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.
To solve a quadratic equation, we use the quadratic formula which is expressed as:
x1 = [-b + √(b²-4ac)] / 2a]
x2 = [-b - √(b²-4ac)] / 2a]
These formulas are used to solve for the values of the roots of the equation given. From the equation,
a = 1
b = -5
c = -4
Substituting the values, we have:
x1 = 5.70
x2 = -0.70
Therefore, the negative real number is -0.70.
The value -0.07 is the negative real number in the quadratic equation 0 = x² – 5x – 4
Any equation of the form \(\rm ax^2+bx+c=0\) where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.
As we know, the formula for the roots of the quadratic equation is given by:
\(\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}\)
We have a quadratic equation;
0 = x² – 5x – 4 or
x² – 5x – 4 = 0
Here a = 1, b = -5, and c = -4
\(\rm x = \dfrac{-(-5) \pm\sqrt{-5^2-4(1)(-4)}}{2(1)}\)
\(\rm x = \dfrac{5 \pm\sqrt{41}}{2}\)
After simplification:
x = 5.70 or x = -0.70
The value x = -0.70 is the negative real number.
Thus, the value -0.07 is the negative real number in the quadratic equation 0 = x² – 5x – 4
Learn more about quadratic equations here:
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