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

1.	If 4ax^2-y^2=(7x+1÷2)(7×-1÷2)What is the value of y
Answer» $4ax^2-y^2=(7x+1\div 2) (7\times -1\div 2)$ $4ax^2-y^2=(7x+ \frac{1}{2}) \times(-\frac{7}{2} )$ $4ax^2-y^2=-\frac{49}{4} x-\frac{7}{4}$ $4ax^2+\frac{49}{4} x+\frac{7}{4}-y^2=0$ 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.

Answer»

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