Solving the quadratic equation with the quadratic formula

Solve the quadratic equation with the quadratic formula

The method of completing the square can be used to develop a formula that enables us to solve any quadratic equation. Consider the general quadratic equation :

a^{2}+bx+c=0\text{ and }a\neq 0

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By applying the method of completing the square, we can obtain the solutions in terms of the coefficients a, b, and c.

ax^2+bx+c=0

x^2+\frac{b}{a}x+\frac{c}{a}=0

x^2+\frac{b}{a}x=-\frac{c}{a}

Since the coefficient of x is b/a, the square of one – half the coefficient of x is (\frac{b}{2a})^2 to each side.

x^{2}+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^{2}=-\frac{c}{a}+\left ( \frac{b}{2a} \right )^{2}

\left ( x+\frac{b}{2a} \right )^{2}=-\frac{4ac}{4a^{2}}+\frac{b^{2}}{4a^{2}}

\left ( x+\frac{b}{2a} \right )^{2}=\frac{b^{2}-4ac}{4a^{2}}

x+\frac{b}{2a}=\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}

x+\frac{b}{2a}=\pm \frac{\sqrt{b^{2}-4ac}}{2a}

x=-\frac{b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}

\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

We now have a formula for solving any quadratic equation. It is usually stated as follows :

The Quadratic Formula

The solutions of the quadratic equation

a^{2}+bx+c=0\text{ and }a\neq 0

Are given by

\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

Example 1 :

Solve

3x^2-4x=-1

Using the quadratic formula.

Solution :

We first put the equation in standard form in order to identify a, b, and c.

3x^2-4x=-1

3x^2-4x+1=0

a = 3, b = – 4, c = 1.

Substituting these values for a, b, and c into the quadratic formula, we have :

\text{x}=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(3)(1)}}{2(3)}

\text{x}=\frac{4\pm\sqrt{16-12}}{6}

\text{x}=\frac{4\pm\sqrt{4}}{6}

\text{x}=\frac{4\pm2}{6}

Therefore,

\text{x}=\frac{4+2}{6}=1\text{ or }\text{x}=\frac{4-2}{6}=\frac{1}{3}

The solution are 1 and 1/3.

Example 2 # :

Solve

x^2+x-1=0

Using the quadratic formula.

Solution :

x^2+x-1=0

a = 1, b = 1, c = -1

\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

\text{x}=\frac{-1\pm\sqrt{(1)^{2}-4(1)(-1)}}{2(1)}

\text{x}=\frac{-1\pm\sqrt{1+4}}{2}

\text{x}=\frac{-1\pm\sqrt{5}}{2}

The solution are

\frac{-1+\sqrt{5}}{2}\text{ and }\frac{-1-\sqrt{5}}{2}

Example 3 # :

Solve

\frac{x^2}{6}+\frac{x}{3}-\frac{1}{3}=0

Using the quadratic formula.

Solution :

We first simplify the equation by multiplying both sides by the lcd 6.

\frac{x^2}{6}+\frac{x}{3}-\frac{1}{3}=0

6\left ( \frac{x^2}{6}+\frac{x}{3}-\frac{1}{3} \right )=6.0

x^2+2x-2=0

a = 1, b = 2, c = – 2

\text{x}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

\text{x}=\frac{-2\pm\sqrt{(2)^{2}-4(1)(-2)}}{2(1)}

\text{x}=\frac{-2\pm\sqrt{12}}{2}

\text{x}=\frac{-2\pm2\sqrt{3}}{2}

Dividing the denominator 2 into both terms in the numerator simplifies the solution to

\text{x}=\frac{-2}{2}\pm\frac{2\sqrt{3}}{2}

\text{x}=-1\pm\sqrt{3}

The solutions are

-1+\sqrt{3}\text{ and }-1-\sqrt{3}

Example 4# :

Solve

x^2-7x=0

Using the quadratic formula.

Solution :

x^2-7x=0

a = 1, b = -7 , c = 0

\text{x}=\frac{-(-7)\pm\sqrt{(2)^{2}-4(1)(-2)}}{2(-2)}

\text{x}=\frac{7\pm\sqrt{49-0}}{2}

\text{x}=\frac{7\pm7}{2}

\text{x}=\frac{7+7}{2}\text{ or }\text{x}=\frac{7-7}{2}

x = 7 or x = 0

The solutions are 7 and 0.

Example 5# :

Solve

x^2+2x+5=0

Using the quadratic formula.

Solution :

x^2+2x+5=0

a = 1, b = 2, c = 5

\text{x}=\frac{-2\pm\sqrt{(2)^2-4(1)(5)}}{2(1)}

\text{x}=\frac{-2\pm\sqrt{4-20}}{2}

\text{x}=\frac{-2\pm\sqrt{-16}}{2}

But \sqrt{-16} is not a real number. Therefore this quadratic equation has no real number solution.

The method of solving a quadratic equation by completing the square can be rather tedious, and for that reason it is seldom used to solve a quadratic equation. The operation of completing the square does, however, have a variety of other application in algebra.

 

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