Question 13 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$2\sqrt3\text{x}^2-\text{5x}+\sqrt3$
AnswerWe have,
$\text{f}(\text{x})=2\sqrt3\text{x}^2-\text{5x}+\sqrt3$
$=2\sqrt3\text{x}^2-\text{2x}-\text{3x}+\sqrt3$
$=2\text{x}\big(\sqrt3\text{x}-1\big)-\sqrt3\big(\sqrt3\text{x}-1\big)$
$=\big(\sqrt3-1\big)\big(2\text{x}-\sqrt3\big)$
$\therefore\text{f}(\text{x})=0$
$ \Rightarrow\big(\sqrt3\text{x}-1\big)\big(\text{2x}-\sqrt3\big)=0$
$\Rightarrow\sqrt3-1=0$ or $\text{2x}-\sqrt3=0$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}$ or $\text{x}=\frac{\sqrt3}{2}$
So, the zeros of $f(x)$ are $\frac{1}{\sqrt3}$ and $\frac{\sqrt3}{2}$
Sum of zeros $=\frac{1}{\sqrt3}+\frac{\sqrt3}{2}=\frac{2+3}{2\sqrt3}$
$=\frac{5}{2\sqrt3}=\frac{-(\text{Coefficient of x})}{(\text{Coeficient of }\text{x}^2)}$
Product of zeros $=\frac{1}{\sqrt3}\times\frac{\sqrt3}{2}$
$=\frac{\sqrt3}{2\sqrt3}=\frac{\text{Constant term}}{(\text{Coefficient of }\text{x}^2)}$
View full question & answer→Question 23 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$5x^2- 4 - 8x$
AnswerWe have,
$ f(x)=x^2-4-8 x=5 x^2-8 x-4 $
$ =5 x^2-10 x+2 x-4 $
$ =5 x(x-2)+2(x-2) $
$ =(x-2)(5 x+2) $
$ \therefore f(x)=0 $
$ \Rightarrow(x-2)(5 x+2)=0 $
$ \Rightarrow x-2=0 \text { or } 5 x+2=0$
$\therefore\text{x}=2,\frac{-2}{5}$
So, the zeros of $f(x)$ are $2$ and $\frac{-2}{5}$
Sum of zeros $=2+\Big(\frac{-2}{5}\Big)=\frac{10-2}{5}$
$=\frac{8}{5} =\frac{-(\text{Cofficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $2\times\Big(\frac{-2}{5}\Big)$
$=\frac{-4}5{}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 33 Marks
Very-Short-Answer Question:
If $(x + a)$ is a factor of $(2 x^2+2 a x+5 x+10),$ find the value of $a.$
AnswerLet $f(x)=2 x^2+2 a x+5 x+10$
Since $(x+a)$ is a factor of $f(x)$, we have
$ f(-a)=0 $
$ \Rightarrow 2(-a)^2+2 a(-a)+5(-a)+10=0 $
$ \Rightarrow 2 a^2-2 a^2-5 a+10=0 $
$ \Rightarrow 5 a=10 $
$ \Rightarrow a=2$
View full question & answer→Question 43 Marks
Find the zeros of the polynomial $x^2+2 x-195$
AnswerHere, $p(x)=x^2+2 x-195$
Let $p(x)=0$
$ \Rightarrow x^2+(15-13) x-195=0 $
$ \Rightarrow x^2+15 x-13 x-195=0 $
$ \Rightarrow x(x+15)-13(x+15)=0 $
$ \Rightarrow(x+15)(x-13)=0 $
$ \Rightarrow x=-15,13$
Hence, the zeros are $-15$ and $13.$
View full question & answer→Question 53 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$x^2- 5$
AnswerWe have,
$\text{f}(\text{x})=\text{x}^2-5$
$=(\text{x})^2-\big(\sqrt5\big)^2$
$=\big(\text{x}-\sqrt5\big)\big(\text{x}+\sqrt5\big)$ $\big[\therefore\text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\big(\text{x}-\sqrt5\big)\big(\text{x}+\sqrt5\big)=0$
$\therefore\text{x}-\sqrt5=0$ or $\text{x}+\sqrt5=0$
$\Rightarrow\text{x}=\sqrt5$ or $\text{x}=-\sqrt5$
So the zeros of f(x) are $\sqrt5$ and $-\sqrt5$
Sum of zeros $=\sqrt5+(-\sqrt5)=0$
$=-\frac{\text{Coefficient of x}}{\text{Coefficient of }\text{x}^2}$
Product of zeros $=\big(\sqrt5\big)\big(-\sqrt5\big)=-5$
$=\frac{-5}{1}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 63 Marks
Find the quadratic polynomial, sum of whose zeros is $\Big(\frac{5}{2}\Big)$ and their product is $1.$ Hence, find the zeros of the polynomial.
