How do you solve $x^2 - 3x - 28 = 0$ by completing the square?

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Answer 1 · 2016-06-19T05:40:29

x is either 7 or $-$ 4

We have, $x^2 -3x-28=0$

This can be written as $x^2 -3x = 28$

$x^2 - (2)(3/2)x = 28$ (the trick is to take out 2 from the coefficient of x)

Now adding the square of 3/2 on both sides.

$x^2 - (2)(3/2)x + (3/2)^2 = 28 + (3/2)^2$

Carefully notice that the LHS becomes a square

$(x-3/2)^2 = 28 + 9/4$

[A side note: if $(f(x))^2 = g(x)$ then $f(x) = +-sqrt(g(x))$ ]

So, $x-3/2 = +-sqrt(28 + 9/4)$

$x-3/2 = +-sqrt(121/4)$

$x-3/2 = +-11/2$

$x= 3/2 +-11/2$

So x is either 7 or $-$ 4

How do you solve x^2 - 3x - 28 = 0x^2 - 3x - 28 = 0 by completing the square?