How do you solve $4 (x^{2} - 1) < 4 x - 1$ $4( x ^ { 2} - 1) < 4x - 1$ ?

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How do you solve $4 (x^{2} - 1) < 4 x - 1$ $4( x ^ { 2} - 1) < 4x - 1$ ?

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Answer 1 · 2017-08-28T19:16:22

$- \frac{1}{2} < x < \frac{3}{2}$ $-1/2< x<3/2$

Explanation:

$4 (x^{2} - 1) < 4 x - 1 \Leftrightarrow 4 x^{2} - 4 < 4 x - 1 \Leftrightarrow$ $4(x^2-1)<4x-1iff4x^2-4<4x-1iff$

$4 x^{2} - 4 x - 3 < 0 \Leftrightarrow 4 (x + \frac{1}{2}) (x - \frac{3}{2}) < 0 \Leftrightarrow$ $4x^2-4x-3<0iff4(x+1/2)(x-3/2)<0iff$

$(x + \frac{1}{2}) (x - \frac{3}{2}) < 0 \Leftrightarrow - \frac{1}{2} < x < \frac{3}{2}$ $(x+1/2)(x-3/2)<0iff-1/2< x<3/2$

Answer 2 · 2017-08-28T19:19:58

$- \frac{1}{2} < x < \frac{3}{2}$ $-1/2 < x < 3/2$

Explanation:

$4 (x^{2} - 1) < 4 x - 1$ $4 (x^2 - 1) < 4x - 1$
$4 x^{2} - 4 = 4 x - 1$ $4x^2 - 4 = 4x - 1$
$4 x^{2} - 4 x - 4 + 3 = 0$ $4x^2 - 4x - 4 + 3 = 0$
$4 x^{2} - 4 x - 3 = 0$ $4x^2 - 4x - 3 = 0$
Factorize
$(2 x + 1) (2 x - 3) = 0$ $(2x + 1)(2x - 3) = 0$
$2 x + 1 = 0 or 2 x - 3 = 0$ $2x + 1 = 0 or 2x - 3 = 0$
$x = \frac{- 1}{2} or x = \frac{3}{2}$ $x = (-1)/2 or x = 3/2$
We have
$x < \frac{- 1}{2}$ $x < (-1)/2$ which doesn't work in original inequality
$\frac{- 1}{2} < x < \frac{3}{2}$ $(-1)/2 < x < 3/2$ which works in original inequality
$x > \frac{3}{2}$ $x > 3/2$ it doesn't work in original inequality

Thus,
The final answer is
$\frac{- 1}{2} < x < \frac{3}{2}$ $(-1)/2 < x < 3/2$

I hope I helped!

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How do you solve $4 (x^{2} - 1) < 4 x - 1$ $4( x ^ { 2} - 1) < 4x - 1$ ?

2 Answers

Explanation:

Explanation:

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