Solving Quadratic Equations

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In this lesson we will learn to how solve the quadratic equation: $ax^{2}+bx+c=0$ for $x$ where all coefficients $a$, $b$ and $c$ are real numbers. In addition, we suppose that $a$ is different from zero, otherwise the equation would be linear.

First, we compute the determinant $\Delta =b^{2}-4ac$. We distinguish three cases:

If the determinant is positive, then equation admits two solutions:

$x_{1,2}={\frac {-b\pm {\sqrt {\Delta }}}{2a}}$

If the determinant is zero, then equation admits the single solution:

$x={\frac {-b}{2a}}$

If the determinant is negative, there are no real value satisfying the quadratic equation.

Example:

Solve $2x^{2}-8x+6=0$ for $x$. The determinant is,

$\Delta =b^{2}-4ac=8^{2}-4\cdot 2\cdot 6=16$

The equation admits therefore two solutions, namely:

$x_{1,2}={\frac {8\pm {\sqrt {16}}}{2\cdot 2}}={\frac {8\pm 4}{4}}=2\pm 1$

The two solutions are thus $x_{1}=1$ and $x_{2}=3$ (the order is not important).

Verification

Indeed, this can be verified substituting $x_{1}$ and $x_{2}$ in the original equation:

$2\cdot 1^{2}-8\cdot 1+6=2-8+6=0$

and

$2\cdot 3^{2}-8\cdot 3+6=18-24+6=0$

Alternative solution

The equation can also be solved by "completing the square":

$2x^{2}-8x+6=0$

Divide by 2 so that $x^{2}$ is alone:

$x^{2}-4x+3=0$

Subtract $3$

$x^{2}-4x=-3$

We want to apply the binomial formula $x^{2}+2xy+y^{2}=(x+y)^{2}$ on the left side of the equation. If $-4x$ is $2xy$, then $y$ is $-2$ and $y^{2}$ is $4$:

$x^{2}-4x+4=-3+4$

Apply the binomial formula:

$(x-2)^{2}=-3+4$

Or

$(x-2)^{2}=1$

Let's take the square root:

$x-2=\pm 1$

Solve for $x$

$x=2\pm 1$

The solutions are thus $1$ and $3$.

The previous approach can be uses to proof the general formula.