Math  /  Algebra

QuestionSolve the rational inequality. 7) 4x+75x2+6>0\frac{-4 x+7}{5 x^{2}+6}>0

Studdy Solution

STEP 1

What is this asking? We need to find the values of xx that make the expression 4x+75x2+6\frac{-4x + 7}{5x^2 + 6} greater than zero! Watch out! Don't forget to consider the denominator; it can't be equal to zero!
Also, be careful with flipping the inequality sign when multiplying or dividing by a negative number.

STEP 2

1. Analyze the denominator
2. Analyze the numerator
3. Combine and conquer

STEP 3

Let's **investigate** the denominator: 5x2+65x^2 + 6.
Since x2x^2 is *always* non-negative (meaning it's zero or positive), 5x25x^2 will also *always* be non-negative.

STEP 4

Now, if we add **6** to something that's already non-negative (5x25x^2), the entire expression 5x2+65x^2 + 6 will *always* be positive!
This is **super important** because it means the denominator will *never* be zero, and it will *always* be positive!
So we don't need to worry about any values of xx making the denominator zero, which is awesome!

STEP 5

Now, let's **focus** on the numerator: 4x+7-4x + 7.
We want the entire fraction to be greater than zero, and we *know* the denominator is *always* positive.
Therefore, we need the numerator to be positive too!

STEP 6

So, we set up the inequality: 4x+7>0-4x + 7 > 0.

STEP 7

Let's **solve** for xx!
First, subtract 77 from both sides: 4x>7-4x > -7.

STEP 8

Now, **divide** both sides by 4-4. *Remember* to flip the inequality sign when dividing by a negative number: x<74x < \frac{-7}{-4}.

STEP 9

Simplifying, we get x<74x < \frac{7}{4}.

STEP 10

We found that the denominator is *always* positive, and the numerator is positive when x<74x < \frac{7}{4}.

STEP 11

Therefore, the entire expression 4x+75x2+6\frac{-4x + 7}{5x^2 + 6} is greater than zero when x<74x < \frac{7}{4}.
That's our **final solution**!

STEP 12

The solution to the inequality 4x+75x2+6>0\frac{-4x + 7}{5x^2 + 6} > 0 is x<74x < \frac{7}{4}.

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