##### Worked Examples – Quadratic equation

Methods to solve a quadratic equation are:

Square Root Method

We have four types of quadratic equations

1. Quadratic equations in the standard form $\displaystyle a{{x}^{2}}+bx+c=0$

We can solve them by using the quadratic formula, graphing or factoring and completing the square.

2. Quadratic equations lacking the linear term bx. $\displaystyle a{{x}^{2}}+c=0$

We can solve them by using the square root method or graphing.

3. Quadratic equations lacking the constant term c. $\displaystyle a{{x}^{2}}+bx=0$

We can solve them by factoring or graphing.

4. Quadratic equations lacking both bx and c which have always 0 as a solution.$\displaystyle a{{x}^{2}}=0$

Be careful! If the quadratic equation is lacking the term ax2 then is not anymore a quadratic equation because it has the form $\displaystyle bx+c=0$ which is a linear equation.

Example 1: Solve the quadratic equations. Our quadratic equations are of type 2, where the term bx is missing.
a) $\displaystyle {{x}^{2}}-49=0$

Using the square root method we obtain:

$\displaystyle {{x}^{2}}-49=0$

$\displaystyle {{x}^{2}}=49$

$\displaystyle x=\sqrt{{49}}$

$\displaystyle x=\pm 7$

So our solutions are $\displaystyle {{x}_{1}}=7$ and $\displaystyle {{x}_{2}}=-7$

Using the Graphing method we obtain:

The solutions of our equation are the values of x where the graph crosses the x-axis.

b) $\displaystyle 3{{x}^{2}}-36=0$

Using the square root method:

$\displaystyle 3{{x}^{2}}-36=0$

$\displaystyle 3{{x}^{2}}=36$

$\displaystyle {{x}^{2}}=\frac{{36}}{3}=12$

$\displaystyle x=\sqrt{{12}}$

$\displaystyle x=\pm 2\sqrt{3}$

So our solutions are: $\displaystyle {{x}_{1}}=2\sqrt{3}$ and $\displaystyle {{x}_{2}}=-2\sqrt{3}$

c) $\displaystyle {{x}^{2}}+4=0$

Using the square root method:

$\displaystyle {{x}^{2}}+4=0$

$\displaystyle {{x}^{2}}=-4$

$\displaystyle x=\sqrt{{-4}}$

We see that we have complex roots.

$\displaystyle x=\sqrt{{(-1)(4)}}=$

$\displaystyle \sqrt{{(-1)}}\sqrt{{(4)}}=\pm 2i$

So our solutions are: $\displaystyle {{x}_{1}}=2i$ and $\displaystyle {{x}_{2}}=-2i$

TIP! To learn how to graph a quadratic equation see this Example

Worked example 2: Solve the quadratic equations. Our quadratic equations are of type 3 where the constant c is missing.

a) $\displaystyle {{x}^{2}}-3x=0$

Using the factoring method

Factor the common term, in our case the common term is the variable x.

$\displaystyle {{x}^{2}}-3x=0$

$\displaystyle x(x-3)=0$

Then make each factor equal to zero.

$\displaystyle x=0$  or$\displaystyle x-3=0$

So our solutions are $\displaystyle {{x}_{1}}=0$ and $\displaystyle {{x}_{2}}=3$a.

Using the graphing method we obtain:

The solutions of our equation are the values of x where the graph crosses the x-axis.

b) $\displaystyle -6{{x}^{2}}-2x=0$

Using the factoring method

Factor the common term, in our case the common term is the variable x and the constant -2 so the term -2x.

$\displaystyle -6{{x}^{2}}-2x=0$

$\displaystyle -2x(3x-1)=0$

Then make each factor equal to zero

$\displaystyle -2x=0$  or $\displaystyle 3x-1=0$

We solve them:

$\displaystyle -2x=0$ or $\displaystyle 3x-1=0$

$\displaystyle x=0$ or $\displaystyle x=\frac{1}{3}$

So our solutions are: $\displaystyle {{x}_{1}}=0$ and $\displaystyle {{x}_{2}}=\frac{1}{3}$

c) $\displaystyle 11{{x}^{2}}+7x=0$

Using the factoring method

Factor the common term, in our case the common term is the variable x.

$\displaystyle 11{{x}^{2}}+7x=0$

$\displaystyle x(11x+7)=0$

Then make each factor equal to zero

$\displaystyle x=0$  or $\displaystyle 11x+7=0$

Solve them:

$\displaystyle 11x+7=0$

$\displaystyle x=-\frac{7}{{11}}$

So our solutions are: $\displaystyle {{x}_{1}}=0$ and $\displaystyle {{x}_{2}}=-\frac{7}{{11}}$

Worked example 3: Solve the equations.

