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A polynomial equation with a degree equal to two is known as a quadratic equation. While "quad" means four, "quadratic" means to make a square. A quadratic equation in standard form is represented as:
${ax}^{2}+bx+c=0$
where b and c are real numbers such that a does not equal 0.
Since the degree of the above-written equation is two, it will have two roots, or solutions. The roots of polynomials are the values of x that satisfy the equation. There are a handful of methods you can use to find the roots of a quadratic equation.
One of them is called completing the square. Using this method, you have to convert the given equation into a perfect square.
To solve ${ax}^{2}+bx+c=0$ by completing the square:
The above steps can be implemented as shown below.
Consider the quadratic equation
${ax}^{2}+bx+c=0$ (a not equal to 0)
Get c by itself on the right-hand side
${ax}^{2}+bx=-c$
Dividing both sides by a, we get
${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
Add ${\left(\frac{b}{2a}\right)}^{2}$ to both sides
${x}^{2}+\left(\frac{b}{a}\right)x+{\left(\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$
Factor the left-hand side
${\left(\frac{2ax+b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$
Square root both sides
$\frac{2ax+b}{2a}=\sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}$
This might look overwhelming, but recall that a, b, and c are just numbers, so the whole right side is just some constant.
Example 1
Solve ${x}^{2}-6x-3=0$ by completing the square.
${x}^{2}-6x=3$
${x}^{2}-6x+(-3){}^{2}=3+9$
${(x-3)}^{2}=12$
$x-3=\pm \sqrt{12}$
$x-3=\pm 2\sqrt{3}$
$x=3\pm 2\sqrt{3}$
Example 2
Solve $7{x}^{2}-8x+3=0$ by completing the square.
$7{x}^{2}-8x=-3$
${x}^{2}-\frac{8x}{7}=-\frac{3}{7}$
${x}^{2}-\frac{8x}{7}+{(-\frac{4}{7})}^{2}=-\frac{3}{7}+\frac{16}{49}$
${\left(x-\frac{4}{7}\right)}^{2}=\frac{-5}{49}$
$x-\frac{4}{7}=\frac{\sqrt{5}}{7}i$
$x=\frac{4}{7}\pm \sqrt{\frac{\sqrt{5}}{7}i}$
The above steps can also be written by defining 2 more terms d and e.
$d=\frac{b}{2a}$
$e=c-\frac{{b}^{2}}{4a}$
We can then write:
$a(x+d{)}^{2}+e=0$
and solve for x in the following way
${(x+d)}^{2}=-\frac{e}{a}$
$x+d=\pm \sqrt{\frac{-e}{a}}$
$x=-d\pm \sqrt{\frac{-e}{a}}$
Then we can either put in our values for a,b, and c, or d and e to get the answers for x.
$a(x+\frac{b}{2a}{)}^{2}+c-\frac{{b}^{2}}{4a}=0$
Further simplification of this will give you the quadratic formula, which is another method for solving quadratic equations.
1. Find the roots of the quadratic equation ${x}^{2}+4x-5=0$ by the method of completing the square.
The given quadratic equation is:
${x}^{2}+4x-5=0$
Comparing the equation with the standard form,
$b=4,c=-5$
${(x+\frac{b}{2})}^{2}=-(c-\frac{{b}^{2}}{4})$
So, ${(x+\frac{4}{2})}^{2}=-(5-\frac{{4}^{2}}{4})$
${(x+2)}^{2}=5+4$
${(x+2)}^{2}=9$
$(x+2)=\pm \sqrt{9}$
$(x+2)=\pm 3$
$x+2=3,x+2=-3$
$x=1,-5$
2. Find the roots of the quadratic equation $3{x}^{2}-5x+2=0$ by completing the square.
The given quadratic equation is:
$3{x}^{2}-5x+2=0$
The given equation is not in the form to which we apply the method of completing the square because the coefficient of ${x}^{2}$ is not 1. To make it 1, we need to divide the whole equation by 3:
${x}^{2}-\frac{5}{3}x+\frac{2}{3}=0$
Comparing with the standard form,
$b=\frac{5}{3},c=\frac{2}{3}$
$c-\frac{{b}^{2}}{4}=\frac{2}{3}-\frac{{(-\frac{5}{3})}^{2}}{4}=\frac{2}{3}-\frac{25}{36}=\frac{24-25}{36}=-\frac{1}{36}$
Substituting these values in the equation ${(x+\frac{b}{2})}^{2}=-(c-\frac{{b}^{2}}{4})$ , we get
${(x-\frac{5}{6})}^{2}=\frac{1}{36}$
$(x-\frac{5}{6})=\pm \sqrt{\frac{1}{36}}$
$(x-\frac{5}{6})=\pm \frac{1}{6}$
$x=1,\frac{2}{3}$
College Algebra Diagnostic Tests
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