 - 4K - 8 = 0$
$\Rightarrow\text{K}=\frac{4\pm\sqrt{16-4(1)(-8)}}{2(1)}$
$\Rightarrow\text{K}=\frac{4\pm2\sqrt{12}}{2}$
$\Rightarrow\text{K}=2\pm\sqrt{12}$
View full question & answer→MCQ 71 Mark
The number of solutions of $ x^2+ |x−1| = 1$ is:
Answer
$x^2 + |x - 1| = x^2 + x - , x > 1$
$= x^2 - x + 1, x < 1$
$x^2 + x − 1 = 1$
$\Rightarrow x^2+ x - 2 = 0$
$\Rightarrow x^2 + 2x - x - 2 = 0$
$\Rightarrow x(x + 2) -1 (x + 2) = 0$
$\Rightarrow x + 2 = 0$ or $ x - 1 = 0$
$\Rightarrow x = -2$ or $x = 1$
Since $−2$ does not satisfy the condition $x \geq 1$
$x^2 - x + 1 = 1$
$\Rightarrow x^2 - x = 0$
$\Rightarrow x (x - 1) = 0$
$\Rightarrow x = 0 or (x - 1) = 0$
$\Rightarrow x = 0, x = 1$
$x = 1$ does not satisfy the condition $x < 1$
So, there are two solutions.
View full question & answer→MCQ 81 Mark
If the difference of the roots of $x^2− px + q = 0$ is unity, then:
- A
$p^2+ 4q = 1$
- ✓
$p^2− 4q = 1$
- C
$p^2+ 4q^2 = (1 + 2q)^2$
- D
$4p^2+ q^2= (1 + 2p)^2$
AnswerCorrect option: B. $p^2− 4q = 1$
Given equation: $x^2 + px + q = 0$
Also, $\alpha$ and $\beta$ are the roots of the equation such that $\alpha-\beta=1$
Sum of the roots $=\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\Big(\frac{-\text{p}}{1}\Big)=\text{p}$
Product of the roots $=\alpha\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2}=\text{q}$
$\therefore(\alpha+\beta)^2-(\alpha-\beta)^2=4\alpha\beta$
$\Rightarrow p^2 - 1 = 4q$
$\Rightarrow p^2 - 4q =1.$
View full question & answer→MCQ 91 Mark
The values of $k$ for which the quadratic equation $kx^2 + 1 = kx + 3x - 11x^2$ has real and equal roots are:
- A
$-11, -3$
- B
$5, 7$
- ✓
$5, -7$
- D
AnswerCorrect option: C. $5, -7$
The given equation is $kx^2 + 1 = kx + 3x - 11x^2 $ which can be written as.
$kx^2 + 11x^2 - kx - 3x + 1 =$
$\Rightarrow (k + 11)x^2 - (k + 3)x + 1 = 0$
For equal and real roots, the discriminant of $(k + 11)x^2 - (k + 3)x + 1 = 0.$
$\therefore (k + 3)^2 -4(k + 11) = 0$
$\Rightarrow k^2+ 2k - 35 = 0$
$\Rightarrow (k - 5)(k + 7) = 0$
$\Rightarrow k = 5, -7$
Hence, the equation has real and equal roots when $k = 5, -7.$
View full question & answer→MCQ 101 Mark
The value of $p$ and $q(\text{P}\neq0,\ \text{q}\neq0)$ for which $p, q$ are the roots of the equation $x^2 + px + q = 0$ are:
- ✓
$p = 1, q = −2$
- B
$p = −1, q = −2$
- C
$p = −1, q = 2$
- D
$p = 1, q = 2$
AnswerCorrect option: A. $p = 1, q = −2$
It is given that, $p$ and $q (\text{P}\neq0,\ \text{q}\neq0)$are the roots of the equation $x2 + px + q = 0$
$\therefore$ Sum of roots $= p + q = −p$
$\Rightarrow 2p + q = 0 ...(1)$
Product of roots $= pq = q$
$\Rightarrow q (p − 1) = 0$
$\Rightarrow p = 1, q = 0$
Now, substituting $p = 1$ in $(1),$ we get,
$2 + q = 0$
$\Rightarrow q = −2$
View full question & answer→MCQ 111 Mark
The set of all values of $m$ for which both the roots of the equation $x^2− (m + 1) x + m + 4 = 0$ are real and negative, is:
AnswerCorrect option: C. $[-4, -3]$
The roots of the quadratic equation $x^2− (m + 1) x + m + 4 = 0$ will be real, if its discriminant is greater than or equal to zero.
