# Quadratic Equation A7

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https://algebrica.org/exercises/quadratic-equation-a-7/
Fetched from algebrica.org test 3473; source modified 2025-03-06T16:12:45.

Solve the [quadratic equation](../quadratic-equations.md):

\\[
(4x + 8)\\left(\\frac{1}{2}x - 6\\right) = 0
\\]

* * *

We can use the distributive property to expand the equation and obtain the following:

\\begin{align\*} \\frac{4}{2}x^2 -24x + \\frac{8}{2}x-48 = 0 \\[0.6em] 2x^2 - 24x +4x-48 = 0 \\[1em] 2x^2 - 20x-48 = 0 \\end{align\*}

* * *

The coefficients \\(a, b\\) and \\(c\\) have \\(2\\) as common multiplier. We can factor out the number and obtain:

\\[
x^2 - 10x - 24 = 0
\\]

* * *

The equation is now reduced to the standard form \\(ax^2+bx+c= 0\\). We can substitute the coefficients \\(a=2, b=20, c=48\\) into the [quadratic formula](../quadratic-formula.md):

\\[
x\_{1,2} = \\frac{{-b \\pm \\sqrt{{b^2 - 4ac}}}}{{2a}}
\\]

We obtain:

\\[
x\_{1,2}= \\frac{{-(-10) \\pm \\sqrt{{(-10)^2-4(1)(-24)}}}}{{2(1)}}
\\]

* * *

In this case, the discriminant \\(\\Delta\\) is \\(\\geq 0\\) so the equation admits two distinct real solutions.

\\begin{align\*} x\_{1,2} &= \\frac{{10 \\pm \\sqrt{{100 + 96}}}}{2}\\[1em] x\_{1,2} &= \\frac{{10 \\pm \\sqrt{{196}}}}{2} \\end{align\*}

* * *

Finally, by performing the calculations, we obtain:

\\[
x\_1 = \\frac{{10 + 14}}{2} = 12
\\]



\\[
x\_2 = \\frac{{10 - 14}}{2} = -2
\\]

The solution to the equation is:

\\[
x =12 \\quad \\quad x =-2
\\]

* * *

Remember that the [discriminant](../quadratic-formula.md) is crucial in determining the nature and number of solutions of quadratic equations.

-   If \\( b^2 - 4ac > 0\\), the quadratic equation has two distinct real solutions.

\\[
S = \\{x\_1, x\_2\\} \\quad x\_1, x\_2 \\in \\mathbb{R} \\quad x\_1 \\neq x\_2
\\]


-   If \\( b^2 - 4ac = 0\\), the quadratic equation has two coincident real solutions.

\\[
S = \\{x\\} \\quad x \\in \\mathbb{R} \\quad x = x\_1 = x\_2
\\]


-   If \\( b^2 - 4ac = < 0\\), the quadratic equation has no real solutions. Instead, it gives rise to complex solutions characterized by imaginary components.

\\[
\\nexists \\hspace{10px} x \\in \\mathbb{R}
\\]