Question

If the two quadratic equation $a{x}^2+bx+c=0$ and $p{x}^2+qx+r=0$ have exactly one root $"\alpha "$ in common then select the correct statement.

• A. $\frac{{\alpha }^2}{br-qc}=\frac{\alpha }{ar-pc}=\frac{1}{aq-pb}$
• B. $\frac{{\alpha }^2}{br-qc}=\frac{-\alpha }{ar-pc}=\frac{1}{aq-pb}$
• C. $\frac{{\alpha }^2}{br-qc}=\frac{\alpha }{ar-pc}=\frac{-1}{aq-pb}$
• D. $\frac{-{\alpha }^2}{br-qc}=\frac{\alpha }{ar-pc}=\frac{1}{aq-pb}$
  

Answer 1

The correct option is B $\frac{{\alpha }^2}{br-qc}=\frac{-\alpha }{ar-pc}=\frac{1}{aq-pb}$

Given: $a{x}^2+bx+c=0\cdots \left(i\right)$
$p{x}^2+qx+r=0\cdots \left(ii\right)$ have a root $"\alpha "$ in common.
Now, For two quadratic equations $a_1{x}^2+b_{1}x+c_1=0\cdots \left(A\right)\&a_2{x}^2+b_{2}x+c_2=0\cdots \left(B\right)$ to have exactly one root $\alpha$ in common:
$\frac{{\alpha }^2}{b_1{c}_2-b_2{c}_1}=\frac{-\alpha }{a_1{c}_2-a_2{c}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Now, comaring $\left(i\right)$ with $\left(A\right)$ and $\left(ii\right)$ with $\left(B\right),$ we get:
$a_1=a;b_1=b;c_1=c;a_2=p,b_2=q;c_2=r$
Thus, we get:
$\frac{{\alpha }^2}{br-qc}=\frac{-\alpha }{ar-pc}=\frac{1}{aq-pb}$