$\therefore (m + 1)^2 - 4 (m + 4) > 0$
$\Rightarrow (m - 5) (m + 3) > 0$
$\Rightarrow m < -3$ or $m > 5 ...(1)$
It is also given that, the roots of $x^2− (m + 1) x + m + 4 = 0$ are negative.
So, the sum of the roots will be negative.
$\therefore$ Sum of the roots $< 0$
$\Rightarrow m + 1 < 0 ....(2)$
and product of zeros $> 0$
$\Rightarrow m + 4 > 0 ...(3)$
From $(1), (2)$ and $(3),$ we get,
$[−4, −3]$
View full question & answer→MCQ 121 Mark
The number of real solutions of $∣2x − x^2−3∣=1$ is:
AnswerGiven equation: $∣2x − x^2 − 3∣ = 1$
$2x − x^2 − 3 = 1$
$\Rightarrow 2x − x^2 − 4 = 0$
$\Rightarrow x^2 − 2x + 4 = 0$
$\Rightarrow (x − 2)^2 = 0$
$\Rightarrow x = 2, 2$
$- 2x + x^2 + 3 = 1$
$\Rightarrow x^2 - 2x + 2 = 0$
$\Rightarrow x^2 - 2x + 1 + 1 = 0$
$\Rightarrow (x - 1 + i) (x - 1 - i) = 0$
$\Rightarrow x = 1 - i, 1 + i$
Hence, the real solutions are $2, 2.$
View full question & answer→MCQ 131 Mark
If the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$ have a non-zero common roots, then $\lambda=$
AnswerLet a be the common roots of the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$
Therefore
$\alpha^2+2\text{a}+3\lambda=0\ ...(1)$
$2\alpha^2+3\alpha+5\lambda=0\ ...(2)$
Solving (1) and (2) by cross multiplication, we get
$\frac{\alpha^2}{10\lambda-9\lambda}=\frac{\alpha}{6\lambda-5\lambda}=\frac{1}{3-4}$
$\Rightarrow\text{a}^2=-\lambda,\alpha=-\lambda$
$\Rightarrow-\lambda=\lambda^2$
$\Rightarrow\lambda=-1$
View full question & answer→MCQ 141 Mark
If $\alpha,\beta$ are the roots of the equation $ax^2+ bx + c = 0,$ then $\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}$
- A
$\frac{\text{c}}{\text{ab}}$
- B
$\frac{\text{a}}{\text{bc}}$
- ✓
$\frac{\text{b}}{\text{ac}}$
- D
AnswerCorrect option: C. $\frac{\text{b}}{\text{ac}}$
Given equation: $ax^2 + bx + c = 0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots $=\alpha+\beta=-\frac{\text{b}}{\text{a}}$
Product of the roots $=\alpha\beta=\frac{\text{c}}{\text{a}}$
$\therefore\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}$
$=\frac{\text{a}\beta+\text{b}+\text{a}\alpha+\text{b}}{(\text{a}\alpha+\text{b})(\text{a}\beta+\text{b})}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}\alpha+\text{ab}\beta+\text{b}^2}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}(\alpha+\beta)+\text{b}^2}$
$=\frac{\text{a}\big(-\frac{\text{b}}{\text{a}}\big)+2\text{b}}{\text{a}^2\big(\frac{\text{c}}{\text{a}}\big)+\text{ab}\big(-\frac{\text{b}}{\text{a}}\big)+\text{b}^2}$
$=\frac{\text{b}}{\text{ac}}$
View full question & answer→MCQ 151 Mark
The equation of the smallest degree with real coefficients having $1 + i$ as one of the roots is:
- A
$x^2 + x + 1 = 0$
- ✓
$x^2− 2x + 2 = 0$
- C
$x^2 + 2x + 2 = 0$
- D
$x^2 + 2x − 2 = 0$
AnswerCorrect option: B. $x^2− 2x + 2 = 0$
We know that, imaginary roots of a quadratic equation occur in conjugate pair.
