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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

Best Answer 

 #2
avatar+177 
+1

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
avatar+121 
0

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
avatar+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
avatar+177 
+1
Best Answer

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 \(\emptyset\).

The3Mathketeers Aug 16, 2023

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