AnswerLet $\alpha,\beta$ be the zeros of required quadratic polynomial $f(x).$
We have,
$\alpha+\beta=\frac{5}{2}$ and $\alpha\beta=1$
Now, $\text{f}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-\frac{5}{2}\text{x}+1$
$=\frac{1}{2}(2\text{x}^2-5+2)$
The polynomial whose zeros are $\alpha,\beta$ is $2x^2- 5x + 2$
Further, $\text{f}(\text{x})=\frac{1}{2}(\text{2x}^2-\text{5x}+2)$
$=\frac{1}{2}(\text{2x}^2-\text{4x}-\text{x}+2)$
$=\frac{1}{2}[\text{2x}(\text{x}-2)-(\text{x}-2)]$
$=\frac{1}{2}(\text{x}-2)(\text{2x}-1)$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\frac{1}{2}(\text{x}-2)(\text{2x}-1)=0$
$\therefore$ for that $x - 2 = 0$ or $2x - 1 = 0$
i.e., Either $x = 2$ or $\text{x}=\frac{1}{2}$
$\therefore$ Zeros of polynomial are $2$ and $\frac{1}{2}$
View full question & answer→Question 73 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$x^2+3 x-10 $
AnswerWe have
$ f(x)=x^2+3 x-10 $
$ =x^2+5 x-2 x-10 $
$ =x(x+5)-2(x+5) $
$ =(x+5)(x-2) $
$ \therefore f(x)=0 $
$ \Rightarrow(x+5)(x-2)=0 $
$ \Rightarrow x+5=0 \text { or } x-2=0 $
$ \Rightarrow x=-5 \text { or } x=2$
Sum of zeros $= (-5) + 2 = -3$
$=\frac{-3}{1} =\frac{-(\text{Cofficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $= (-5) × (2) = -10$
$=\frac{-10}{1}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 83 Marks
Very-Short-Answer Question:
Find the zeros of the polynomial $x^2-3 x-m(m+3)$.
AnswerWe have,
$ f(x)=x^2-3 x-m(m+3)$
$=x^2-(m+3) x+m x-m(m+3)$
$=x[x-(m+3)]+m[x-(m+3)]$
$=(x+m)[x-(m+3)]$
$\therefore f(x)=0$
$\Rightarrow(x+m)[x-(m+3)]=0$
$\Rightarrow x+m=0 \text { or } x-(m+3)=0$
$\Rightarrow x=-m \text { or } x=m+3$
so, the zeros of $f(x)$ are $-m$ and $(m + 3).$
View full question & answer→Question 93 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$x^2+ 7x + 12$
Answer$ x^2+7 x+12=0 $
$x^2+4 x+3 x+12=0 $
$x(x+4)+3(x+4)=0 $
$(x+4)(x+3)=0 $
$(x+4)=0 \text { or }(x+3)=0 $
$x=-4 \text { or } x=-3$
$\text { Sum of zeros }=-4+(-3)$
$=\frac{-7}{1} =\frac{-(\text{Cofficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $= (-4)(-3)$
$=\frac{12}{1}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 103 Marks
Using remainder theorem, find the remainder when $p(x)=x^3+3 x^2-5 x+4$ is divided by $(x - 2).$
AnswerGiven: $p(x)=x^3+3 x^2-5 x+4$
Now, $p(2)=2^3+3(2)^2-5(2)+4$
$ =8+12-10+4 $
$ =14$
View full question & answer→Question 113 Marks
Very-Short-Answer Question:
If the product of the zeros of the quadratic polynomial $x^2- 4x + k$ is $3$ then write the value of k.
