How do you write the quadratic equation in standard form $x^{2} - 12 + 4 x = {(x - 2)}^{2} + 2$ $x^2 - 12 + 4x = (x - 2)^2 + 2$ ?

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How do you write the quadratic equation in standard form $x^{2} - 12 + 4 x = {(x - 2)}^{2} + 2$ $x^2 - 12 + 4x = (x - 2)^2 + 2$ ?

1 Answer

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Answer 1 · 2016-10-22T05:46:02

you have to expand the left hand side and rearrange to have 0 on one side.

Explanation:

so first expand ${(x - 2)}^{2} + 2$ $(x-2)^2 +2$ to $x^{2} - 4 x + 4$ $x^2-4x+4$
leaving us with
$x^{2} + 4 x - 12 = x^{2} - 4 x + 4$ $x^2+4x-12=x^2-4x+4$
$8 x - 18 = 0$ $8x-18=0$

which is in standard form.

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How do you write the quadratic equation in standard form $x^{2} - 12 + 4 x = {(x - 2)}^{2} + 2$ $x^2 - 12 + 4x = (x - 2)^2 + 2$ ?

1 Answer

Explanation:

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How do you write the quadratic equation in standard form x2−12+4x=(x−2)2+2x^2 - 12 + 4x = (x - 2)^2 + 2?

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Explanation:

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How do you write the quadratic equation in standard form $x^{2} - 12 + 4 x = {(x - 2)}^{2} + 2$ $x^2 - 12 + 4x = (x - 2)^2 + 2$ ?