MCQ 11 Mark
Which of the following is a true statement?
AnswerCorrect option: D. $5 x^3$ is a monomial.
$5 x^3$ is a monomial as it contains only one term.
View full question & answer→MCQ 21 Mark
If $\alpha$ and $\beta$ are the zero of $2x^2+ 5x − 8,$ then the value of $(\alpha\beta)$ is:
- A
$\frac{-5}{2}$
- B
$\frac{5}{2}$
- ✓
$\frac{-9}{2}$
- D
$\frac{9}{2}$
AnswerCorrect option: C. $\frac{-9}{2}$
Let $\alpha$ and $\beta$ be the zeros of the $2x^2+ 5x - 9$
Then, we have
$\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{-9}{2}$
View full question & answer→MCQ 31 Mark
The zeros of the polynomial $\text{x}^2+\frac{1}{6}\text{x}-2$ are:
- A
$-3,\ 4$
- ✓
$\frac{-3}{2},\ \frac{4}{3}$
- C
$\frac{-4}{3},\ \frac{3}{2}$
- D
AnswerCorrect option: B. $\frac{-3}{2},\ \frac{4}{3}$
$\text{f}(\text{x})=\text{x}^2+\frac{1}{6}\text{x}-2$
Now, f(x) = 0
$\Rightarrow\text{x}^2+\frac{1}{6}\text{x}-2=0$
$ \Rightarrow 6 x^2+x-12=0 $
$ \Rightarrow 6 x^2+9 x-8 x-12=0 $
$ \Rightarrow 3 x(2 x+3)-4(2 x+3)=0 $
$ \Rightarrow(2 x+3)(3 x-4)=0 $
$ \Rightarrow 2 x+3=0 \text { or } 3 x-4=0$
$\Rightarrow\text{x}=-\frac{3}{2}$ or $\text{x}=\frac{4}{3}$
So, the zeros of given polynomial are $-\frac{3}{2}$ and $-\frac{4}{3}$
View full question & answer→MCQ 41 Mark
If $\alpha,\ \beta$ are the zeros of $kx^2- 2x + 3k$ such that $\alpha+\beta=\alpha\beta,$ then k = ?
- A
$\frac{1}{3}$
- B
$\frac{-1}{3}$
- ✓
$\frac{2}{3}$
- D
$\frac{-2}{3}$
AnswerCorrect option: C. $\frac{2}{3}$
Here, $p(x) = x^2- 2x + 3k$
Comparing the given polynomial with $ax^2+ bx + c,$ we get:
$a = 1, b = -2$ and $c = 3k$
It is given that $\alpha$ and $\beta$ are the roots of the polynomial.
$\therefore\alpha+\beta=-\frac{\text{b}}{\text{a}}$
$\Rightarrow\alpha+\beta=-\Big(\frac{-2}{1}\Big)$
$\Rightarrow\alpha+\beta=2\dots(\text{i})$
Also,
$\alpha\beta=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\beta=\frac{\text{5k}}1{}$
$\Rightarrow\alpha\beta=\text{3k}\dots(\text{ii})$
Now, $\alpha+\beta=\alpha\beta$
$\Rightarrow2=\text{3k}$ [Using (i) and (ii)]
$\Rightarrow\text{k}=\frac{2}{3}$
View full question & answer→MCQ 51 Mark
The zeros of the polynomial $x^2- 2x - 34$ are:
- A
$-3, 1$
- B
$-3, -1$
- ✓
$3, -1$
- D
$3, 1$
AnswerCorrect option: C. $3, -1$
$ f(x)=x^2-2 x-3 $
$ =x^2-3 x+x-3 $
$ =x(x-3)+1(x-3)$
$= (x - 3)(x + 1)$
$\therefore f(x) = 0$
$(x - 3)(x + 1) = 0$
$x - 3 = 0$ or $x + 1 = 0$
$x = 3$ or $x = - 1$
So, the zeros of given polynomial are $3$ and $-1$
View full question & answer→MCQ 61 Mark
The zeros of the quadratic polynomial $x^2+ 88x + 125$ are:
- A
- ✓
- C
One positive and one negative.
