# Quadratic Equation A6

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https://algebrica.org/exercises/quadratic-equation-a-6/
Fetched from algebrica.org test 3470; source modified 2025-03-06T16:13:05.

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

\\[
(x - 4)^2 - 9 = 0
\\]

* * *

The term, \\((x - 4)^2\\), represents the [square of a binomial](../notable-products.md) of the form \\((a-b)^2\\) and can be expanded as \\(a^2-2ab+b^2\\). Thus, the given equation can be rewritten as:

\\[
(x^2 - 8x+16)-9 = 0
\\]

* * *

After performing the necessary calculations, the equation can be written in the following form:

\\[
x^2 - 8x+7 = 0
\\]

* * *

We can substitute the coefficients \\(a= 1, b=-8, c=7\\) into the [quadratic formula](../quadratic-formula.md):

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

We obtain:

\\[
x\_{1,2} = \\frac{{-(-8) \\pm \\sqrt{{(-8)^2 - 4(1)(7)}}}}{{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{{8 \\pm \\sqrt{{64 - 28}}}}{2}\\[0.6em] &= \\frac{{8 \\pm \\sqrt{{36}}}}{2} \\end{align\*}

* * *

Finally, by performing the calculations, we obtain:

\\[
x\_1 = \\frac{8 - 6}{2} = \\frac{2}{2} = 1
\\]



\\[
x\_2 = \\frac{8 + 6}{2} = \\frac{14}{2} = 7
\\]

The solution to the equation is:

\\[
x\_1 =1 \\quad x\_2=7
\\]

* * *

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}
\\]