Question

Find the zeroes of the following polynomial by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:

$2x^2+\frac{7}{2}x+\frac{3}{4}$.

• A. $\frac{-1}{2},\frac{-1}{4}$
• B. $\frac{-3}{2},\frac{-1}{4}$
• C. $\frac{-7}{2},\frac{-1}{4}$
• D. $\frac{-9}{2},\frac{-1}{4}$

Answer 1

The correct option is B $\frac{-3}{2},\frac{-1}{4}$

Let $f\left(x\right)=2x^2+\frac{7}{2}x+\frac{3}{4}$.
Comparing it with the standard quadratic polynomial $a{x}^2+bx+c$, we get,
$a=2$, $b=\frac{7}{2}$, $c=\frac{3}{4}$.
Now, $2x^2+\frac{7}{2}x+\frac{3}{4}$
$=2x^2+\frac{6}{2}x+\frac{1}{2}x+\frac{3}{4}$
$=2x(x+\frac{3}{2})+\frac{1}{2}(x+\frac{3}{2})$
$=(x+\frac{3}{2})(2x+\frac{1}{2})$.
The zeros of $f\left(x\right)$ are given by $f\left(x\right)=0$.
$=>(x+\frac{3}{2})(2x+\frac{1}{2})=0$
$=>(x+\frac{3}{2})=0$ or $(2x+\frac{1}{2})=0$
$=>x=-\frac{3}{2}$ or $x=-\frac{1}{4}$.
Hence the zeros of the given quadratic polynomial are $-\frac{3}{2}$, $-\frac{1}{4}$.

Verification of the relationship between the roots and the coefficients:
Sum of the roots $=(-\frac{3}{2})+(-\frac{1}{4}$)
$=-\frac{7}{4}$
$=-\frac{7}{2\times 2}$
$=\frac{-coefficientofx}{coefficientof{x}^2}$.
Product of the roots $=(-\frac{3}{2})(-\frac{1}{4}$)
$=-\frac{3}{8}$
$=-\frac{3}{4\times 2}$
$=\frac{constantterm}{coefficientof{x}^2}$.

Hence, verified.

Therefore, option $B$ is correct.