Question

question 1 of 10
which values of x are solutions to the equation below?
check all that apply.
10x² - 56 = 88 - 6x²
a. x = -3
b. x = -12
c. x = √3
d. x = 12
e. x = 3
f. x = -√3

Explanation:

Step1: Combine like terms

First, we want to get all the \(x^2\) terms on one side and the constant terms on the other side. So we add \(6x^2\) to both sides and add \(56\) to both sides of the equation \(10x^2 - 56 = 88 - 6x^2\).

$$10x^2+6x^2- 56 + 56=88 + 56- 6x^2+6x^2$$

Simplifying both sides, we have:

$$16x^2=144$$

Step2: Solve for \(x^2\)

Divide both sides of the equation \(16x^2 = 144\) by \(16\) to isolate \(x^2\).

$$\frac{16x^2}{16}=\frac{144}{16}$$

Simplifying, we get:

$$x^2 = 9$$

Step3: Solve for \(x\)

Take the square root of both sides of the equation \(x^2=9\). Remember that when we take the square root of a number, we get both a positive and a negative root.

$$x=\pm\sqrt{9}=\pm 3$$

Answer:

A. \(x = - 3\), E. \(x = 3\), F. \(x=-\sqrt{3}\) (Wait, no, from our calculation \(x^2 = 9\), so \(x=\pm3\). Wait, let's re - check the calculation.

Wait, in step 1: \(10x^2-56 = 88 - 6x^2\)

Adding \(6x^2\) to both sides: \(10x^2+6x^2-56=88\)

Adding \(56\) to both sides: \(16x^2=88 + 56=144\)

Then \(x^2=\frac{144}{16} = 9\), so \(x=\pm\sqrt{9}=\pm3\). So the correct solutions are \(x = 3\), \(x=-3\) and also, wait, if we made a mistake in the coefficient? Wait no, \(10x^2+6x^2 = 16x^2\), \(88 + 56=144\), \(144\div16 = 9\), so \(x^2 = 9\), so \(x = 3\) or \(x=-3\), and also, wait, if we consider \(x^2=9\), then \(x=\pm3\), but also, if we have \(x^2 = 9\), is there another solution? No, because \((\sqrt{3})^2=3
eq9\), \((-\sqrt{3})^2 = 3
eq9\), \(12^2=144
eq9\), \((-12)^2 = 144
eq9\). So the correct options are A (\(x=-3\)), E (\(x = 3\)). Wait, but also, wait, did I make a mistake in the equation?

Wait the original equation is \(10x^2-56=88 - 6x^2\)

Let's check \(x = 3\): Left side: \(10\times(3)^2-56=10\times9 - 56=90 - 56 = 34\). Right side: \(88-6\times(3)^2=88 - 54 = 34\). So \(x = 3\) is a solution.

\(x=-3\): Left side: \(10\times(-3)^2-56=10\times9 - 56 = 34\). Right side: \(88-6\times(-3)^2=88 - 54 = 34\). So \(x=-3\) is a solution.

\(x=\sqrt{3}\): Left side: \(10\times(\sqrt{3})^2-56=10\times3 - 56=30 - 56=-26\). Right side: \(88-6\times(\sqrt{3})^2=88 - 18 = 70\). \(-26
eq70\), so not a solution.

\(x = -\sqrt{3}\): Left side: \(10\times(-\sqrt{3})^2-56=10\times3 - 56=-26\). Right side: \(88-6\times(-\sqrt{3})^2=88 - 18 = 70\). \(-26
eq70\), so not a solution.

\(x = 12\): Left side: \(10\times12^2-56=10\times144-56 = 1440 - 56 = 1384\). Right side: \(88-6\times12^2=88 - 864=-776\). \(1384
eq - 776\), not a solution.

\(x=-12\): Left side: \(10\times(-12)^2-56=10\times144-56 = 1384\). Right side: \(88-6\times(-12)^2=88 - 864=-776\). \(1384
eq - 776\), not a solution.

So the correct answers are A. \(x=-3\) and E. \(x = 3\), and also, wait, when we solved \(x^2=9\), we get \(x = 3\) or \(x=-3\), so the correct options are A and E, and also, is there a mistake in the problem? Wait no, the options F is \(x = -\sqrt{3}\), which we saw is not a solution. So the correct solutions are \(x = 3\) and \(x=-3\), so the correct options are A. \(x=-3\), E. \(x = 3\) and also, wait, if we consider that maybe I made a mistake in the coefficient:

Wait, let's re - do the equation:

\(10x^2-56=88 - 6x^2\)

Bring all \(x^2\) terms to left: \(10x^2 + 6x^2=88 + 56\)

\(16x^2=144\)

\(x^2 = 9\)

\(x=\pm3\). So the correct options are A (\(x=-3\)) and E (\(x = 3\)), and also, wait, the option F is \(x=-\sqrt{3}\), which is not a solution, and option C is \(x=\sqrt{3}\), which is not a solution. So the correct answers are A and E.

Wait, but in the initial calculation, I thought maybe F, but no, because \((-\sqrt{3})^2=3
eq9\). So the correct solutions are \(x = 3\) and \(x=-3\), so the correct options are A. \(x=-3\) and E. \(x = 3\).

So the final answer is A. \(x=-3\), E. \(x = 3\)