- D
AnswerLet $\alpha$ and $\beta$ be the zeros of the $x^2+ 88x + 125$
Then, we have
$\alpha+\beta=-\frac{\text{b}}{\text{a}}=-88$ and $\alpha\beta=\frac{\text{c}}{\text{a}}=125$
Now, this is applicable only if $\alpha$ and $\beta$ are both negative.
View full question & answer→MCQ 71 Mark
If $\alpha$ and $\beta$ are the zero of $x^2+ 5x + 8$, then the value of $(\alpha+\beta)$ is:
Answer$-5$
Let $\alpha$ and $\beta$ be the zeros of the $x^2+ 5x + 8$
Then, we have
$\alpha+\beta=-\frac{\text{b}}{\text{a}}$
$=-\frac{5}{1}=-5$
View full question & answer→MCQ 81 Mark
Which of the following is not a polynomial?
- A
$\sqrt3\text{x}^2-2\sqrt3\text{x}+5$
- B
$\text{9x}^2-\text{4x}+\sqrt2$
- C
$\frac{3}{2}\text{x}^3+\text{6x}^2-\frac{1}{\sqrt2}\text{x}-8$
- ✓
$\text{x}+\frac{3}{\text{x}}$
AnswerCorrect option: D. $\text{x}+\frac{3}{\text{x}}$
An expression of the form $p(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$, where $a_n \neq 0$, is called a polynomial in x of degree n.
Here, $a_0, a_1, a_2, \ldots, a_n$ are real numbers and each power of x is a non-negative integer.
Hence, $\text{x}+\frac{3}{\text{x}}$ is not a polynomial.
View full question & answer→MCQ 91 Mark
The zeros of the polynomial $7\text{x}^2-\frac{11\text{x}}{3}-\frac{2}{3}$ are:
- ✓
$\frac{2}{3},\ \frac{-1}{7}$
- B
$\frac{2}{7},\ \frac{-1}{3}$
- C
$\frac{-2}{3},\ \frac{1}{7}$
- D
AnswerCorrect option: A. $\frac{2}{3},\ \frac{-1}{7}$
$\text{f}(\text{x})=7\text{x}^2-\frac{11\text{x}}{3}-\frac{2}{3}$
Now, f(x) = 0
$\Rightarrow7\text{x}^2-\frac{11\text{x}}{3}-\frac{2}{3}=0$
$ \Rightarrow 21 x^2-11 x-2=0 $
$ \Rightarrow 21 x^2-14 x+3 x-2=0 $
$ \Rightarrow 7 x(3 x-2)+1(3 x-2)=0 $
$ \Rightarrow(3 x-2)(7 x+1)=0 $
$ \Rightarrow 3 x-2=0 \text { or } 7 x+1=0$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}{7}$
So, the zeros of given polynomial are $\frac{2}{3}$ and $-\frac{1}{7}$
View full question & answer→MCQ 101 Mark
The zeros of the polynomial $\text{x}^2-\sqrt2\text{x}-12$ are:
- A
$\sqrt2,\ -\sqrt2$
- ✓
$3\sqrt2,\ -2\sqrt2$
- C
$-3\sqrt2,\ 2\sqrt2$
- D
$3\sqrt2,\ 2\sqrt2$
AnswerCorrect option: B. $3\sqrt2,\ -2\sqrt2$
$\text{f}(\text{x})=\text{x}^2-\sqrt2\text{x}-12$
$=\text{x}^2-3\sqrt2\text{x}+2\sqrt2\text{x}-12$
$=\text{x}\big(\text{x}-3\sqrt2\big)+2\sqrt2\big(\text{x}-3\sqrt2\big)$
$=\big(\text{x}-3\sqrt2\big)\big(\text{x}+2\sqrt2\big)$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\big(\text{x}-3\sqrt2\big)\big(\text{x}+2\sqrt2\big)=0$
$\Rightarrow\text{x}-3\sqrt2=0$ or $\text{x}+2\sqrt2=0$
$\Rightarrow\text{x}=3\sqrt2$ or $\text{x}=-2\sqrt2$
