Question
Solve the following equation by using formula :
$10ax^2 – 6x + 15ax – 9 = 0,a\neq 0$
$10ax^2 – 6x + 15ax – 9 = 0,a\neq 0$
✓
Answer
$10 a x^2-6 x+15 a x-9=0$
$\text { Here } a=10 a, b=-(6-15 a), c=-9$
$D=b^2-4 a c$
$=[-(6-15 a)]^2-4 \times 10 a(-9)$
$=36-180 a+225 a^2+360 a$
$=36+180 a+225 a^2=(6+15 a)^2$
$\therefore x=\frac{-b \pm \sqrt{D}}{2 a}$
$=\frac{-[-(6-15 a)] \pm \sqrt{(6+15 a)^2}}{2 \times 10 a}$
$=\frac{(6-15 a) \pm(6+15 a)}{20 a}$
$\therefore x_1=\frac{6-15 a+6+15 a}{20 a}$
$=\frac{12}{20 a}$
$=\frac{3}{5 a}$
$x_2=\frac{6-15 a-6-15 a}{20 a}$
$=\frac{-30 a}{20 a}$
$=\frac{-3}{2}$
(Hence $\left.x=\frac{3}{5} a\right), \frac{-3}{2}$.
$\text { Here } a=10 a, b=-(6-15 a), c=-9$
$D=b^2-4 a c$
$=[-(6-15 a)]^2-4 \times 10 a(-9)$
$=36-180 a+225 a^2+360 a$
$=36+180 a+225 a^2=(6+15 a)^2$
$\therefore x=\frac{-b \pm \sqrt{D}}{2 a}$
$=\frac{-[-(6-15 a)] \pm \sqrt{(6+15 a)^2}}{2 \times 10 a}$
$=\frac{(6-15 a) \pm(6+15 a)}{20 a}$
$\therefore x_1=\frac{6-15 a+6+15 a}{20 a}$
$=\frac{12}{20 a}$
$=\frac{3}{5 a}$
$x_2=\frac{6-15 a-6-15 a}{20 a}$
$=\frac{-30 a}{20 a}$
$=\frac{-3}{2}$
(Hence $\left.x=\frac{3}{5} a\right), \frac{-3}{2}$.
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