Question

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

$5t^2+12t+7=0$.

• A. $-1,-\frac{7}{5}$
• B. $-1,-\frac{5}{7}$
• C. $-2,-\frac{7}{5}$
• D. $-1,-\frac{4}{5}$

Answer 1

The correct option is A $-1,-\frac{7}{5}$

Let $f\left(t\right)=5t^2+12t+7$.
Comparing it with the standard quadratic polynomial $a{x}^2+bx+c$, we get,
$a=5$, $b=12$, $c=7$.
Now, $5t^2+12t+7$
$=5t^2+5t+7t+7$
$=5t(t+1)+7(t+1)$
$=(t+1)(5t+7)$.
The zeros of $f\left(t\right)$ are given by $f\left(t\right)=0$.
$=>(t+1)(5t+7)=0$
$=>t+1=0$ or $5t+7=0$
$=>t=-1$ or $t=\frac{-7}{5}$.
Hence the zeros of the given quadratic polynomial are $-1$, $\frac{-7}{5}$.

Verification of the relationship between the roots and the coefficients:
Sum of the roots $=-1+\left(\frac{-7}{5}\right)$
$=\frac{-5-7}{5}$
$=\frac{-12}{5}$
$=\frac{-coefficientofx}{coefficientof{x}^2}$.
Product of the roots $=-1\times \left(\frac{-7}{5}\right)$
$=\frac{7}{5}$
$=\frac{constantterm}{coefficientof{x}^2}$.

Hence verified.

Therefore, option $A$ is correct.