Question

Need help with finding the following with the given equation.

Answer 1

Let's answer question 3 first, because by solving it through the quadratic formula, we actually get the discriminant for free and it's easy to get the axis of symmetry.

In this quadratic, 2x²- 12x + 7 the values for the quadratic formula are a = 2, b = -12, c = 7. (Read the coefficients with a for the squared term, b for the x term, c for the number).

The axis of symmetry is at x = $\frac{-b}{2a} = \frac{-(-12)}{2*2} = 3$

Remember that $\frac{-b}{2a}$ part? That's half of the quadratic formula.

x = $x = \frac{-b + \sqrt{b^{2}-4ac}}{2a} or x = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

(You often see the plus or minus part, but this chat box doesn't have that key.) Also, the discriminant is the other half - the b²-4ac part. That is 12²-4(2)(7) = 144 - 56 = 88.

$x = \frac{-(-12) + \sqrt{((-12)^{2}-4(2)(7))}}{2(2)} or x = \frac{-(-12) - \sqrt{((-12)^{2}-4(2)(7)})}{2(2)}$

$x = \frac{12 + \sqrt{144-56}}{4} or x = \frac{12 - \sqrt{144-56}}{4}$

$x = \frac{12+\sqrt{88}}{4} or x = \frac{12-\sqrt{88}}{4}$

$x = \frac{12 + \sqrt{4}\sqrt{22}}{4} or x = \frac{12 - \sqrt{4}\sqrt{22}}{4}$

$x = \frac{12+ 2\sqrt{22}}{4} or x = \frac{12- 2\sqrt{22}}{4}$

$x = \frac{6+\sqrt{22}}{2} or x = \frac{6-\sqrt{22}}{2}$

Thus, $x = \frac{6+\sqrt{22}}{2} or x = \frac{6-\sqrt{22}}{2}$ are the solutions, the axis of symmetry is at x =3, and the discriminant is 88.