So, the zeros of given polynomial are $3\sqrt2$ and $-2\sqrt2$
View full question & answer→MCQ 111 Mark
If $-2$ and $3$ are the zeros of the quadratic polynomial $x^2+ (a + 1) x + b,$ then:
- A
$a = -2, b = 6$
- B
$a = 2, b = -6$
- ✓
$a = -2, b = -6$
- D
$a = 2, b = 6$
AnswerCorrect option: C. $a = -2, b = -6$
Since $-2$ and $3$ are zeros of $x^2+ (a + 1)x + b$, we have
Sum of roots $= -2 + 3 = 1$
$\Rightarrow\frac{-(\text{a}+1)}{1}=1$
$\Rightarrow a + 1 = -1$
$\Rightarrow a = -2$
Product of roots $= -2 × 3 = -6$
$\Rightarrow\frac{{\text{b}}}{1}=-6$
$\Rightarrow\text{b}=-6$
View full question & answer→MCQ 121 Mark
If the sum of the zeros of the quadratic polynomial $kx^2+ 2x + 3k$ is equal to the product of its zeros, then $k = ?$
- A
$\frac{1}{3}$
- B
$\frac{-1}{3}$
- C
$\frac{2}{3}$
- ✓
$\frac{-2}{3}$
AnswerCorrect option: D. $\frac{-2}{3}$
Let $\alpha$ and $\beta$ be the roots of $kx^2+ 2x + 3k.$
Then, we have
$\alpha+\beta=\alpha\beta$
$\Rightarrow-\frac{2}{\text{k}}=\frac{\text{3k}}{\text{k}}$
$\Rightarrow-\frac{2}{\text{k}}=3$
$\Rightarrow\text{k}=-\frac{2}{3}$
View full question & answer→MCQ 131 Mark
On dividing a polynomial p(x) by a non-zero polynomial q(x), let g(x) be the quotient and r(x) be the remainder, than p(x) = q(x).g(x) + r(x), where
- A
- B
deg r(x) < deg g(x) always.
- ✓
either r(x) = 0 or deg r(x) < deg g(x).
- D
AnswerCorrect option: C. either r(x) = 0 or deg r(x) < deg g(x).
By Division Algorithm on polynomials, we have
either r(x) = 0 or deg r(x) < deg g(x).
View full question & answer→MCQ 141 Mark
Zeros of $p(x) = x^2- 2x - 3$ are:
- A
$1, -3$
- ✓
$3, -1$
- C
$-3, -1$
- D
$1, 3$
AnswerCorrect option: B. $3, -1$
Here, $p(x)=x^2-2 x-3$
Let $\mathrm{x}^2-2 \mathrm{x}-3=0$
$\Rightarrow \mathrm{x}^2-(3-1) \mathrm{x}-3=0$
$\Rightarrow \mathrm{x}^2-3 \mathrm{x}+\mathrm{x}-3=0$
$\Rightarrow x(x-3)+1(x-3)=0$
$\Rightarrow(x-3)(x+1)=0$
$\Rightarrow x=3,-1$
View full question & answer→MCQ 151 Mark
If one of the zeros of the cubic polynomial $a x^3+b x^2+c x+d$ is 0, then the product of the other two zeros is:
- A
$\frac{-\text{c}}{\text{a}}$
- ✓
$\frac{\text{c}}{\text{a}}$
- C
$0$
- D
$\frac{-\text{b}}{\text{a}}$
AnswerCorrect option: B. $\frac{\text{c}}{\text{a}}$
Let $0,\ \beta,\ \gamma$ be the zeros of the cubic polynomial $a x^3+b x^2+c x+d.$
Now, sum of the products of zeros, taken two at a time is given by
$\Rightarrow0\times\beta+\beta\gamma+\gamma\times0=\frac{\text{c}}{\text{a}}$
$\Rightarrow\beta\gamma=\frac{\text{c}}{\text{a}}$
View full question & answer→MCQ 161 Mark
