Worked Examples – Quadratic equation
Methods to solve a quadratic equation are:
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.
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
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$
Using the Quadratic Formula Method
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}$
b) $ \displaystyle {{x}^{2}}+9x+18=0$
Using the Quadratic Formula Method
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$
Using the Quadratic Formula Method
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$
