# Quadratic Equation B9
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-b-9/
Fetched from algebrica.org test 5010; source modified 2025-03-06T18:07:34.
Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).
\\[
\\sqrt{2}x^2 +7x +5\\sqrt{2} = 0
\\]
* * *
The equations is in the standard form \\(ax^2+bx+c=0\\). It is essential to verify the its [discriminant](../quadratic-formula.md) \\(\\Delta = b^2 - 4ac\\) is \\(\\geq0\\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \\(\\Delta\\), we get:
\\[
\\Delta = (7)^2 - 4(\\sqrt{2})(5) = 49 + 20\\sqrt{2} \\geq 0
\\]
\\(\\Delta \\gt 0\\) means the equation has real solutions.
* * *
Now, we need to [factorize](../factoring-quadratic-equations.md) the [polynomial](../polynomials.md). We must find two numbers, \\(r\_1, r\_2\\) whose sum \\(S = r\_1 + r\_2\\) equals \\(b = 7\\) and whose product \\(P = r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = \\sqrt{2} \\cdot 5\\sqrt{2}= 10\\). We can use this simple table to find the numbers that satisfy our constraints.
\\begin{array}{rrrr} & r\_1 & r\_2 & P & S \\\\ \\hline & 2 & 5 & 10 & 7 \\\\ & -2 & -5 & 10 & -7\\\\ \\end{array}
The numbers \\(r\_1, r\_2\\) satisfying the constraint are 2 and 5. Then we need to rewrite the polynomial as \\(ax^2 + r\_{1}x + r\_{2}x + c\\).
* * *
The equation becomes:
\\[
\\sqrt{2}x^2+2x++5x+5\\sqrt{2} = 0
\\]
Factoring common terms, we get:
\\begin{align\*} & \\sqrt{2}x^2+2x+5(x+\\sqrt{2}) = 0 \\[1em] & \\sqrt{2}x^2+\\sqrt{2}\\cdot\\sqrt{2}x^2x+5(x+\\sqrt{2}) = 0 \\[1em] & \\sqrt{2}x(x+\\sqrt{2})+5(x+\\sqrt{2}) = 0 \\[1em] & (\\sqrt{2}x+5)(x+\\sqrt{2}) = 0 \\end{align\*}
The solutions are the values of \\(x\\) for which \\(\\sqrt{2}x+5 = 0\\) and \\(x+\\sqrt{2} = 0\\).
\\[
\\sqrt{2}x+5 = 0 \\to \\sqrt{2}x+5 \\to \\sqrt{2}x = -5 \\to x = -\\frac{5}{\\sqrt{2}}
\\]
\\[
x+\\sqrt{2} = 0 \\to x=-\\sqrt{2}
\\]
The solution to the equation is:
The solution to the equation is:
\\[
x=-\\frac{5}{\\sqrt{2}} \\quad\\quad x=-\\sqrt{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}
\\]