AnswerLet $f(x) = x^2- 4x + k$
Product of zeros $=\frac{\text{Constant term}}{\text{Coefficient of x}}$
$=\frac{\text{k}}{\text{1}}=\text{k}$
Given, Product of zeros $= 3$
$\therefore\text{k}=3$
View full question & answer→Question 123 Marks
Very-Short-Answer Question:
If $f(x) = (x^3+ x^2- ax + b)$ is divisible by $(x^2- x)$, write the values of $a$ and $b.$
AnswerSince $f(x) = (x^3+ x^2- ax + b)$ is divisible by $(x^2- x)$, we have
$x^2- x = 0$
$⇒ x(x - 1) = 0$
$⇒ x = 0$ or $x = 1$
Hence,
$f(0) = 0$
$⇒ 0^3+ 0^2- a(0) + b = 0$
$⇒ b = 0$
Also,
$f(1) = 0$
$⇒ 1^3+ 1^2- a(1) + 0 = 0$
$⇒ 1 + 1 - a = 0$
$⇒ a = 2$
$\therefore a = 2$ and $b = 0$
View full question & answer→Question 133 Marks
Very-Short-Answer Question:
If $-4$ is a zero of the quadratic polynomial $x^2- x - (2k + 2)$ then find the value of $k.$
Answer$\text { Since }-4 \text { is a zero of } f(x)=x^2-x-(2 k+2) \text {, we have }$
$ f(-4)=0 $
$ \Rightarrow(-4)^2-(-4)-2 k-2=0 $
$ \Rightarrow 16+4-2 k-2=0 $
$ \Rightarrow 18-2 k=0 $
$ \Rightarrow 2 k=18 $
$ \Rightarrow k=9$
View full question & answer→Question 143 Marks
Find the quadratic polynomial, the sum of whose roots is $\sqrt2$ and their product is $\frac{1}{3}.$
AnswerLet $\alpha$ and $\beta$ be the zeros of required quadratic polynomial.
Then, we have
$\alpha+\beta=\sqrt2$ and $\alpha\beta=\frac{1}{3}$
Now, a quadratic polynomial whose zeros are $\alpha$ and $\beta$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\therefore$ Required quadratic polynomial,
$\text{p}(\text{x})=\text{x}^2-\sqrt2\text{x}+\frac{1}{3}$
View full question & answer→Question 153 Marks
Find a quadratic polynomial whose zeros are $2$ and $-5.$
AnswerIt is given that the two roots of the polynomial are $2$ and $-5.$
Let $\alpha=2$ and $\beta=-5$
Now, sum of the zeroes, $\alpha+\beta=2+(-5)=-3$
Product of the zeroes, $\alpha\beta=2\times-5=-10$
$\therefore$ Required polynomial $\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-(-3)\text{x}+(-10)$
$=\text{x}^2+\text{3x}-10$
View full question & answer→Question 163 Marks
Very-Short-Answer Question:
If $\alpha,\ \beta$ are the zeros of a polynomial such that $\alpha+\beta=6$ and $\alpha\beta=4$ the write the polynomial.
Answer$\alpha$ and $\beta$ be the zeros of the required quadratic polynomial.