It is given that, $1 + i$ is one of the roots.
So, the other root will be $1−i1 - i.$
Thus, the quadratic equation having roots $1 + i $ and $1 - i$ is,
$x^2 - (1 + i + 1 - i)x + (1 + i)(1 - i) = 0$
$\Rightarrow x^2 - 2x + 2 = 0$
View full question & answer→MCQ 161 Mark
If $\alpha,\beta$ are the roots of the equation $x2 + px + 1 = 0; \gamma,\delta$ the roots of the equation $x^2+ qx + 1 = 0$, then $(\alpha-\gamma)(\alpha+\delta)(\beta-\delta)(\beta+\delta)=$
- ✓
$q^2− p^2$
- B
$p^2 − q^2$
- C
$p^2 + q^2$
- D
AnswerCorrect option: A. $q^2− p^2$
Given: $\alpha$ and $\beta$ are the roots of the equation $x^2 + px + 1 = 0.$
Also, $\gamma$ and $\delta\gamma$ and $\delta$ are the roots of the equation $x^2 + qx + 1 = 0$
Then, the sum and the product of the roots of the given equation are as follows:
$\alpha+\beta=-\frac{\text{p}}{1}=-\text{p}$
$\alpha\beta=\frac{1}{1}=1$
$\gamma+\delta=-\frac{\text{q}}{1}=-\text{P}$
$\gamma\delta=\frac{1}{1}=1$
Moreover, $(\gamma-\delta)^2=\gamma^2+\delta^2+2\gamma\delta$
$\Rightarrow\gamma^2+\delta^2=\text{q}^2-2$
$\therefore(\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta+\delta)=(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)$
$=(\alpha\beta-\alpha\gamma-\beta\gamma+\gamma^2)(\alpha\beta+\alpha\gamma+\beta\delta+\delta^2)$
$=[\alpha\beta-\gamma(\alpha+\beta)+\gamma^2][\alpha\beta+\delta(\alpha+\beta)+\delta^2])$
$=(1-\gamma(-\text{p})+\gamma^2)(1+\delta(-\text{p}+\delta^2))$
$=(1+\gamma\text{p}+\gamma^2)(1-\delta\text{p}+\delta^2)$
$=(1+\gamma\text{p}+\gamma^2)(1-\delta\text{p}+\delta^2)$
$=1-\text{p}\delta+\delta^2+\text{p}\gamma-\text{p}^2\gamma\delta+\text{p}\gamma\delta^2+\gamma^2-\text{p}\delta\text{y}^2+\gamma^2\delta^2$
$=1-\text{p}\delta+\text{p}\gamma+\delta^2-\text{p}^2\gamma\delta+\text{p}\gamma\delta^2+\gamma^2=\text{p}\delta\gamma^2+\gamma^2\delta^2$
$=1-\text{p}(\delta-\gamma)-\text{p}^2\gamma\delta+\text{p}\gamma\delta(\delta-\gamma)+(\gamma^2+\delta)+1$
$=1-\text{p}^2+(\delta-\gamma)\text{p}(\delta-\gamma)+(\gamma^2+\delta^2)+1$
$=-\text{P}^22+(\delta-\gamma)\text{p}(1-1)+\text{q}^2$
$=\text{q}^2-\text{p}^2$
View full question & answer→MCQ 171 Mark
The value of a such that $x^2 - 11x + a = 0$ and $x^2 - 14x + 2a = 0$ may have a common root is:
AnswerLet $\alpha$ be the common roots of the equations
$x^2 - 11x + a = 0$ and $x^2 - 14x + 2a = 0$
Therefore,
$\alpha^2-11\alpha+\alpha=0\ ...(1)$
$\alpha^2-14\alpha+2\alpha=0\ ...(2)$
Solving $(1)$ and $(2)$ by cross multiplication, we get,
$\frac{\alpha^2}{-22\alpha+14\alpha}=\frac{\alpha}{\alpha-2\alpha}=\frac{1}{-14+11}$
$\Rightarrow\alpha^2=\frac{-22\alpha+14\alpha}{-14+11},\alpha=\frac{\alpha-2\alpha}{-14+11}$
