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# Inequality

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For what values of $x$ is $2x^2+8x\le-6+6x+4$? Express your answer in interval notation.

Aug 16, 2023

#2
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If you manipulate the inequality by completing the square, the x-values that satisfy and do not satisfy this inequality become clear rather quickly.

$$2x^2 + 8x < -6 + 6x + 4 \\ 2x^2 + 8x < 6x - 2 \\ 2x^2 + 2x < -2 \\ 2(x^2 + x) < -2 \\ 2\left(x^2 + x + \frac{1}{4}\right) < -2 + 2 * \frac{1}{4} \\ 2\left(x + \frac{1}{2}\right)^2 < -\frac{3}{2}$$

Now, analyze the inequality carefully. I noticed that the left-hand side is the product of two nonnegative quantities. 2 is positive, and $$\left(x + \frac{1}{2}\right)^2$$ has a square, which guarantees that the result is either 0 or positive in nature. If we multiply 2 by a quantity that is either 0 or positive, then it is impossible for that quantity to be less than $$-\frac{3}{2}$$. This indicates that there is no solution to this inequality. In interval notation, that is typically denoted as $$\emptyset$$.

Aug 16, 2023

#1
+121
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To solve the inequality $$2x^2 + 8x \le -6 + 6x + 4$$, we need to simplify the right side first:

$$-6 + 6x + 4 = 6x - 2.$$

Now, the inequality becomes:

$$2x^2 + 8x \le 6x - 2.$$

Subtract $$6x - 2$$ from both sides:

$$2x^2 + 8x - 6x - 2 \le 0.$$

Simplify further:

$$2x^2 + 2x - 2 \le 0.$$

Now, we have a quadratic inequality. To solve for $$x$$, we can factor the quadratic:

$$2(x^2 + x - 1) \le 0.$$

Using the quadratic formula, we get

$$\frac{-1-\sqrt5}2 \le x \le \frac{-1+\sqrt5}2.$$

In interval notation, that is $$[\frac{-1-\sqrt5}2, \frac{-1\sqrt5}2]$$.

Aug 16, 2023
#3
+177
+1

Greetings, SpectraSynth!

You made a great attempt to solve this inequality, but I noticed you made an algebraic error along the way. You subtracted the quantity $$6x-2$$ incorrectly from both sides of the inequality, which lead your answer astray.

The3Mathketeers  Aug 16, 2023
#2
+177
+1
$$2x^2 + 8x < -6 + 6x + 4 \\ 2x^2 + 8x < 6x - 2 \\ 2x^2 + 2x < -2 \\ 2(x^2 + x) < -2 \\ 2\left(x^2 + x + \frac{1}{4}\right) < -2 + 2 * \frac{1}{4} \\ 2\left(x + \frac{1}{2}\right)^2 < -\frac{3}{2}$$
Now, analyze the inequality carefully. I noticed that the left-hand side is the product of two nonnegative quantities. 2 is positive, and $$\left(x + \frac{1}{2}\right)^2$$ has a square, which guarantees that the result is either 0 or positive in nature. If we multiply 2 by a quantity that is either 0 or positive, then it is impossible for that quantity to be less than $$-\frac{3}{2}$$. This indicates that there is no solution to this inequality. In interval notation, that is typically denoted as $$\emptyset$$.