nmargot wrote:

If 3x^2 + 2x + 9 = 2x^2 + 9x + 3, what all the possible values of x ?

A. -6 and -1

B. -6 and 1

C. -1 and 6

D. 1 and 6

E. it cannot be determined from the information given.

Could someone expand on this pls?

I will try, but I'm not sure exactly where you are confused.

\(3x^2 + 2x + 9 = 2x^2 + 9x + 3\) gives you an equation from which to work. Get everything on one side, with 0 on the other. You are trying to "solve for x," to find out what x equals.

\((3x^2 - 2x^2) + (2x - 9x) + (9 - 3) = 0\)

\(x^2 - 7x + 6 = 0\) is just a quadratic equation. Find the roots, the solutions, that satisfy the quadratic equation.

"What [are] all the possible values of x" means "find the roots," which requires that you factor the quadratic.

\(x^2 - 7x + 6 = 0\)

\((x - 6)(x - 1) = 0\)

By the zero product rule (if a*b = 0, then a = 0 or b = 0, out both a and b = 0), where the two expressions in parentheses are equivalent to the "a" and "b" in the rule:

\(x - 6 = 0, x = 6\) OR

\(x - 1 = 0, x = 1\)

or both

When you plug x = 1 and x = 6 into the original two quadratics, both numbers work. For (x = 1) 14 = 14, and for (x = 6), 129 = 129.

There can be zero, one, or two solutions to a quadratic equation, depending on what is called the "discriminant."

This time there are two roots, two possible values of x.

They are 1 and 6.

Answer D

Does that help?

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