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

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Answer 1 · 2015-07-04T15:07:28

( $-∞, 1/3$ ) and ( $2, ∞$ )

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

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

Step 1. Write the inequality in standard form.

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

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

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

Step 3. Find the critical values.

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

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

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

$x = 1/3$ or $x = 2$

The critical values are $x = 1/3$ and $x = 2$ .

Step 4. Identify the intervals.

The intervals to consider: ( $-∞, +1/3$ ), ( $1/3, 2$ ), and ( $2, ∞$ ).

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: $x < 1/3$ or $x > 2$

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