Question

SOMEONE PLS HELP MEEEE

Answer 1

The solutions to the quadratic equations are:

a) $x = \frac{3 \pm \sqrt{19}}{5}$

b) x = -8, x = 5.

A quadratic function is given according to the following rule:

$y = ax^2 + bx + c$

The solutions are:

$x_1 = \frac{-b + \sqrt{\Delta}}{2a}$

$x_2 = \frac{-b - \sqrt{\Delta}}{2a}$

In which:

$\Delta = b^2 - 4ac$

Item a:

The coefficients are a = 5, b = -6, c = -2, hence:

• $\Delta = (-6)^2 - 4(5)(-2) = 76$

• $x_1 = \frac{6 + \sqrt{76}}{10} = \frac{6 + 2\sqrt{19}}{10} = \frac{3 + \sqrt{19}}{5}$

• $x_2 = \frac{6 - \sqrt{76}}{10} = \frac{6 - 2\sqrt{19}}{10} = \frac{3 - \sqrt{19}}{5}$

Item b:

The equation is:

x² + 3x = 40

x² + 3x - 40 = 0.

It can be simplified as follows:

(x - 5)(x + 8) = 0.

Hence the solutions are:

• x - 5 = 0 -> x = 5.

• x + 8 = 0 -> x = -8.

More can be learned about quadratic equations at brainly.com/question/24737967

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