Our quadratic equations are of type 1 on the standard form.

a) $\displaystyle {{x}^{2}}-x-3=0$

The coefficients are a=1, b=-1, c=-3

Finding the discriminant $\displaystyle D={{b}^{2}}-4ac$

$\displaystyle D={{(-1)}^{2}}-4\cdot 1\cdot (-3)=1+12=13$

Since D<0 then we have two solutions

$\displaystyle {{x}_{1}}=\frac{{-b+\sqrt{D}}}{{2a}}=\frac{{1+\sqrt{{13}}}}{2}$

$\displaystyle {{x}_{2}}=\frac{{-b-\sqrt{D}}}{{2a}}=\frac{{1-\sqrt{{13}}}}{2}$

Completing the Square  Method

Since a=1 then we don’t divide the terms with the coefficient a

Move the term c on the right of the equation

$\displaystyle {{x}^{2}}-x-3=0$

$\displaystyle {{x}^{2}}-x=3$

Complete the square on the left of the equation by adding $\displaystyle \frac{1}{4}$ on both sides

$\displaystyle {{x}^{2}}-x+\frac{1}{4}=3+\frac{1}{4}$

$\displaystyle {{x}^{2}}-x+\frac{1}{4}=\frac{{12}}{4}+\frac{1}{4}$

$\displaystyle {{x}^{2}}-x+\frac{1}{4}=\frac{{13}}{4}$

$\displaystyle {{(x-\frac{1}{2})}^{2}}=\frac{{13}}{4}$

Take the square root on both sides to find x

$\displaystyle \sqrt{{{{{(x-\frac{1}{2})}}^{2}}}}=\sqrt{{\frac{{13}}{4}}}$

$\displaystyle (x-\frac{1}{2})=\pm \frac{{\sqrt{{13}}}}{2}$

$\displaystyle {{x}_{1}}=\frac{{\sqrt{{13}}}}{2}+\frac{1}{2}=\frac{{\sqrt{{13}}+1}}{2}$

$\displaystyle {{x}_{2}}=-\frac{{\sqrt{{13}}}}{2}+\frac{1}{2}=\frac{{1-\sqrt{{13}}}}{2}$

Note! In this case we can’t use the Factoring method and we can use the Graphing method but since the roots are in a rational way the crossing with the x-axis won´t be precise if we graph it by ourselves.

b) $\displaystyle {{x}^{2}}+9x+18=0$

The coefficients are a=1, b=9, c=18

Finding the discriminant $\displaystyle D={{b}^{2}}-4ac$

$\displaystyle D={{(9)}^{2}}-4\cdot 1\cdot 18=81-72=9$

Since D>0 we have two solutions

$\displaystyle {{x}_{1}}=\frac{{-9-\sqrt{9}}}{2}=\frac{{-12}}{2}=-6$

$\displaystyle {{x}_{2}}=\frac{{-9+\sqrt{9}}}{2}=\frac{{-6}}{2}=-3$

Factoring method

Determine the coefficients a=1, b=9, c=18.
Find two numbers that multiply gives ac and add to give b.
The numbers are 6 and 3 because 18 = 6 x 3 and 9 = 6 + 3.
Then we write the equation on the form:

$\displaystyle {{x}^{2}}+9x+18=(x-6)(x-3)$

Make each factor equal to zero and find the roots.

$\displaystyle (x+6)=0$ or $\displaystyle (x3)=0$

$\displaystyle {{x}_{1}}=-6$or $\displaystyle {{x}_{2}}=-3$

Completing the Square  Method

Since a=1 then we don’t divide the terms with the coefficient a.
Move the term c on the right of the equation
$\displaystyle {{x}^{2}}+9x=-18$
Complete the square on the left of the equation by adding 81/4 on both sides

$\displaystyle {{x}^{2}}+9x+\frac{{81}}{4}=-18+\frac{{81}}{4}$

$\displaystyle {{(x+\frac{9}{2})}^{2}}=\frac{{-72+81}}{4}$

$\displaystyle {{(x+\frac{9}{2})}^{2}}=\frac{9}{4}$

Take the square root on both sides to find x

$\displaystyle \sqrt{{{{{(x+\frac{9}{2})}}^{2}}}}=\sqrt{{\frac{9}{4}}}$

$\displaystyle (x+\frac{9}{2})=\pm \frac{3}{2}$

$\displaystyle {{x}_{1}}=\frac{3}{2}-\frac{9}{2}=-\frac{6}{2}=-3$

$\displaystyle {{x}_{2}}=-\frac{3}{2}-\frac{9}{2}=-\frac{{12}}{2}=-6$

So the solutions are $\displaystyle {{x}_{1}}=-3$ and $\displaystyle {{x}_{2}}=-6$

### Graphing method

The solutions of our equation are the values of x where the graph crosses the x-axis.

c) $\displaystyle 4{{x}^{2}}-36x+81=0$

The coefficients are a=4, b=-36, c=81

Finding the discriminant $\displaystyle D={{b}^{2}}-4ac$

$\displaystyle D={{(-36)}^{2}}-(4\cdot 4\cdot 81)$

$\displaystyle =1296-1296=0$

Since D=0 we have 1 solution

$\displaystyle x=-\frac{b}{{2a}}=-\frac{{36}}{8}=-4.5$

Graphing method

The solution of our equation is the value of x where the graph crosses the x-axis.

