Question

Solve the quadratic equation $x^2+6x-7=0$ by completing the square.

Answer 1

Given: $x^2+6x-7=0$

Now by using the identity $(a+b{)}^2=a^2+2ab+b^2$

Let us now compare the given equation to the identity and we see that,

$a=x$

and $2\times a\times b=6x$

$\Rightarrow 2\times x\times b=6x(\because a=x)$

$\therefore b=\frac{6x}{2x}$

$\Rightarrow b=3$

$\Rightarrow b^2=9$

Hence we can now complete the square for the above quadratic equation by adding $9$ to both the sides

$x^2+6x-7+9=0+9$

$\Rightarrow x^2+6x+9=9+7$ ...{By transposing $7$ to R.H.S}

$\Rightarrow x^2+6x+9=16$

This can be written as

$(x+3{)}^2=(4{)}^2$

$\therefore x+3=\pm 4$ ....{Taking square root on both sides}

Now if,

$x+3=4$
$\Rightarrow x=4-3$
$\Rightarrow x=1$

or,

$x+3=-4$

$\Rightarrow x=-4-3$

$\Rightarrow x=-7$

$\therefore$ $x=1$ or $-7$ are the roots of $x^2+6x-7=0$