The zeros of the polynomial $4\text{x}^2+5\sqrt2\text{x}-3$ are:
- A
$-3\sqrt2,\ \sqrt2$
- B
$-3\sqrt2,\ \frac{\sqrt2}{2}$
- ✓
$\frac{-3\sqrt2}{2},\ \frac{\sqrt2}{4}$
- D
AnswerCorrect option: C. $\frac{-3\sqrt2}{2},\ \frac{\sqrt2}{4}$
$\text{f}(\text{x})=4\text{x}^2+5\sqrt2\text{x}-3$
$=4\text{x}^2+6\sqrt2\text{x}-\sqrt2\text{x}-3$
$=2\sqrt2\text{x}\big(\sqrt2\text{x}+3\big)-1\big(\sqrt2\text{x}+3\big)$
$=\big(2\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+3\big)$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\big(2\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+3\big)=0$
$\Rightarrow2\sqrt2\text{x}-1=0$ or $\sqrt2\text{x}+3=0$
$\Rightarrow\text{x}=\frac{1}{2\sqrt2}$ or $\text{x}=-\frac{3}{\sqrt2}$
$\Rightarrow\text{x}=\frac{\sqrt2}{4}$ or $\text{x}=-\frac{3\sqrt2}{2}$
So, the zeros of given polynomial are $\frac{\sqrt2}{4}$ and $-\frac{3\sqrt2}{2}$
View full question & answer→MCQ 171 Mark
If $\alpha,\ \beta,\ \gamma$ be the zeros of the polynomial p(x) such that $(\alpha+\beta+\gamma)=3,\ (\alpha\beta+\beta\gamma+\gamma\alpha)=-10$ and $\alpha\beta\gamma=-24$ then $p(x) = ?$
- A
$ x^3+3 x^2-10 x+24$
- B
$ x^3+3 x^2+10 x-24$
- ✓
$ x^3-3 x^2-10 x+24$
- D
AnswerCorrect option: C. $ x^3-3 x^2-10 x+24$
A cubic polynomial whose zeros are $\alpha,\ \beta$ and $\gamma$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta+\gamma)\text{x}^2+\$\alpha\beta+\beta\gamma+\gamma\alpha)\text{x}-\alpha\beta\gamma$
Given, $(\alpha+\beta+\gamma)=3,\ (\alpha\beta+\beta\gamma+\gamma\alpha)=-10$ and $\alpha\beta\gamma=-24$
Thus, $\text{p}(\text{x})=\text{x}^3-\text{3x}^2-\text{10x}+24$
View full question & answer→MCQ 181 Mark
If $\alpha,\ \beta,\ \gamma$ are the zeros of the polynomial $2 x^3+x^2-13 x+6$, then $\alpha\beta\gamma=?$
- ✓
$-3$
- B
$3$
- C
$\frac{-1}{2}$
- D
$\frac{-13}2{}$
AnswerSince $\alpha,\ \beta,\ \gamma$ are the zeros of $2 x^3+x^2-13 x+6$, we have
$\alpha\beta\gamma=-\frac{\text{d}}{\text{a}}$
$=-\frac{6}{2}=-3$
View full question & answer→MCQ 191 Mark
If one zero of the quadratic polynomial $(k - 1) x^2+ kx + 1 is -4,$ then the value of k is:
- A
$\frac{-5}{4}$
- ✓
$\frac{5}{4}$
- C
$\frac{-4}{3}$
- D
$\frac{4}{3}$
AnswerCorrect option: B. $\frac{5}{4}$
Since $-4$ is a zero of $f(x)=(k-1) x^2+k x+1$, we have
$ f(-4)=0 $
$ \Rightarrow(k-1)(-4)^2+k(-4)+1=0 $
$ \Rightarrow(k-1) 16-4 k+1=0 $
$ \Rightarrow 16 k-16-4 k+1=0 $
$ \Rightarrow 12 k-15=0 $
$ \Rightarrow 12 k=15$
$\Rightarrow\text{k}=-\frac{15}{12}=\frac{5}{4}$
View full question & answer→MCQ 201 Mark
Which of the following is a polynomial?
