MCQ 11 Mark
If the roots of $x^2− bx + c = 0$ are two consecutive integers, then $b^2 − 4 c$ is:
Answer
- 1
Solution:
Given 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
- 2
Solution:
$\left(x^2+2 x\right)^2-(x+1)^2-55=0$
$\Rightarrow\left(x^2+2 x+1-1\right)^2-(x+1)^2-55=0$
$\Rightarrow\left\{(x+1)^2-1\right\}^2-(x+1)^2-55=0$
$\Rightarrow\left\{(x+1)^2\right\}^2+1-3(x+1)^2-55=0$
$\Rightarrow\left\{(x+1)^2\right\}^2-3(x+1)^2-54=0$
Let $\mathrm{p}=(\mathrm{x}+1)^2$
$\Rightarrow p^2-3 p-54=0$
$\Rightarrow p^2-9 p+6 p-54=0$
$\Rightarrow(p+6)(p-9)=0$
$\Rightarrow p=9 \text { or } p=-6$
Rejecting $p=-6$
$\Rightarrow(x+1)^2=9$
$\Rightarrow x^2+2 x-8=0$
$\Rightarrow(x+4)(x-2)=0$
$\Rightarrow x=2, x=-4$ 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. $-\frac{3}{7}$
- $-\frac{3}{7}$
Solution:
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. $\frac{49}{4}$
- $\frac{49}{4}$
Solution:
It is given that, 4 is the root of the equation $\mathrm{x}^2+\mathrm{px}+12=0$.
$\therefore 16+4 p+12=0$
$\Rightarrow p=-7$
It is also given that, the equation $x^2+p x+q=0$ has equal roots. So, the discriminant of: $x^2+p x+q=0$ will be zero.
$\therefore p^2-4 q=0$
$\Rightarrow 4 q=(-7)^2=49$
$\Rightarrow q=\frac{49}{4}$ View full question & answer→MCQ 51 Mark
The values of x satisfying $log_3 (x^2+ 4x + 12) = 2$ are:
Answer
- −1, −3
Solution:
The given equation is $\log _3\left(x^2+4 x+12\right)=2$
$\Rightarrow x^2+4 x+12=3^2=9$
$\Rightarrow x^2+4 x+3=0$
$\Rightarrow(x+1)(x+3)=0$
$\Rightarrow x=-1,-3$ 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. $2\pm\sqrt{12}$
- $2\pm\sqrt{12}$
Solution:
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
- 2
Solution:
$x^2+|x-1|=x^2+x-, x>1$
$=x^2-x+1, x<1$
i. $x^2+x-1=1$
$\Rightarrow x^2+x-2=0$
$\Rightarrow x^2+2 x-x-2=0$
$\Rightarrow x(x+2)-1(x+2)=0$
$\Rightarrow x+2=0 \text { or } x-1=0$
$\Rightarrow x=-2 \text { or } x=1$
Since -2 does not satisfy the condition $x \geq 1$
ii. $x^2-x+1=1$
$\Rightarrow x^2-\mathrm{x}=0$
$\Rightarrow \mathrm{x}(\mathrm{x}-1)=0$
$\Rightarrow \mathrm{x}=0 \text { or }(\mathrm{x}-1)=0$
$\Rightarrow \mathrm{x}=0, \mathrm{x}=1$
$x=1$ 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 $x^2-p x+q=0$ is unity, then:
- A
$p^2+4 q=1$
- ✓
$p^2-4 q=1$
- C
$p^2+4 q^2=(1+2 q)^2$
- D
$4 p^2+q^2=(1+2 p)^2$
AnswerCorrect option: B. $p^2-4 q=1$
- $p^2-4 q=1$
Solution:
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:
Answer
- 5, -7
Solution:
The given equation is $k x^2+1=k x+3 x-11 x^2$ which can be written as.
$k x^2+11 x^2-k x-3 x+1=$
$\Rightarrow(k+11) x^2-(k+3) x+1=0$
For equal and real roots, the discriminant of $(k+11) x^2-(k+3) x+1=0$.
$\therefore(k+3)^2-4(k+11)=0$
$\Rightarrow k^2+2 k-35=0$
$\Rightarrow(k-5)(k+7)=0$
$\Rightarrow k=5,-7$
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:
Answer
- p = 1, q = −2
Solution:
It is given that, p and q $(\text{P}\neq0,\ \text{q}\neq0)$are the roots of the equation $x2 + px + q = 0$
$\therefore \text { Sum of roots }=p+q=-p$
$\Rightarrow 2 p+q=0 \ldots(1)$
$\text { Product of roots }=p q=q$
$\Rightarrow q(p-1)=0$
$\Rightarrow p=1, q=0$
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:
Answer
- [-4, -3]
Solution:
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.
