How do you solve $16x^2 - 8x + 1 = 49$ ?

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Answer 1 · 2015-10-29T02:44:44

Solve $16x^2 - 8x + 1 = 47$

Ans: 2 and $-3/2$

The two sides are both perfect squares.
$16x^2 - 8x + 1 = (4x - 1)^2 = 7^2$
$(4x - 1) = +- 7$

4x - 1 = 7 --> 4x = 8 --> $x = 2$

4x - 1 = -7 --> 4x = -6 --> $x = -6/4 = -3/2$

How do you solve 16x^2 - 8x + 1 = 4916x^2 - 8x + 1 = 49?