- A
$\text{x}^2-\text{5x}+4\sqrt{\text{x}}+3$
- B
$\text{x}^{\frac{3}{2}}-\text{x}+\text{x}^{\frac{1}{2}}+1$
- C
$\sqrt{\text{x}}+\frac{1}{\sqrt{ \text{x}}}$
- ✓
$\sqrt2\text{x}^2-3\sqrt3\text{x}+\sqrt6$
AnswerCorrect option: D. $\sqrt2\text{x}^2-3\sqrt3\text{x}+\sqrt6$
An expression of the form $p(x)=a_0+a_1 x+a_2 x^2+\ldots .+a_n x^n$, where $a_n \neq 0$, is called a polynomial in $x$ of degree $n$.
Here, $a_0, a_1, a_2, \ldots, a_n$ are real numbers and each power of $x$ is a non-negative integer.
Hence, $\sqrt2\text{x}^2-3\sqrt3\text{x}+\sqrt6$ is a polynomial.
View full question & answer→MCQ 211 Mark
If two of the zeros of the cubic polynomial $ax^3+ bx^2+ cx + d$ is $0$, then the third zeros is:
- ✓
$\frac{-\text{b}}{\text{a}}$
- B
$\frac{\text{b}}{\text{a}}$
- C
$\frac{\text{c}}{\text{a}}$
- D
$\frac{-\text{d}}{\text{a}}$
AnswerCorrect option: A. $\frac{-\text{b}}{\text{a}}$
Let $0,\ 0,\ \gamma$ be the zeros of the cubic polynomial $ax^3+ bx^2+ cx + d.$
Now, sum of zeros $=-\frac{\text{b}}{\text{a}}$
$\Rightarrow0+0+\gamma=-\frac{\text{b}}{\text{a}}$
$\Rightarrow\gamma=-\frac{\text{b}}{\text{a}}$
View full question & answer→MCQ 221 Mark
It is given that the difference between the zeroes of $4x^2- 8kx + 9$ is 4 and $k > 0.$ Then, $k = ?$
- A
$\frac{1}{2}$
- B
$\frac{3}2{}$
- ✓
$\frac{5}{2}$
- D
$\frac{7}{2}$
AnswerCorrect option: C. $\frac{5}{2}$
Let the zeroes of the polynomial be $\alpha$ and $\alpha+4.$
Here, $p(x) = 4x^2- 8kx + 9$
Comparing the given polynomial with $ax^2+ bx + c,$ we get:
$a = 4, b = -8k$ and $c = 9$
Now, sum of the roots $=-\frac{\text{b}}{\text{a}}$
$\Rightarrow\alpha+\alpha+4=\frac{-(-8\text{k})}{4}$
$\Rightarrow2\alpha=4=\text{2k}$
$\Rightarrow\alpha+2=\text{k}$
$\Rightarrow\alpha=(\text{k}-2)\dots(\text{i})$
Also, product of the roots, $\alpha\beta=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha(\alpha+4)=\frac{9}{4}$
$\Rightarrow(\text{k}-2)(\text{k}-2+4)=\frac{9}{4}$
$\Rightarrow(\text{k}-2)(\text{k}+2)=\frac{9}{2}$
$\Rightarrow\text{k}^2-4=\frac{9}{4}$
$\Rightarrow\text{4k}^2-16=9$
$\Rightarrow\text{4k}^2=25$
$\Rightarrow\text{k}^2=\frac{25}{4}$
$\Rightarrow\text{k}=\frac{5}{2}$ $(\because\text{k}>0)$
View full question & answer→MCQ 231 Mark
If $\alpha,\ \beta,\ \gamma$ are the zeros of the polynomial $x^3- 6x^2- x + 30,$ then the value of $(\alpha\beta+\beta\gamma+\gamma\alpha)$ is:
AnswerHere, $p(x) = x^3- 6x^2- x + 30$
Comparing the given polynomial with
$\text{x}^ 3-(\alpha+\beta+\gamma)\text{x}^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\text{x}-\alpha\beta\gamma,$
we get:
$(\alpha\beta+\beta\gamma+\gamma\alpha)=-1$
View full question & answer→MCQ 241 Mark
A quadratic polynomial whose zeros are $5$ and $-3$, is:
- A
$ x^2+2 x-15 $
- B
$ x^2-2 x+15 $
- ✓
$ x^2-2 x-15$
- D
AnswerCorrect option: C. $ x^2-2 x-15$
Let $\alpha$ and $\beta$ be the zeros of the required quadratic polynomial.
