Answer:
3x^2 - 48 = 0
Explanation:
The roots of a quadratic equation whose roots are -4 and 4, means that valid solutions for the quadratic equation are -4 and 4.
The two forms we know for the quadratic equation are the factored binomials:
(ax + cy)(bx + dz) = 0
and the full equation
ax^2 + bx + c = 0
We will start with the factored form:
(x + 4) ( x - 4) = 0
Note that if we solve this, we get the roots: x + 4 = 0 ==> -4 && x - 4 = 0 ==> 4
Now we apply foil to get the unfactored form:
(x + 4) (x - 4) = 0
x * x + x * -4 + 4 * x + 4 * -4 = 0
x^2 + -4x + 4x + -16 = 0
x^2 - 16 = 0
And now we want the leading coefficient to be 3.
3x^2 - 16 = 0
This looks like what we want; however, now with the leading 3, our solutions are no long +4 and -4. So, we need to re-evaluate.
Going back to the binomials with the idea of a leading 3, let's rewrite them to still give us the solutions 4 and -4, but with the 3 included:
(3x + 12) (x - 4) = 0
Note, we simply multipled one binomial by 3 to maintain the roots of the original, but also incorporate the leading 3.
3x + 12 = 0 x - 4 = 0
3x = -12 x = 4
x = -4
So now, we simply apply foil to this new equation of binomials to get our final quadratic equation:
(3x + 12) (x - 4) = 0
3x * x + 3x * -4 + 12 * x + 12 * -4 = 0
3x^2 + -12x + 12x + -48 = 0
3x^2 - 48 = 0
This is our quadratic equation we wish to use.
Note, we could have gotten this same equation by simply multiplying the original (x^2 - 16) = 0 by 3.
Cheers.