How do you solve $- 2 + 7 x - 3 x^{2} < 0$ $-2 + 7x - 3x^2<0$ ?

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Answer 1 · 2015-07-04T15:07:28

( $- \infty, \frac{1}{3}$ $-∞, 1/3$ ) and ( $2, \infty$ $2, ∞$ )

$- 2 + 7 x - 3 x^{2} < 0$ $-2 + 7x -3x^2 < 0$

One way to solve this inequality is to use a sign chart.

Step 1. Write the inequality in standard form.

$- 3 x^{2} + 7 x - 2 < 0$ $-3x^2 +7x -2 < 0$

Step 2. Multiply the inequality by $- 1$ $-1$ .

$3 x^{2} - 7 x + 2 > 0$ $3x^2 -7x +2 > 0$

Step 3. Find the critical values.

Set $f (x) = 3 x^{2} - 7 x + 2 = 0$ $f(x) =3x^2 -7x +2 = 0$ and solve for $x$ $x$ .

$(3 x - 1) (x - 2) = 0$ $(3x-1)(x-2) = 0$

$3 x - 1 = 0$ $3x-1 = 0$ or $x - 2 = 0$ $x-2 = 0$

$x = \frac{1}{3}$ $x = 1/3$ or $x = 2$ $x = 2$

The critical values are $x = \frac{1}{3}$ $x = 1/3$ and $x = 2$ $x = 2$ .

Step 4. Identify the intervals.

The intervals to consider: ( $- \infty, + \frac{1}{3}$ $-∞, +1/3$ ), ( $\frac{1}{3}, 2$ $1/3, 2$ ), and ( $2, \infty$ $2, ∞$ ).

Step 5. Evaluate the intervals.

We pick a test number and evaluate the function and its sign at that number.

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Step 6. Create a sign chart for the function.

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The only intervals for which the signs are positive are.

Solution: $x < \frac{1}{3}$ $x < 1/3$ or $x > 2$ $x > 2$

How do you solve -2 + 7x - 3x^2<0−2+7x−3x2<0-2 + 7x - 3x^2<0?