Question

Rohan and Sohan were attempting to solve the quadratic equation $x^2-ax+b=0$. Rohan copied the coefficient of x wrongly and obtained the roots as 4 and 12 . Sohan copied the constant term wrongly and obtained the roots as -19 and 3. Find the correct roots

• A. -8, -10
• B. -8, -6
• C. -4, -12
• D. 4, 12

Answer 1

The correct option is C -4, -12

With the roots $4,12$, the equation was $(x-4)\left)\right(x-12)=x^2-4x-12x+48=x^2-16x+48$
As Rohan made a mistake in noting the coffecient of $x$ , in the original equation, coefficient of $x^2=1$ and constant $=48$
Now, with the roots $-19,3$, the equation was $(x-(-19\left)\right)(x-3)=(x+19)(x-3)=x^2+19x-3x-57=x^2+16x-57$
As Sohan made a mistake in noting just the constant term in the original equation, coefficient of $x^2=1$ and of $x=16$
So, we get the original equation as $x^2+16x+48=0$
Solving it, we get $x^2+4x+12x+48=0$
$=>x(x+4)+12(x+4)=0$
$=>(x+4)(x+12)=0$
$=>x=-4,-12$