Then, we have
$\alpha+\beta=5+(-3)=2$ and $\alpha\beta=5\times(-3)=-15$
Now, a quadratic polynomial whose zeros are $\alpha$ and $\beta$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\Rightarrow\text{p}(\text{x})=\text{x}^2-(2)\text{x}+(-15)$
$=\text{x}^2-\text{2x}-15$
View full question & answer→MCQ 251 Mark
If one zero of the quadratic polynomial $kx^2+ 3x + k$ is $2$, then the value of k is:
- A
$\frac{5}{6}$
- B
$\frac{-5}{6}$
- C
$\frac{6}{5}$
- ✓
$\frac{-6}{5}$
AnswerCorrect option: D. $\frac{-6}{5}$
Since $2$ is a zero of $f(x) = kx^2+ 3x + k,$ we have
$f(2) = 0$
$\Rightarrow k(2)^2+ 3(2) + k = 0$
$\Rightarrow 4k + 6 + k = 0$
$\Rightarrow 5k + 6 = 0$
$\Rightarrow 5k = -6$
$\Rightarrow\text{k}=-\frac{6}{5}$
View full question & answer→MCQ 261 Mark
A quadratic polynomial whose zeros are $\frac{3}{5}$ and $\frac{-1}{2}$ is:
- A
$ 10 x^2+x+3 $
- B
$ 10 x^2+x-3 $
- C
$ 10 x^2-x+3 $
- ✓
$ 10 x^2-x-3 $
AnswerCorrect option: D. $ 10 x^2-x-3 $
Let $\alpha$ and $\beta$ be the zeros of the required quadratic polynomial.
Then, we have
$\alpha+\beta=\frac{3}{5}+\Big(-\frac{1}{2}\Big)$
$=\frac{6-5}{10}=\frac{1}{10}$
$\alpha\beta=\frac{3}{5}\times\Big(-\frac{1}{2}\Big)=-\frac{3}{10}$
Now, a quadratic polynomial whose zeros are $\alpha$ and $\beta$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\Rightarrow\text{p}(\text{x})=\text{x}^2-\Big(\frac{1}{10}\Big)\text{x}+\Big(-\frac{3}{10}\Big)$
$=\text{x}^2-\frac{1}{10}\text{x}-\frac{3}{10}$ or $10\text{x}^2-\text{x}-3$
View full question & answer→MCQ 271 Mark
If $\alpha,\ \beta$ be the zero of the polynomial $2x^2+ 5x + k$ such that $\alpha^2+\beta^2+\gamma^2=\frac{21}{4}$ then $k = ?$
AnswerSince $\alpha$ and $\beta$ are the zeros of $2x^2+ 5x + k,$ we have
$\alpha+\beta=-\frac{5}{2}$ and $\alpha\beta=\frac{\text{k}}{2}$
Now, $\alpha^2+\beta^2+\alpha\beta=\frac{21}{4}$
$\Rightarrow(\alpha+\beta)^2-\alpha\beta=\frac{21}{4}$
$\Rightarrow\Big(\frac{-5}{2}\Big)^2-\frac{\text{k}}{2}=\frac{21}{4}$
$\Rightarrow\frac{25}{4}-\frac{\text{k}}2{}=\frac{21}{4}$
$\Rightarrow\frac{\text{k}}2{}=\frac{25}{4}-\frac{21}{4}=1$
$\Rightarrow\text{k}=2$
View full question & answer→MCQ 281 Mark
