Question

The coefficient of $x$ in the quadratic equation $a{x}^2+bx+c=0$ was wrongly taken as $17$ in place of $13$ and its roots were found to be $-2$ and $-15$, the actual roots of the equation are

• A. $-2$ and $15$
• B. $-3$ and $-10$
• C. $-4$ and $-9$
• D. $-5$ and $-6$

Answer 1

The correct option is B $-3$ and $-10$

$a{x}^2+bx+c=0$

when $a=17$

roots $=-2,-15$

$x^2-(a+b)+a.b=0$

$x^2-(a+b)x+a.b=0$

$x^2-(-2-15)x+30$

$x^2+13+30=0$

$x^2(10+3)x+30=0$

$x^2+10x+3x+30=0$

$x(x+10)+3(x+10)=0$

$x(x+10)+3(x+10)=0$

$x=-3$

$x=-10$