$\therefore(m+1)^2-4(m+4)>0$
$\Rightarrow(m-5)(m+3)>0$
$\Rightarrow m<-3 \text { or } m>5 \ldots(1)$
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 $<0$
$\Rightarrow \mathrm{m}+1<0$
and product of zeros $>0$
$\Rightarrow m+4>0 \ldots(3)$
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:
Answer
- 2
Solution:
i. Given equation: $\left|2 x-x^2-3\right|=1$
$2 x-x^2-3=1$
$\Rightarrow 2 x-x^2-4=0$
$\Rightarrow x^2-2 x+4=0$
$\Rightarrow(x-2)^2=0$
$\Rightarrow x=2,2$
ii. $-2 x+x^2+3=1$
$\Rightarrow x^2-2 x+2=0$
$\Rightarrow x^2-2 x+1+1=0$
$\Rightarrow(x-1+i)(x-1-i)=0$
$\Rightarrow x=1-i, 1+i$
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. $\frac{\text{b}}{\text{ac}}$
- $\frac{\text{b}}{\text{ac}}$
Solution:
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
$x^2+x+1=0$
- ✓
$x^2-2 x+2=0$
- C
$x^2+2 x+2=0$
- D
$x^2+2 x-2=0$
AnswerCorrect option: B. $x^2-2 x+2=0$
- $x^2-2 x+2=0$
Solution:
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,
$x^2-(1+i+1-i) x+(1+i)(1-i)=0$
$\Rightarrow x^2-2 x+2=0$ 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. $q^2-p^2$
- $q^2-p^2$
Solution:
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:
Answer
- 24
Solution:
Let $\alpha$ 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:
Answer
- 0
Solution:
Let $P=|x|$
$\Rightarrow \mathrm{p}^2+\mathrm{p}-6=0$
$\Rightarrow \mathrm{p}^2+3 \mathrm{p}-2 \mathrm{p}-6=0$
$\Rightarrow(\mathrm{p}+3)(\mathrm{p}-2)=0$
$\Rightarrow p=-3,2$
Also, $|x|=p$
$\Rightarrow|x|=2$, or $|x|=-3$
Modules can not be negative,
$\therefore|x|=2$
$\Rightarrow x= \pm 2$
$\Rightarrow x=2 \text { or }-2$
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. $\text{K}\in\Big[\frac{1}{3,3}\Big]$
- $\text{K}\in\Big[\frac{1}{3,3}\Big]$
Solution:
$\text{K}=\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1}$
$\Rightarrow kx^2 + kx + K = x^2 - x + 1$
$\Rightarrow (k - 1)x^2 + (k + 1) x + k - 1 = 0$
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. $\frac{-3}{7}$
- $\frac{-3}{7}$
Solution:
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
- 1
Solution:
$\frac{(\text{x}+2)(\text{x}-5)}{(\text{x}+3)(\text{x}+6)}=\frac{\text{x}-2}{\text{x}+4}$
$\Rightarrow\left(x^2-3 x-10\right)(x+4)=\left(x^2+3 x-18\right)(x-2)$
$\Rightarrow x^3+4 x^2-3 x^2-12 x-10 x-40=x^3-2 x^2+3 x^2-6 x-18 x+36$
$\Rightarrow x^2-22 x-40=x^2-24 x+36$
$\Rightarrow 2 x=76$
$\Rightarrow x=38$
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
$x^2-p x+q=0$
- B
$x^2+p x+q=0$
- ✓
$q x^2+p x+1=0$
- D
$q x^2-p x+1=0$
AnswerCorrect option: C. $q x^2+p x+1=0$
- $q x^2+p x+1=0$
Solution:
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 $(\alpha+1)(\beta+1)=$
Answer
- 1 - c
Solution:
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 $a^2 + b^2=$
Answer
- -1
Solution:
Given 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 $=\text{ab}=\frac{\text{constant term}}{\text{Coefficient of x}}=\frac{1}{1}=1$
$\therefore(a+b)^2=a^2+b^2+2 a b$
$\Rightarrow(-1)^2=a^2+b^2+2 \times 1$
$\Rightarrow 1-2=a^2+b^2$
$\Rightarrow a^2+b^2=-1$ 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:
Answer
- 7
Solution:
The 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→