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

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

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Answer 1 · 2015-10-29T02:44:44

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

Ans: 2 and $- \frac{3}{2}$ $-3/2$

Explanation:

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

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

4x - 1 = -7 --> 4x = -6 --> $x = - \frac{6}{4} = - \frac{3}{2}$ $x = -6/4 = -3/2$

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

1 Answer

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How do you solve 16x2−8x+1=4916x^2 - 8x + 1 = 49?

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