And,
$\alpha+\beta=6$ and $\alpha\beta=4$
Now, a quadratic polynomial whose zeros are $\alpha$ and $\beta$ is given by:
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\Rightarrow\text{p}(\text{x})=\text{x}^2-\text{6x}+4$
View full question & answer→Question 173 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$4x^2- 4x - 3$
Answer$ 4 x^2-4 x-3 $
$ =4 x^2-6 x+2 x-3 $
$ =2 x(2 x-3)+1(2 x-3) $
$ =(2 x-3)(2 x+1) $
$ \therefore f(x)=0$
$\text { So, } 4 x^2-4 x-3=0$
$ \Rightarrow(2 x-3)(2 x+1)=0 $
$ \Rightarrow 2 x-3=0 \text { or } 2 x+1=0$
$\Rightarrow\text{x}=\frac{3}{2}$ or $\text{x}=\frac{-1}{2}$
$\therefore$ Zeros of $4x^2- 4x - 3$ are $\frac{3}{2}$ and $-\frac{1}{2}$
Sum of zeros $=\frac{3}{2}+\Big(-\frac{1}{2}\Big)=1$
$=\frac{-(-4)}{4} =-\frac{\text{b}}{\text{a}}=\frac{-(\text{Cofficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $=\frac{3}{2}\times\Big(-\frac{1}{2}\Big)=\frac{-3}{4}$
$=\frac{\text{c}}{\text{a}}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 183 Marks
Find the quadratic polynomial, sum of whose zeros is $0$ and their product is $-1.$ Hence, find the zeros of the polynomial.
AnswerLet $\alpha,\beta$ be the zeros of required quadratic polynomial $f(x).$
We have,
$\alpha+\beta=0$ and $\alpha\beta=-1$
$\therefore$ Polynomial whose zeros are $\alpha,\beta$ is
$\text{f}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-0.\text{x}+(-1)$
$=\text{x}^2-1$
$\therefore$ Required polynomial is $x^2-1$
Now $f(x)=x^2-1$
$=(x-1)(x+1)$
$\therefore f(x)=0$
$?(x-1)(x+1)=0$
$\therefore$ Either $x-1=0$ or $x+1=0$
i.e. Either $x=1$ or $x=-1$
$\therefore$ Zeros of polynomial are $1$ and $-1$
View full question & answer→Question 193 Marks
If $\alpha,\ \beta$ are the zeros of the polynomial $f(x)=x^2-5 x+k$ such that $\alpha-\beta=1,$ find the value of $k.$
AnswerGiven:$f(x)=x^2-5 x+k$
The co-efficients are $a = 1, b = -5$ and $c = k.$
$\therefore\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\Rightarrow\alpha+\beta=-\frac{-\text{5}}{\text{1}}$
$\Rightarrow\alpha+\beta=5\ \dots(1)$
Also, $\alpha-\beta=1\ \dots(2)$
From $(1)\ \&\ (2)$, we get:
$2\alpha=6$
$\Rightarrow\alpha=3$
Putting the value of $\alpha$ in $(1),$ we get $\beta=2$
Now, $\alpha\beta=\frac{\text{c}}{\text{a}}$
$\Rightarrow3\times2=\frac{\text{k}}1{}$
$\therefore\text{k}=6$
View full question & answer→Question 203 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$x^2- 2x - 8$
AnswerWe have,
$ f(x)=x^2-2 x-8 $
$ =x^2-4 x+2 x-8 $
$ =x(x-4)+2(x-4) $
$ =(x-4)(x+2) $
$ \therefore f(x)=0 $
$ (x-4)(x+2)=0 $
$ x-4=0 \text { or } x+2=0 $
$ x=4 \text { or } x=-2$
So, the zeros of $f(x)$ are $4$ and $-2$
Sum of zeros $= 4 + (-2) = 2$
$=\frac{2}{1} =\frac{-(\text{Cofficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $= 4 × (-2) = -8$
$=\frac{-8}{1}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→