How do you solve using the completing the square method $x^{2} + 4 x = 6$ $x^2 + 4x = 6$ ?

Question

How do you solve using the completing the square method $x^{2} + 4 x = 6$ $x^2 + 4x = 6$ ?

1 Answer

Impact of this question

2076 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-01T22:17:00

$x = - 2 \pm \sqrt{10}$ $x=-2+-sqrt(10)$

Explanation:

When wanting to complete the square for a quadratic in the form

$a x^{2} + b x = c$ $ax^2+bx=c$ , we add ${(\frac{b}{2})}^{2}$ $(b/2)^2$ to each side and factor the left side. To solve, we can take the root of both sides, accounting for positive and negative answers on the right side.

In this case, $b = 4, {(\frac{b}{2})}^{2} = {(\frac{4}{2})}^{2} = 2^{2} = 4$ $b=4, (b/2)^2=(4/2)^2=2^2=4$ , so we add $4$ $4$ to each side.

$x^{4} + 4 x + 4 = 6 + 4$ $x^4+4x+4=6+4$

Factoring $x^{4} + 4 x + 4$ $x^4+4x+4$ yields ${(x + 2)}^{2}$ $(x+2)^2$

${(x + 2)}^{2} = 10$ $(x+2)^2=10$

So, take the root of both sides, account for positive and negative:

$\sqrt{{(x + 2)}^{2}} = \pm \sqrt{10}$ $sqrt((x+2)^2)=+-sqrt(10)$

The root of a squared term is just that term:

$x + 2 = \pm \sqrt{10}$ $x+2=+-sqrt(10)$

$x = - 2 \pm \sqrt{10}$ $x=-2+-sqrt(10)$

Science

Math

Humanities

... and beyond

How do you solve using the completing the square method $x^{2} + 4 x = 6$ $x^2 + 4x = 6$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve using the completing the square method x2+4x=6x^2 + 4x = 6?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve using the completing the square method $x^{2} + 4 x = 6$ $x^2 + 4x = 6$ ?