Question

If $x^2-2x+c=0$ and $x^2-ax+b=0$ have a root in common and the second equation has equal roots, then $b+c$ is equal to

• A. a
• B. 2a
• C. 3a
• D. None of these

Answer 1

Answer: (A)

a

Answer 2

The second equation $x^2-ax+b=0$ having equal roots implies its discriminant is zero, so $a^2=4b$. The repeated root is $x=\frac{a}{2}$. Since this is the common root, it must satisfy the first equation $x^2-2x+c=0$. Substituting $x=\frac{a}{2}$ into the first equation gives $(\frac{a}{2})^2 - 2(\frac{a}{2}) + c = 0$, which simplifies to $\frac{a^2}{4} - a + c = 0$. Substituting $a^2=4b$ into this equation yields $\frac{4b}{4} - a + c = 0$, which simplifies to $b - a + c = 0$. Rearranging this gives $b+c=a$.