Question

`7x^2+3x-4=0` 

Answer 1

`7x^2+3x-4=0` 

⇒`49x^2+21x-28=0`               (Multiplying both sides by 7)

⇒`49x^2+21x=28` 

⇒`(7x)^2+xx7x xx3/2+(3/2)^2=28+(3/2)^2`        [Adding (3/2)^2 on both sides]

⇒`(7x+3/2)^2=28+9/4=121/4=(11/2)^2` 

⇒`7x+3/2=+-11/2`                (Taking square root on both sides)

⇒`7x+3/2=11/2  or  7x+3/2=-11/2` 

⇒`7x=11/2-3/2=8/2=4  or 7x=11/2-3/2=-14/2=-7` 

⇒`x=4/7  or  x=-1` 

Hence, `4/7  and -1` are the roots of the given equation.