Completing the Square Method

Divide all the terms with a=4

$\displaystyle \frac{{4{{x}^{2}}-36x+81}}{4}=0$

$\displaystyle {{x}^{2}}-9x+\frac{{81}}{4}=0$

Move the term c/a on the right of the equation

$\displaystyle {{x}^{2}}-9x=-\frac{{81}}{4}$

Complete the square on the left of the equation by adding 81/4 on both sides

$\displaystyle {{x}^{2}}-9x+\frac{{81}}{4}=-\frac{{81}}{4}+\frac{{81}}{4}$

$\displaystyle {{x}^{2}}-9x+\frac{{81}}{4}=0$

$\displaystyle {{(x-\frac{9}{2})}^{2}}=0$

Take the square root on both sides to find x

$\displaystyle \sqrt{{{{{(x-\frac{9}{2})}}^{2}}}}=\sqrt{0}$

$\displaystyle (x-\frac{9}{2})=0$

$\displaystyle x=\frac{9}{2}=4.5$

Example 4: Write the quadratic equation in the vertex form.

A long method to convert the quadratic equation into the vertex form is by completing the square .

Steps to convert $\displaystyle a{{x}^{2}}+bx+c$ into the form $\displaystyle a{{(x+p)}^{2}}+q$

1. Determine the coefficients.

2. Find p and q by using the formula $\displaystyle p=\frac{b}{{2a}}$ and $\displaystyle q=c-\frac{{{{b}^{2}}}}{{4a}}$

3. Write it on the vertex form $\displaystyle y=a{{(x+p)}^{2}}+q$

a) $\displaystyle {{x}^{2}}-2x=0$

The coefficients a=1, b=-2, c=0

Finding p and q:

$\displaystyle p=\frac{b}{{2a}}$

$\displaystyle p=-\frac{2}{{2\cdot 1}}=-1$

$\displaystyle q=c-\frac{{{{b}^{2}}}}{{4a}}$

$\displaystyle q=0-\frac{{{{{(-2)}}^{2}}}}{{4\cdot 1}}=\frac{4}{4}=1$

Substitute to write it on the vertex form

$\displaystyle y=a{{(x+p)}^{2}}+q$

$\displaystyle y=1{{(x-1)}^{2}}+1$

$\displaystyle y={{(x-1)}^{2}}+1$

b) $\displaystyle {{x}^{2}}+11x+3=0$

The coefficients a=1, b=11, c=3

Finding p and q:

$\displaystyle p=\frac{b}{{2a}}$

$\displaystyle p=\frac{{11}}{{2\cdot 1}}=\frac{{11}}{2}$

$\displaystyle q=c-\frac{{{{b}^{2}}}}{{4a}}$

$\displaystyle q=3-\frac{{{{{(11)}}^{2}}}}{{4\cdot 1}}=3-\frac{{121}}{4}=$

$\displaystyle \frac{{12}}{4}-\frac{{121}}{4}=-\frac{{109}}{4}$

Substitute to write it on the vertex form

$\displaystyle y=a{{(x+p)}^{2}}+q$

$\displaystyle y=1{{(x+\frac{{11}}{2})}^{2}}-\frac{{109}}{4}$

$\displaystyle y={{(x+\frac{{11}}{2})}^{2}}-\frac{{109}}{4}$

c) $\displaystyle 3-4x-{{x}^{2}}=0$

Firstly multiply with minus one to turn it on a regular form.

$\displaystyle -1(3-4x-{{x}^{2}})=(-1\cdot 0)$

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

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

The coefficients a=1, b=4, c=-3

Finding p and q:

$\displaystyle p=\frac{b}{{2a}}$

$\displaystyle p=\frac{4}{{2\cdot 1}}=2$

$\displaystyle q=c-\frac{{{{b}^{2}}}}{{4a}}$

$\displaystyle q=-3-\frac{{{{{(4)}}^{2}}}}{{4\cdot 1}}=-3-\frac{{16}}{4}=$

$\displaystyle -\frac{{12}}{4}-\frac{{16}}{4}=-\frac{{28}}{4}=7$

Substitute to write it on the vertex form

$\displaystyle y=a{{(x+p)}^{2}}+q$

$\displaystyle y=1{{(x+2)}^{2}}+7$

$\displaystyle y={{(x+2)}^{2}}+7$

d) $\displaystyle 9{{x}^{2}}+12x-2=0$

The coefficients a=9, b=12, c=-2

Finding p and q:

$\displaystyle p=\frac{b}{{2a}}$

$\displaystyle p=\frac{{12}}{{2\cdot 9}}=\frac{{12}}{{18}}=\frac{2}{3}$

$\displaystyle q=c-\frac{{{{b}^{2}}}}{{4a}}$

$\displaystyle q=-2-\frac{{{{{(12)}}^{2}}}}{{4\cdot 9}}=$

$\displaystyle -2-\frac{{144}}{{36}}=$

$\displaystyle -2-4=-6$

Substitute to write it on the vertex form

$\displaystyle y=a{{(x+p)}^{2}}+q$

$\displaystyle y=9{{(x+\frac{2}{3})}^{2}}-6$