Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2-2x-8$

(ii) $4s^2-4s+1$

(iii) $6x^2-3-7x$
(iv) $4u^2+8u$

(v) $t^2-15$

(vi) $3x^2-x-4$

Answer 1

(i) $x^2-2x-8$

$x^2-2x-8$

$=x^2-4x+2x-8$

$=x(x-4)+2(x-4)$

$=(x+2)(x-4)$

Factorize the equation, we got $(x+2)(x-4)$
So, the value of $x^2-2x-8$ is zero when $x+2=0,x-4=0,$ i.e., when $x=-2$ or $x=4$.
Therefore, the zeros of $x^2-2x-8$ are -2 and 4.
Now,

$\Rightarrow$Sum of zeroes $=-2+4=2$=$-\frac{2}{1}=-\frac{Coefficientofx}{Coefficientof{x}^2}$

$\Rightarrow$Product of zeros $=(-2)\times \left(4\right)=-8$ = $\frac{-8}{1}=\frac{Constantterm}{Coefficientof{x}^2}$

(ii) $4s^2-4s+1$
$4s^2-4s+1$

$=4s^2-2s-2s+1$

$=2s(2s-1)-2(2s-1)$

$=(2s-1)(2s-1)$

Factorize the equation, we got $(2s-1)(2s-1)$
So, the value of $4s^2-4s+1$ is zero when $2s-1=0,2s-1=0$, i.e., when $s=\frac{1}{2}$ or $s=\frac{1}{2}$.
Therefore, the zeros of $4s^2-4s+1$ are $\frac{1}{2}$ and $\frac{1}{2}$.
Now,

$\Rightarrow$Sum of zeroes = $\frac{1}{2}+\frac{1}{2}=1=-\frac{-4}{4}=-\frac{Coefficientofs}{Coefficientof{s}^2}$

$\Rightarrow$Product of zeros = $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}=\frac{1}{4}=\frac{Constantterm}{Coefficientof{s}^2}$

(iii) $6x^2-3-7x$

$=6x^2-7x-3$

$=6x^2-9x+2x-3$

$=3x(2x-3)+1(2x-3)$

$=(3x+1)(2x-3)$

Factorize the equation, we got $(3x+1)(2x-3)$
So, the value of $6x^2-3-7x$ is zero when $3x+1=0,2x-3=0$, i.e., when $x=-\frac{1}{3}$ or $x=\frac{3}{2}$.
Therefore, the zeros of $6x^2-3-7x$ are $-\frac{1}{3}$ and $\frac{3}{2}$.
Now,

$\Rightarrow$Sum of zeroes = $-\frac{1}{3}+\frac{3}{2}=\frac{7}{6}=-\frac{-7}{6}=-\frac{Coefficientofx}{Coefficientof{x}^2}$

$\Rightarrow$Product of zeros = $-\frac{1}{3}\times \frac{3}{2}=-\frac{1}{2}=\frac{-3}{6}=\frac{-1}{2}=\frac{Constantterm}{Coefficientof{x}^2}$

(iv) $4u^2+8u$

$=4u(u+2)$

Factorize the equation, we got $4u(u+2)$
So, the value of $4u^2+8u$ is zero when $4u=0,u+2=0$, i.e., when $u=0$ or $u=-2$.
Therefore, the zeros of $4u^2+8u$ are $0$ and $-2.$
Now,

$\Rightarrow$Sum of zeroes = $0-2=-2=-\frac{8}{4}=-2=-\frac{Coefficientofu}{Coefficientof{u}^2}$

$\Rightarrow$Product of zeros = $-0x-2=0=\frac{0}{4}=0=\frac{Constantterm}{Coefficientof{u}^2}$

(v) $t^2-15$
$=t^2-(\sqrt{15}{)}^2$

$=(t+\sqrt{15})(t-\sqrt{15})$

So, the value of $t^2-15$ is zero when $t+\sqrt{15}=0,t-\sqrt{15}=0$, i.e., when $t=\sqrt{15}$ or $t=-\sqrt{15}$.
Therefore, the zeros of $t^2-15$ are $\pm \sqrt{15}$.
Now,

$\Rightarrow$Sum of zeroes = $\sqrt{15}-\sqrt{15}=0=-\frac{0}{1}=0=-\frac{Coefficientoft}{Coefficientof{t}^2}$

$\Rightarrow$Product of zeros = $\sqrt{15}\times \sqrt{-15}=-15=\frac{-15}{1}=\frac{Constantterm}{Coefficientof{t}^2}$

(vi) $3x^2-x-4$

$=3x^2-x-4$

$=3x^2+3x-4x-4$

$=3x(x+1)-4(x+1)$

$=(x+1)(3x-4)$

Factorize the equation, we get$(x+1)(3x-4)$
So, the value of $3x^2-x-4$ is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x = $\frac{4}{3}$.
Therefore, the zeros of $3x^2-x-4$ are -1 and $\frac{4}{3}$.
Now,

$\Rightarrow$Sum of zeroes = $-1+\frac{4}{3}=\frac{1}{3}=-\frac{-1}{3}=-\frac{Coefficientofx}{Coefficientof{x}^2}$

$\Rightarrow$Product of zeros = $-1\times \frac{4}{3}=-\frac{4}{3}=\frac{-4}{3}=\frac{Constantterm}{Coefficientof{x}^2}$