Question

Given, $\frac{q^2-q-42}{q+6}=\sqrt{q-5}$
Calculate the value of the extraneous solution of the above equation.

• A. $5$
• B. $6$
• C. $8$
• D. $9$

Answer 1

Answer: (B), (D)

The correct options are
B $6$
D $9$

Given, $\frac{q^2-q-42}{q-6}=\sqrt{q-5}$

$\Rightarrow \frac{q^2-7q+6q-42}{q+6}=\sqrt{q-5}$

$\Rightarrow \frac{q(q-7)+6(q-7)}{q+6}=\sqrt{q-5}$

$\Rightarrow \frac{(q-7)(q+6)}{q+6}=\sqrt{q-5}$

$\Rightarrow (q-7)=\sqrt{q-5}$

On squaring both the sides, we get

$(q-7{)}^2=q-5$

$\Rightarrow q^2-14q+49=q-5$

$\Rightarrow q^2-14q-q+49+5=0$

$\Rightarrow q^2-15q+54=0$

$\Rightarrow q^2-6q-9q+54=0$

$\Rightarrow q(q-6)-9(q-6)=0$

$\Rightarrow (q-6)(q-9)=0$

$\therefore$ $q-6=0$ $\Rightarrow q=6$

and $q-9=0$ $\Rightarrow q=9$

Hence, $q=9$ and $q=6$.