$\Rightarrow\alpha^2=\frac{-8\alpha}{-3}=\frac{8\alpha}{3},\alpha=\frac{-\alpha}{-3}=\frac{\alpha}{3}$
$\Rightarrow\Big(\frac{\alpha}{3}\Big)^2=\frac{8\alpha}{3}$
$\Rightarrow\alpha^2=24\alpha$
$\Rightarrow\alpha^2-24\alpha=0$
$\Rightarrow\alpha(\alpha-24)=0$
$\Rightarrow\alpha=0$ or $\alpha=24$
View full question & answer→MCQ 181 Mark
For the equation $|x|^2+ |x| - 6 = 0,$ the sum of the real roots is:
AnswerLet $P = |x|$
$\Rightarrow p^2 + p - 6 = 0$
$\Rightarrow p^2 + 3p - 2p - 6 = 0$
$\Rightarrow (p + 3) (p - 2) = 0$
$\Rightarrow p = -3, 2$
Also,$ |x| = p$
$\Rightarrow |x| = 2,$ or $|x| = -3$
Modules can not be negative,
$\therefore|\text{x}| = 2$
$\Rightarrow\text{x} = \pm 2$
$\Rightarrow x = 2$ or $−2$
Sum of the roots of $x$ is $0.$
View full question & answer→MCQ 191 Mark
If $x$ is real and $\text{K}=\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1},$ then
AnswerCorrect option: A. $\text{K}\in\Big[\frac{1}{3,3}\Big]$
$\text{K}=\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1}$
$\Rightarrow kx^2 + kx + K = x^2 - x + 1$
$\Rightarrow (k - 1)x^2 + (k + 1) x + k - 1 = 0$
For real values of $x$, the discriminant of $(k - 1)x^2 + (k + 1) x + k - 1 = 0$ should be greater than or equal to zero.
$\therefore$ If $\text{k}\neq1$
$(k + 1)^2 - 4(k - 1) (k - 1) > 0$
$\Rightarrow(\text{k}+1)^2-\big\{2(\text{k}-1)\big\}^2>0$
$\Rightarrow (3k - 1) (k - 3) < 0$
$\Rightarrow\frac{1}{3}<\text{K}<3$
And if $k = 1,$ then,
$x = 0,$ which is real $...(ii)$
So, from $(i)$ and $(ii)$, we get,
$\text{k}\in\Big[\frac{1}{3},3\Big]$
View full question & answer→MCQ 201 Mark
If $\alpha,\beta$ are roots of the equation $4x^2+ 3x + 7 = 0,$ then $\frac{1}{\alpha}+\frac{1}{\beta}$ is equal to:
- A
$\frac{7}{3}$
- B
$\frac{-7}{3}$
- C
$\frac{3}{7}$
- ✓
$\frac{-3}{7}$
AnswerCorrect option: D. $\frac{-3}{7}$
Given equation: $4x^2 + 3x + 7 = 0$
Also, $\alpha$ and $\beta$ are the roots of the equation.
Sum of the roots $=\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\frac{3}{4}$
Product of the roots $=\alpha\beta=\frac{\text{Coefficient term}}{\text{Coefficient of x}^2}=\frac{7}{4}$
$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-\frac{3}{4}}{\frac{7}{4}}=-\frac{3}{7}$
View full question & answer→MCQ 211 Mark
The number of roots of the equation $\frac{(\text{x}+2)(\text{x}-5)}{(\text{x}+3)(\text{x}+6)}=\frac{\text{x}-2}{\text{x}+4}$ is:
Answer$\frac{(\text{x}+2)(\text{x}-5)}{(\text{x}+3)(\text{x}+6)}=\frac{\text{x}-2}{\text{x}+4}$
$\Rightarrow (x^2 - 3x - 10) (x + 4) = (x^2 + 3x - 18) (x - 2)$
$\Rightarrow x^3 + 4x^2- 3x^2 - 12x - 10x - 40 = x^3 - 2x^2 + 3x^2 - 6x - 18x + 36$
$\Rightarrow x^2 - 22x - 40 = x^2 - 24x + 36$
$\Rightarrow 2x = 76$
$\Rightarrow x = 38$
Hence, the equation has only $1$ root.
