Question

Joline is solving the equation 0 = x2 – 5x – 4 using the quadratic formula. Which value is the negative real number solution to her quadratic equation? Round to the nearest tenth if necessary.

Answer 1

Answer:

The negative real number solution to her quadratic equation is -0.70.

Step-by-step explanation:

Consider the provided equation $x^2-5x-4=0$

The standard form of quadratic equation is $ax^2+bx+c=0$

To solve the above quadratic equation, use the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Now compare the provided equation with the standard equation we can concluded that:

a = 1, b = -5, c = -4

Substituting the values in the above formula:

$x_{1,2}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-4)}}{2(1)}$

$x_{1,2}=\frac{5\pm\sqrt{25+16}}{2}$

$x_{1,2}=\frac{5\pm\sqrt{41}}{2}$

$x_{1,2}=\frac{5\pm6.4}{2}$

$x_1 = \frac{5+6.4}{2}$ or $x_2 = \frac{5-6.4}{2}$

$x_1 = \frac{11.4}{2}$ or $x_2 = \frac{-1.4}{2}$

$x_1 = 5.7$ or $x_2 = -0.70$

Therefore, the negative real number solution to her quadratic equation is -0.70.