# Quadratic Equation A10

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https://algebrica.org/exercises/quadratic-equation-a-10/
Fetched from algebrica.org test 3487; source modified 2025-03-06T16:11:23.

Solve the equation:

\\[
9x^2 - 5=0
\\]

* * *

In this case, we have an [incomplete](../incomplete-quadratic-equations.md) quadratic equation, where \\(b\\), the coefficient of the linear term is equal to zero. We can rewrite the equation as:

\\[
x^2=\\frac{c}{a}
\\]

* * *

In this case, \\(a\\) the coefficient of the quadratic term \\(x^2\\) and the constant \\(c\\) have different signs, so the equation admits two distinct real solutions:

\\[
x\_{1,2} = \\pm \\sqrt{\\frac{5}{9}}
\\]

* * *

Taking the \\(1/9\\) out of the root, we get:

\\[
x\_{1,2}=\\pm \\frac{\\sqrt{5}}{3}
\\]

The solution to the equation is:

\\[
x\_1 =-\\frac{\\sqrt{5}}{3} \\quad x\_2 =\\frac{\\sqrt{5}}{3}
\\]

* * *

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}
\\]