View full question & answer→MCQ 221 Mark
If $\alpha\beta$ are the roots of the equation $x^2+ px + q = 0$ then $-\frac{1}{\alpha}+\frac{1}{\beta}$ are the roots of the equation:
- A
$x^2− px + q = 0$
- B
$x^2+ px + q = 0$
- ✓
$qx^2+ px + 1 = 0$
- D
$qx^2− px + 1 = 0$
AnswerCorrect option: C. $qx^2+ px + 1 = 0$
Given equation: $x^2 + px + q = 0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots $=\alpha+\beta=-\text{p}$
Product of the roots $=\alpha\beta=\text{q}$
Now, for roots $-\frac{1}{\alpha,}-\frac{1}{\beta},$ we have:
Sum of the roots = $-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha\beta}=-\Big(\frac{-\text{p}}{\text{q}}\Big)=\frac{\text{p}}{\text{q}}$
Product of the roots = $\frac{1}{\alpha\beta}=\frac{1}{\text{q}}$
Hence, the equation involving the roots $-\frac{1}{\alpha},-\frac{1}{\beta}$ is as follows:
$\text{x}^2+(\alpha+\beta)\text{x}+\alpha\beta=0$
$\Rightarrow\text{x}^2-\frac{\text{p}}{\text{q}}\text{x}+\frac{1}{\text{q}}=0$
$\Rightarrow\text{qx}^2-\text{px}+1=0$
View full question & answer→MCQ 231 Mark
If $\alpha,\beta$ are the roots of the equation $x^2− p(x + 1) − c = 0,$ then $(\alpha+1)(\beta+1)=$
AnswerCorrect option: C. $1 - c$
Given equation:
$x^2 − p(x + 1) − c = 0$
or $x^2 − px − p − c = 0$
Also $\alpha $ and $\beta$ are the roots of the equation.
Sum of the roots $=\alpha+\beta=\text{p}$
Product of the roots $=\alpha\beta=-(\text{c}+\text{p})$
Then, $(\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1$
$=-(\text{c}+\text{p})+\text{p}+1$
$=1-\text{c}$
$=-\text{c}-\text{p}+\text{p}+1$
View full question & answer→MCQ 241 Mark
If $a, b$ are the roots of the equation $x^2+ x + 1 = 0,$ then $a^2 + b^2=$
AnswerGiven equation: $x^2 + x + 1 = 0$
Also, $a$ and $b$ are the roots of the given equation.
Sum of the roots $=\text{a}+\text{b}=\frac{-\text{Co-efficient of x}}{\text{C-oefficient of x}^2}=-\frac{1}{1}=-1$
Product of the roots $=\text{ab}=\frac{\text{constant term}}{\text{Coefficient of x}}=\frac{1}{1}=1$
$\therefore (a + b)^2 = a^2 + b^2 + 2ab$
$\Rightarrow (-1)^2 = a^2 + b^2 + 2 \times 1$
$\Rightarrow 1 - 2 = a^2 + b^2$
$\Rightarrow a^2 + b^2 $
$= -1$
View full question & answer→MCQ 251 Mark
The least value of $k$ which makes the roots of the equation $x^2+ 5x + k = A0$ imaginary is:
AnswerThe roots of the quadratic equation $x^2 + 5x + k=0$ will be imaginary if its discriminant is less than zero.
$\therefore25-4\text{0k}<0$
$\Rightarrow\text{k}>\frac{25}{4}$
Thus, the minimum integral value of $k$ for which the roots are imaginary is $7.$
View full question & answer→