Question

If the equation $x^2+px+q=0$ and $x^2+p^{{}^{\prime }}x+q^{{}^{\prime }}=0$ have a common root, then it must be equal to

• A. $\frac{p{q}^{{}^{\prime }}-p^{{}^{\prime }}q}{q-q^{{}^{\prime }}}$
• B. $\frac{q-q^{{}^{\prime }}}{p^{{}^{\prime }}-p}$
• C. $\frac{p^{{}^{\prime }}-p}{q-q^{{}^{\prime }}}$
• D. $\frac{p{q}^{{}^{\prime }}-p^{{}^{\prime }}q}{p-p^{{}^{\prime }}}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{p{q}^{{}^{\prime }}-p^{{}^{\prime }}q}{q-q^{{}^{\prime }}}$
B $\frac{q-q^{{}^{\prime }}}{p^{{}^{\prime }}-p}$

$x^2+px+q=0$ and $x^2+p^{{}^{\prime }}x+q^{{}^{\prime }}=0$

Let common root be $\alpha$

${\alpha }^2+p\alpha +q=0$ .......$\left(1\right)$

${\alpha }^2+p^{{}^{\prime }}\alpha +q^{{}^{\prime }}=0$ .........$\left(2\right)$

Subtracting $\left(1\right)-\left(2\right)$

$\alpha (p-p^{{}^{\prime }})+q-q^{{}^{\prime }}=0$

$⟹\alpha =\frac{(q-q^{{}^{\prime }})}{(p^{{}^{\prime }}-p)}$

OR

${\alpha }^2+p\alpha +q=0$ .......$\left(1\right)$

${\alpha }^2+p^{{}^{\prime }}\alpha +q^{{}^{\prime }}=0$ .........$\left(2\right)$

Multiply $eq\left(1\right)$ by$q^{{}^{\prime }}$ and $eq\left(2\right)$ by $q$

we get

$q^{{}^{\prime }}{\alpha }^2+p{q}^{{}^{\prime }}\alpha +q^{{}^{\prime }}q=0$ .......$\left(1\right)$

$q{\alpha }^2+p^{{}^{\prime }}q\alpha +q{q}^{{}^{\prime }}=0$ .........$\left(2\right)$

Subtracting $\left(1\right)-\left(2\right)$

${\alpha }^2(q^{{}^{\prime }}-q)+\alpha (p{q}^{{}^{\prime }}-p^{{}^{\prime }}q)=0$

$\alpha =\frac{(p{q}^{{}^{\prime }}-p^{{}^{\prime }}q)}{(q-q^{{}^{\prime }})}$ and $\alpha \ne 0$