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 - 8x $
$= 5x^2 - 8x - 4 $
$= 5x^2 - 10x + 2x - 4 $
$= 5x(x - 2) + 2(x - 2) $
$= (x - 2)(5x + 2)$
$\therefore$ $f(x) = 0 $
$\Rightarrow (x - 2)(5x + 2) = 0$
$ \Rightarrow x - 2 = 0$ or $5x + 2 = 0$
$\therefore\text{x}=2,\frac{-2}{5}$
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 $(2x^2 + 2ax + 5x + 10)$, find the value of a.
AnswerLet $f(x) = 2x^2 + 2ax + 5x + 10$
Since $(x + a)$ is a factor of $f(x),$ we have
$f(-a) = 0$
$\Rightarrow 2(-a)^2 + 2a(-a) + 5(-a) + 10 = 0$
$\Rightarrow 2a^2 - 2a^2 - 5a + 10 = 0$
$\Rightarrow 5a = 10$
$\Rightarrow a = 2$
View full question & answer→Question 43 Marks
Find the zeros of the polynomial $x^2 + 2x - 195$
AnswerHere, $p(x) = x^2 + 2x - 195$
Let $p(x) = 0$
$\Rightarrow x^2 + (15 - 13)x - 195 = 0$
$\Rightarrow x^2 + 15x - 13x - 195 = 0$
$\Rightarrow x(x + 15) - 13(x + 15) = 0$
$\Rightarrow (x + 15)(x - 13) = 0$
$\Rightarrow x = -15, 13$
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:
$x^2 + 3x - 10$
AnswerWe have
$f(x) = x^2 + 3x - 10$
$= x^2 + 5x - 2x - 10$
$= x(x + 5) - 2(x + 5)$
$= (x + 5)(x - 2)$
$\therefore$ $f(x) = 0$
$\Rightarrow (x + 5)(x - 2) = 0$
$\Rightarrow x + 5 = 0$ or $x - 2= 0$
$\Rightarrow x = -5$ 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 $x^2 - 3x - m(m + 3)$.
AnswerWe have,
$f(x) = x^2 - 3x - m(m + 3)$
$= x^2 - (m + 3)x + mx - m(m + 3)$
$= x[x - (m + 3)] + m[x - (m + 3)]$
$= (x + m)[x - (m + 3)]$
$\therefore$$ f(x) = 0$
$\Rightarrow (x + m)[x - (m + 3)] = 0$
$\Rightarrow x + m = 0$ or$ x - (m + 3) = 0$
$\Rightarrow x = -m$ 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$x^2 + 7x + 12 = 0$
$x^2 + 4x + 3x + 12 = 0$
$x(x + 4) + 3(x + 4) = 0$
$(x + 4)(x + 3) = 0$
$(x + 4) = 0$ or $(x + 3) = 0$
$x = -4$ or $x = -3$
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 $p(x) = x^3 + 3x^2 - 5x + 4$ is divided by $(x - 2)$.
AnswerGiven: $p(x) = x^3 + 3x^2 - 5x + 4$
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 $x^3 + x^2 - ax + b$ 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
$x^2 - x = 0$
$\Rightarrow x(x - 1) = 0$
$\Rightarrow x = 0$ or $x = 1$
Hence,
$f(0) = 0$
$\Rightarrow 0^3 + 0^2- a(0) + b = 0$
$\Rightarrow b = 0$
Also,
$f(1) = 0$
$\Rightarrow 1^3 + 1^2 - a(1) + 0 = 0$
$\Rightarrow 1 + 1 - a = 0$
$\Rightarrow a = 2$
$\therefore$ $a = 2$ 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 $k$.
AnswerSince $-4$ is a zero of $f(x) = x^2 - x - (2k + 2)$, we have
$f(-4) = 0$
$\Rightarrow (-4)^2 - (-4) - 2k -2 = 0$
$\Rightarrow 16 + 4 - 2k - 2 = 0$
$\Rightarrow 18 - 2k = 0$
$\Rightarrow 2k = 18$
$\Rightarrow k = 9$
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$4x^2 - 4x - 3$
$= 4x^2 - 6x + 2x - 3$
$= 2x(2x - 3) + 1(2x - 3)$
$= (2x - 3)(2x + 1)$
$\therefore$ $f(x) = 0$
So, $4x^2 - 4x - 3 = 0$
$\Rightarrow (2x - 3)(2x + 1) = 0$
$\Rightarrow 2x - 3 = 0$ or $2x + 1= 0$
$\Rightarrow\text{x}=\frac{3}{2}$ 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 $f(x) = x^2 - 5x + k$ such that $\alpha-\beta=1,$ find the value of $k$.
AnswerGiven: $f(x) = x^2 - 5x + k$
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,
$f(x) = x^2 - 2x - 8$
$= x^2 - 4x + 2x - 8$
$= x(x - 4) + 2(x - 4)$
$= (x - 4)(x + 2)$
$\therefore$ $f(x) = 0$
$(x - 4)(x + 2) = 0$
$x - 4 = 0$ or $x + 2 = 0$
$x = 4$ 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→