Algebra 1 - Quadratic Equations

Introduction

• Quadratic Equation in One Variable is a mathematical sentence of degree 2 that can be written in the standard form $a{x}^2+bx+c=0, where a,b,c\in \mathrm{ℝ} and a\ne 0$.
• The members of the equation are $a{x}^2=$quadratic term, $bx=$linear term, and $c=$constant.
• The equation $3x^2+5x-2=0$ is a complete quadratic equation with $a=3, b=5, c=-2$.
• The equation $2x\left(x-9\right)=0$ is also a quadratic form but it is not simplified in standard form.
• In the case of $\left(4x-3\right)\left(2x+1\right)=0$, the quadratic equation is expressed in factored form. You need to use the FOIL method to write this in standard form.

Solved Examples (Extraction of Square Roots)

• Finding the roots or solutions of a quadratic equation implies finding all permissible values of $x$ in $a{x}^2+bx+c=0$.
• Quadratic equations in the form $x^2=b$ can be solved by applying the following concepts: 
• If $b>0$, then $x^2=b$ has two real solutions or roots, $x=\pm \sqrt{b}$.
• If $b=0$, then $x^2=b$ has one real solution or root, $x=0$.
• If $b<0$, then $x^2=b$ has no real solutions or roots.
• The principle in these concepts is facilitated using the Square Root Property: If $x^2=b and b\ge 0$, then $x=\pm \sqrt{b}$.

 

Example 1. What are the roots of $3{\left(x-5\right)}^2=135?$

Solution:

$3{\left(x-5\right)}^2-135=0\to 3{\left(x-5\right)}^2=135$

$\frac{3{\left(x-5\right)}^2}{3}=\frac{135}{3}\to {\left(x-5\right)}^2=45$

$\sqrt{{\left(x-5\right)}^2}=\sqrt{45}$

$x-5=\pm 3\sqrt{5}$

$x=3\sqrt{5}+5 or x=-3\sqrt{5}+5$

Solved Examples (Factoring and the Po-Shen Loh Method)

Solving quadratic equations using Factoring:

• Apply the different ways of solving a quadratic trinomial of $a{x}^2+bx+c=0 where a=1 or a{x}^2+bx+c=0 where a\ne 1$. 

 

Example 1. Solve for the roots of $a^2-11a-26=0$.

Solution:

Using factoring, we write the trinomial on the left side as $\left(a-13\right)\left(a+2\right)=0$.

Using the Zero Product Property, we have:

$\left(a-13\right)\left(a+2\right)=0$

$a=-2 or a=13$

 

Example 2. Find the roots of $3x^2+11x-20=0$.

Solution:

In cases of $a{x}^2+bx+c=0 where a\ne 1$, we use the ac test method.

To illustrate, we have:

$3x^2+11x-20=0$

$a=3, b=11, c=-20$

$ac=\left(3\right)\left(-20\right)=-60$

Think of factors of $-60$ whose sum is $11$. The two factors are $15 and -4$.

Using the factors of $-60$, we express the linear term of the equation as $15x +\left(-4x\right)$.

$3x^2+15x-4x-20=0$

Take the common factors as $3x\left(x+5\right)-4\left(x+5\right)=0$.

$3x\left(x+5\right)-4\left(x+5\right)=0\to \left(x+5\right)\left(3x-4\right)=0$

Thus, the roots are $x=-5 or x=\frac{4}{3}$

 

Solving quadratic equations using Po-Shen Loh Method:

Example 1. What are the roots of $5x^2+18x-8=0$?

Solution: 

For cases like $a{x}^2+bx+c=0 where a\ne 1$, we divide each term of both sides of the equation by $a$.

$\frac{5x^2}{5}+\frac{18x}{5}-\frac{8}{5}=\frac{0}{5}$

$x^2+\frac{18x}{5}-\frac{8}{5}=0$

The two numbers whose sum is $-\frac{18}{5}$ are $-\frac{9}{5} and -\frac{9}{5}$. 

The two roots of the equation is written as $x_1=-\frac{9}{5}-u and x_2=-\frac{9}{5}+u$

Using the idea of product of $-\frac{8}{5}$, we have $\left(-\frac{9}{5}-u\right)\left(-\frac{9}{5}+u\right)=-\frac{8}{5}$

Solving for $u$ yields $\pm \frac{11}{5}$. Use either $u=-\frac{11}{5} or u=\frac{11}{5}$

$x_1=-\frac{9}{5}-\frac{11}{5}=-4$

$x_2=-\frac{9}{5}+\frac{11}{5}=\frac{2}{5}$

Solved Examples (Completing the Square and Quadratic Formula)

Example 1. Solve the equation $x^2-4x-96=0$ using completing the square or quadratic formula.

Solution 1. Using Completing the Square

Rewrite the equation as $x^2-4x=96$.

Divide the coefficient of the linear term by 2 and square the result. The number obtained must be added to both sides of the equation. 

$\frac{4}{2}=2\to {\left(2\right)}^2=4$

$x^2-4x+4=96+4$

${\left(x-2\right)}^2=100$

$\sqrt{{\left(x-2\right)}^2}=\sqrt{100}$

$x-2=\pm 10$

Thus, the solutions are $x=12 or x=-8$

 

Solution 2. Using the Quadratic Formula

Use the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. By direct substitution of $a=1, b=-4, and c=-96$, we have:

$x=\frac{4\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(-96\right)}}{2}\to x=\frac{4\pm 20}{2}$

The roots are $x=-8 or x=12$

Cheat Sheet

• Use the discriminant $D=b^2-4ac$ to determine the nature of the roots of a quadratic equation.
• If $b^2-4ac>0$ and not a perfect square, then there are two real, irrational roots.
• If $b^2-4ac>0$ and a perfect square, then there are two unequal real rational roots.
• If $b^2-4ac=0$, then there is only one real and rational root.
• if $b^2-4ac<0$, there are no real roots.
• To formulate a quadratic equation with the given roots $x_1 and x_2$, use the model $x-\left(x_1+x_2\right)x+\left(x_1\right)\left(x_2\right)=0$.

Blunder Areas

• Completing the Square, Quadratic Formula, and Po-Shen Loh Method are applicable for quadratic equations having trinomials either factorable or non-factorable.
• Always be mindful of the signs when solving for the roots of the quadratic equation. 
• When $x^2+3x+2=0$ is written as $\left(x+2\right)\left(x+1\right)=0$, it is a common mistake to write the roots as $x=2 or x=1$. The roots must be $x=-2 or x=-1$.
• If you are asked to write the solution or roots of a quadratic equation, then use the term "or" instead of "and." For instance, we have: $x=-2 or x=-1$. It is incorrect to write it as $x=-2 and x=-1$.
• If you are asked to write the solution set of a quadratic equation, then write it as $\left\{x_1, x_2\right\}$.
• Be careful in dealing with signs of the coefficients of $a{x}^2+bx+c=0$ when using the quadratic formula.