How do you solve $x^{2} - 6 x = 16$ by completing the square?

Open in App

Solution

Solve the given equation:
Given that $x^{2} - 6 x = 16$
To complete the square, add square of ( $\frac{1}{2}$ of co-efficient of $x$ ) to both sides i.e. ${(\frac{- 6}{2})}^{2} = 9$
$\Rightarrow$ $x^{2} + 2 (- 3) x + 9 = 16 + 9$
$\Rightarrow$ ${(x - 3)}^{2} = 25$
Take the square root of both sides,
$\Rightarrow$ $\sqrt{{(x - 3)}^{2}} = \sqrt{25}$
$\Rightarrow$ $x - 3 = \pm 5$
$\Rightarrow$ $x = 3 \pm 5$
$\Rightarrow$ $x = 3 - 5 o r x = 3 + 5$
$\Rightarrow$ $x = - 2 o r x = 8$
Hence, the solution of the given equation is $x = - 2 o r x = 8$ .

Suggest Corrections

Similar questions

x^{2} - 6 x + 3 = 0

How do you complete the square to solve $x^{2} + 6 x = 7$ ?

How do you solve the equation $x^{2} - 16 x = - 20$ by completing the square?

How do you solve $2 x^{2} - 7 x + 3 = 0$ by completing the square.

x^{2} + 6 x - 7 = 0

Join BYJU'S Learning Program