Question

If 4ax^2-y^2=(7x+1÷2)
(7×-1÷2)
What is the value of y

Answer 1

$4ax^2-y^2=(7x+1\div 2) (7\times -1\div 2)$

$4ax^2-y^2=(7x+ \frac{1}{2}) \times(-\frac{7}{2} )$

$4ax^2-y^2=-\frac{49}{4} x-\frac{7}{4}$

$4ax^2+\frac{49}{4} x+\frac{7}{4}-y^2=0$

Let $4ax^2+\frac{49}{4} x+\frac{7}{4}=t^2$.

The expression above is expressed as $t^2-y^2=0$.

If we use the identity $a^2-b^2=(a+b)(a-b)$, we can factor it.

$(t+y)(t-y)=0$

∴ $y=\pm t$

Substitute $t=\pm \sqrt{4ax^2+\frac{49}{4} x+\frac{7}{4}}$.

$y=\pm(\pm \sqrt{4ax^2+\frac{49}{4} x+\frac{7}{4}})$

$y=\pm \sqrt{4ax^2+\frac{49}{4} x+\frac{7}{4}}$

Now it's time to simplify the right hand side.

$y=\pm \sqrt{\frac{1}{4} (16ax^2+49x+7)}$

∴ $y=\pm \frac{1}{2} \sqrt{16ax^2+49x+7}$

I could just take the root on the both side,

but in my opinion, this way was easier to understand.