Question

Does anyone know the answer to these two problems?

Answer 1

Answer:

8) -8, -9

9) 4, $\frac{1}{2}$

Step-by-step explanation:

8) x² + 12x + 27

Solutions of a polynomial means the values of $x$ such that the polynomial becomes zero. $i.e., y = 0$.

We use the following formula to determine the solutions of a quadratic equation of the form: $ax^2 + bx + c = 0$

    $\implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this equation, $a = 1; b = 12; c = 27$

Substituting in the formula, we have:

$x = \frac{-12 \pm \sqrt{144 - 108}}{2}$

$\implies x = \frac{-12 \pm 6}{2}$

Taking 2 common outside we get:

$x = - 6 \pm 3$

Therefore, we get two solutions viz, x = -3, -9.

9) 2x² - 9x + 4 = y

X- intercepts of a polynomial means the value of x such that y = 0. It means the roots of the polynomial.

Solving the equation using the formula above, we get:

$x = \frac{-(-9) \pm \sqrt{81 - 32}}{2(2)}$

$\implies x = \frac{9 \pm 7}{4}$

The values of x = 4, $\frac{1}{2}$.

Hence, the answer.