Question

Find the zeroes of the following quadratic polynomial:
$6x^2+x-5$

• A. $\frac{1}{2},1$
• B. 6, 5
• C. $\frac{5}{6},-1$
• D. 2, 1

Answer 1

The correct option is C $\frac{5}{6},-1$

$6x^2+x-5=0$
We need to find two numbers 'a' and 'b' such that
$a\pm b=1,a\times b=-30$
$\Rightarrow 6-5=1,-5\times 6=-30$
$6x^2+6x-5x-5=0$
$6x(x+1)-5(x+1)=0$
$(6x-5)(x+1)=0$
So,
$6x-5=0$
$\Rightarrow x=\frac{5}{6}$

or,
$x+1=0$
$\Rightarrow x=-1$
Thus, the zeroes of the polynomial
$6x^2+x-5$ are $\frac{5}{6}$ and -1.