# How do you solve 3x^2 - x < 8 - 4x?

How do you solve 3x^2 - x < 8 - 4x?

Answer 1

$- 2.208 < x < 1.208$

#### Explanation:

First, simplify the expression:
3x^2 − x < 8 − 4x
$3 {x}^{2} + 3 x - 8 < 0$

Now we can solve for the roots of the "equality" and then check to set the directional inequality limits of the solutions.
${x}_{1} = 1.208$
${x}_{2} = - 2.208$

3x^2 − x < 8 − 4x; 3(1.208^2) − 1.208 < 8 − 4(1.208)

4.38 − 1.208 < 8 − 4.832 : $3.172 < 3.168$ (incorrect), thus the inequality value must be ${x}_{1} < 1.208$
CHECK:
3(1.20^2) − 1.20 < 8 − 4(1.20)
4.32 − 1.20 < 8 − 4.8 : $3.12 < 3.2$ Correct.
AND
3x^2 − x < 8 − 4x; 3(-2.208^2) − -2.208 < 8 − 4(-2.208)

$14.63 + 2.208 < 8 + 8.832$ : $16.838 < 16.832$ (incorrect), thus the inequality value must be ${x}_{1} > - 2.208$
CHECK:
3(-2.2^2) − -2.2 < 8 − 4(-2.2)
$14.52 + 2.20 < 8 + 8.8$ : $16.72 < 16.8$ Correct.

The range is thus:
$- 2.208 < x < 1.208$

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