Question

The quadratic equation $x^2-8 x+a=0$ and $x^2-b x+16=0$ have one root in common and the other roots of $1^{\text {st }}$ and $2^{\text {nd }}$ equations are integers in the ratio $3: 4$, then which of the following is(are) correct?

• A. The common root is a prime number
• B. The value of a is 12 .
• C. The value of b is 10 .
• D. The sum of $( a + b )$ is a rational number.

Answer 1

Answer: (A), (B), (C), (D)

$\alpha+3 \beta=8 \& 3 \alpha \beta= a$
$\alpha+4 \beta=b \& 4 \alpha \beta=16 \Rightarrow \alpha \beta=4 \therefore a=12$
Hence, $x^2-8 x+12=0$
$x =2 \text { or } x =6$
If common root is 2 then $\beta=2$ for which we get $3 \beta$ and $4 \beta$ both integers but if common root is 6 then $\beta=\frac{2}{3}$ for which $4 \beta$ is not integer.
Hence, common root is 2