Question
Solve the following equation by using formula :
$x^2 + (4 – 3a)x – 12a = 0$
$x^2 + (4 – 3a)x – 12a = 0$
✓
Answer
$
x^2+(4-3 a) x-12 a=0
$Here $a=1, b=4-3 a, c=-12 a$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(4-3 a)^2-4 \times 1 \times(-12 a) \\
& =16-24 a+9 a^2+48 a \\
& =16+24 a+9 a^2=(4+3 a)
\end{aligned}
$
$
\begin{aligned}
& \therefore x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-(4-3 a) \pm \sqrt{4+3 a^2}}{2 \times 1} \\
& =\frac{3 a-4 \pm 3 a+4}{2}
\end{aligned}
$
$\therefore x _1=\frac{3 a-4+3 a+4}{2}$
$
=\frac{6 a}{2}
$
$
=3 a
$
and
$
\begin{aligned}
& x_2=\frac{3 a-4-3 a-4}{2} \\
& =\frac{-8}{2} \\
& =-4
\end{aligned}
$
$\therefore$ Roots are $3 a,-4$.
x^2+(4-3 a) x-12 a=0
$Here $a=1, b=4-3 a, c=-12 a$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(4-3 a)^2-4 \times 1 \times(-12 a) \\
& =16-24 a+9 a^2+48 a \\
& =16+24 a+9 a^2=(4+3 a)
\end{aligned}
$
$
\begin{aligned}
& \therefore x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-(4-3 a) \pm \sqrt{4+3 a^2}}{2 \times 1} \\
& =\frac{3 a-4 \pm 3 a+4}{2}
\end{aligned}
$
$\therefore x _1=\frac{3 a-4+3 a+4}{2}$
$
=\frac{6 a}{2}
$
$
=3 a
$
and
$
\begin{aligned}
& x_2=\frac{3 a-4-3 a-4}{2} \\
& =\frac{-8}{2} \\
& =-4
\end{aligned}
$
$\therefore$ Roots are $3 a,-4$.
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