If one of the zeros of the cubic polynomial $x^3+ ax^2+ bx + c$ is $-1$, then the product of the other two zeros is:
- A
$a - b - 1$
- B
$b - a - 1$
- ✓
$1 - a + b$
- D
$1 + a - b$
AnswerCorrect option: C. $1 - a + b$
Since $-1$ is a zero of the cubic polynomial $x^3+a x^2+b x+c$,
$(-1)^3+a(-1)^2+b(-1)+c=0 $
$ \Rightarrow-1+a-b+c=0 $
$ \Rightarrow c=1-a+b$
Now, product of all zeros is given by:
$-1\times\beta\times\gamma=-\text{c}$
$\Rightarrow-\beta\gamma=-(1-\text{a}+\text{b})$
$\Rightarrow\beta\gamma=1-\text{a}+\text{b}$
View full question & answer→MCQ 291 Mark
If one zero of $3x^2+ 8x + k$ be the reciprocal of the other, then $k = ?$
- ✓
$3$
- B
$-3$
- C
$\frac{1}{3}$
- D
$\frac{-1}{3}$
AnswerLet $\alpha$ and $\frac{1}{\alpha}$ be the roots of $3x^2+ 8x + k.$
Then, we have
$\alpha\times\frac{1}{\alpha}=\frac{\text{k}}{3}$
$\Rightarrow1=\frac{\text{k}}{3}$
$\Rightarrow\text{k}=3$
View full question & answer→MCQ 301 Mark
If $\alpha,\ \beta$ are the zeros of the polynomial $x^2+ 6x + 2$, then $\Big(\frac{1}{\alpha}+\frac{1}{\beta}\Big)=?$
AnswerSince $\alpha$ and $\beta$ are the zeros of $x^2+ 6x + 2$, we have
$\alpha+\beta=-\frac{6}{1}=-6$
$\alpha\beta=\frac{2}{1}=2$
$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha\beta}$
$=\frac{-6}{2}=-3$
View full question & answer→MCQ 311 Mark
If $\alpha,\ \beta,\ \gamma$ are the zeros of the polynomial $x^3- 6x^2- x + 30$, then $(\alpha\beta+\beta\gamma+\gamma\alpha)=?$
AnswerSince $\alpha,\ \beta,\ \gamma$ are the zeros of $x^3- 6x^2- 2x + 30,$ we have
$\alpha\beta+\beta\gamma+\gamma\alpha=-\frac{\text{c}}{\text{a}}$
$=-\frac{1}{1}=-1$
View full question & answer→MCQ 321 Mark
The sum and product of the zeros of a quadratic polynomial are $3$ and $-10$ respectively. The quadratic polynomial is:
- A
$ x^2-3 x+10 $
- B
$ x^2+3 x-10 $
- ✓
$ x^2-3 x-10 $
- D
$ x^2+3 x+10 $
AnswerCorrect option: C. $ x^2-3 x-10 $
Let $\alpha$ and $\beta$ be the zeros of the required quadratic polynomial.
Then, we have
$\alpha+\beta=3$ and $\alpha\beta=-10$
Now, a quadratic polynomial whose zeros are $\alpha$ and $\beta$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\Rightarrow\text{p}(\text{x})=\text{x}^2-(3)\text{x}+(-10)$
$=\text{x}^2-\text{3x}-10$
View full question & answer→