Question

Umesh and Varun are solving an equation of the form $x^2+bx+c=0$. In doing so Umesh commits a mistake in noting down the constant term and finds the roots as $-3$ and $-12$. And Varun commits a mistake in noting down the coefficient of $x$ and find the roots as $-27$ and $-2$. If so find the original equation

• A. $x^2-15x+36=0$
• B. $x^2+15x+36=0$
• C. $x^2-15x+54=0$
• D. $x^2+15x+54=0$

Answer 1

The correct option is D $x^2+15x+54=0$

With the roots $-3,-12$, the equation was $(x-(-3\left)\right)(x-(-12)=(x+3\left)\right(x+6)=x^2+3x+12x+36=x^2+15x+36$
As Umesh made a mistake in noting just the constant term, in the original equation, coefficient of $x^2=1$ and of $x=15$
Now, with the roots $-27,-2$, the equation was $(x-(-27\left)\right)(x-(-2\left)\right)=(x+27)(x+2)=x^2+27x+2x+54=x^2+29x+54$
As Varun made a mistake in noting the coffecient of $x$ in the original equation, coefficient of $x^2=1$ and constant $=54$
So, we get the original equation as $x^2+15x+54=0$