QuestionFind values of $x$ such that: $6 x^{2}+17 x \geq 14$ .

Studdy Solution

STEP 1

Assumptions1. We are looking for the set of real number values for $x$ that satisfy the inequality $6x^ +17x \geq14$ .

STEP 2

First, we need to rearrange the inequality to a standard quadratic inequality form, which is $ax^2 + bx + c \geq0$ . We can do this by subtracting14 from both sides of the inequality.
$6x^2 +17x -14 \geq0$

STEP 3

Next, we need to factor the quadratic expression on the left side of the inequality. This can be done by finding two numbers that multiply to $-84$ (which is $6 \times -14$ ) and add to $17$ .

STEP 4

The two numbers that satisfy these conditions are $21$ and $-4$ . So, we can write the quadratic expression as follows $(3x -2)(2x +7) \geq0$

STEP 5

Now, we need to find the roots of the quadratic equation $(3x -2)(2x +7) =0$ . We can do this by setting each factor equal to zero and solving for $x$ .

STEP 6

Setting $3x -2 =0$ and solving for $x$ gives $3x =2$ $x =2/3$

STEP 7

Setting $2x +7 =0$ and solving for $x$ gives $2x = -7$ $x = -7/2$

STEP 8

So, the roots of the quadratic equation are $x =2/3$ and $x = -7/2$ .

STEP 9

Next, we need to determine the intervals of $x$ that satisfy the inequality. We can do this by testing the sign of the quadratic expression at points in the intervals $(-\infty, -7/2)$ , $(-7/2,2/3)$ , and $(2/3, \infty)$ .

STEP 10

For the interval $(-\infty, -7/2)$ , we can choose a test point, say $x = -4$ . Substituting $x = -4$ into the quadratic expression gives a positive value, so the inequality is satisfied in this interval.

STEP 11

For the interval $(-7/,/3)$ , we can choose a test point, say $x =0$ . Substituting $x =0$ into the quadratic expression gives a negative value, so the inequality is not satisfied in this interval.

STEP 12

For the interval $(2/, \infty)$ , we can choose a test point, say $x =$ . Substituting $x =$ into the quadratic expression gives a positive value, so the inequality is satisfied in this interval.

STEP 13

Therefore, the set of values of $x$ that satisfy the inequality $6x^2 +17x \geq$ is $(-\infty, -7/2] \cup [2/3, \infty)$ .

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