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Solving Quadratic Equations

Concept

Quadratic Equations

In standard form, quadratic equations take the form y=ax2+bx+c,y=ax^2+bx+c, and can be solved in various ways. In general, they are solved for the value(s) of xx that make the equation equal to 0.0. Thus, y=ax2+bx+cy=ax^2+bx+c becomes 0=ax2+bx+c. 0=ax^2+bx+c. Graphically, all points with a yy-coordinate of 00 are the xx-intercepts of the function — or the zeros of the parabola. That means, solving a quadratic equation leads to finding the zeros of the parabola. Since a parabola can have 0,1,0, 1, or 22 zeros, a quadratic equation can have 0,1,0, 1, or 22 solutions.

Method

Solving a Quadratic Equation Graphically

A quadratic equation can be solved graphically by expressing the equation as a function and identifying the solutions in the function's graph. Consider the following equation as an example. 2x2+3x5=5x+7 2x^2+3x-5=5x+7 This and other quadratic equations can be solved using this method.

1

If needed, rearrange the equation and simplify

If neither side of the equation equals 0,0, rearrange it so that all terms are on the same side. 2x2+3x5=5x+72x2+3x55x7=0 2x^2+3x-5=5x+7 \quad \Leftrightarrow \quad 2x^2+3x-5-5x-7=0 If there is more than one term of the same degree, simplify the equation by combining like terms. 2x2+3x55x7=02x22x12=0 2x^2+3x-5-5x-7=0 \quad \Leftrightarrow \quad 2x^2-2x-12=0

2

Plot the function

The side with the variables can now be seen as a function, f(x).f(x). Plot that function in a coordinate plane. For the equation 2x22x12=0, 2x^2-2x-12=0, the function is f(x)=2x22x12.f(x) = 2x^2-2x-12.


3

Identify any points with yy-coordinate 00

Now, find all the points on the graph that have the yy-coordinate 0.0.


4

Identify the xx-coordinates

The xx-coordinates of any identified points solve the equation, f(x)=0.f(x)=0.

In the example, the xx-coordinates are -2\text{-} 2 and 3.3. Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into f(x)f(x) and evaluate the expression. Verifying the solution x=-2x = \text{-} 2 is done by evaluating f(-2).f(\text{-} 2).

f(x)=2x22x12f(x) = 2x^2-2x-12
f(-2)=2(-2)22(-2)12f({\color{#0000FF}{\text{-} 2}}) = 2\cdot ({\color{#0000FF}{\text{-} 2}})^2-2\cdot({\color{#0000FF}{\text{-} 2}})-12
f(-2)=242(-2)12f(\text{-} 2) = 2\cdot 4-2\cdot(\text{-} 2)-12
f(-2)=24+2212f(\text{-} 2) = 2\cdot 4+2\cdot2-12
f(-2)=8+412f(\text{-} 2) = 8+4-12
f(-2)=0f(\text{-} 2) = 0
We find that the function value is 0.0. Therefore, x=-2x=\text{-} 2 is an exact solution. The solution x=3x = 3 can now be verified the same way.
f(x)=2x22x12f(x) = 2x^2-2x-12
f(3)=2322312f({\color{#0000FF}{3}}) = 2\cdot {\color{#0000FF}{3}}^2-2\cdot{\color{#0000FF}{3}}-12
f(3)=292312f(3) = 2\cdot 9-2\cdot3-12
f(3)=18612f(3) = 18-6-12
f(3)=0f(3) = 0
Since f(3)=0f(3)=0 then x=3x=3 is also a solution. Thus, the equation 2x2+3x5=5x+72x^2+3x-5=5x+7 has two solutions and they are x=-2andx=3. x=\text{-} 2 \quad \text{and} \quad x=3.
Concept

Simple Quadratic Equations

Simple quadratic equations take the form ax2+c=0 ax^2+c=0 and are solved using inverse operations. Once x2x^2 remains on the left-hand side, the equation can be written as x2=d,x^2=d, where d=-cad=\frac{\text{-} c}{a}. The value of dd gives the number of solutions the equation has.

d>0:2 real solutionsd=0:1 real solutiond<0:0 real solutions\begin{aligned} \mathbf{d>0} \quad &: \quad 2\text{ real solutions}\\ \mathbf{d=0} \quad &: \quad 1\text{ real solution}\\ \mathbf{d<0} \quad &: \quad 0\text{ real solutions} \end{aligned}
Method

Solving Quadratic Equations with Square Roots

Quadratic equations of the form ax2+c=0 ax^2+c=0 can be solved using inverse operations. To undo the exponent of 22 on the x2x^2 term, square roots must be used. Consider 5x2500=0.5x^2-500=0.

1

Isolate x2x^2
To begin, isolate x2x^2 using inverse operations to move cc and then aa to the opposite side of the equation. Here, that means adding 500500 and then dividing by 5.5.
5x2500=05x^2-500=0
5x2=5005x^2=500
x2=100x^2=100

2

Square-root both sides of the equation
Now that x2x^2 has been isolated, it is necessary to undo the exponent. Exponents and radicals of the same index undo each other. Thus, square roots undo powers of 22 and cube roots undo powers of 3,3, etc. To finish isolating x2,x^2, square-root both sides of the equation.
x2=100x^2=100
x=±100x=\pm \sqrt{100}
Using a calculator to determine 100\sqrt{100} gives one value — 10.10. This is because calculators give the principal root of a square root. It is necessary to remember that there are two values that, when raised to the power of 2,2, equal 100.100. 1010=100and(-10)(-10)=100, 10 \cdot 10 = 100 \quad \text{and} \quad (\text{-} 10)(\text{-} 10)=100, Thus, x=10x=10 and x=-10.x=\text{-} 10.
Method

Factoring a Quadratic Trinomial

When factoring an expression in the form ax2+bx+cax^2+bx+c it can be difficult to see its factors. 4x2+13x+3\begin{gathered} 4x^2+13x+3 \end{gathered} This expression can be factored by finding a pair of integers whose product is ac,a\cdot c, which here is 43,4\cdot 3, and whose sum is b,b, which in the example is 13.13.

1

List all pairs of integers whose product is equal to aca\cdot c

The first step is to find all possible pairs of integers that multiply to ac.a\cdot c. In this case, a=4a=4 and c=3.c=3. Thus, their product is 34=12. 3\cdot 4 = 12. To find all factor pairs, start with the pair where one factor is 1.1. The other factor must then be 12.12. Then continue with the pair where one of the factors is 2,2, and so forth. In this case, there are three pairs. 1and122and63and4\begin{aligned} &1 \quad \text{and} \quad 12\\ &2 \quad \text{and} \quad 6\\ &3 \quad \text{and} \quad 4 \end{aligned}

2

Find the pair whose sum is bb

If the given expression is factorable, one of the factor pairs will add to equal b.b. In this case, b=13.b=13. 1+12=132+6=83+4=7\begin{aligned} 1+12&=13 \\ 2+6&=8 \\ 3+4&=7 \end{aligned} Here, 11 and 1212 is the only factor pair that adds to 13.13.

3

Write bxbx as a sum

Now, use the factor pair to rewrite the xx-term of the original expression as a sum. Since the factor pair is 11 and 12,12, the middle term can be written as 13x=12x+x. 13x = 12x + x. This gives the following equivalent expression. 4x2+  13x +34x2+   12x+x+3\begin{aligned} 4x^2 &+ \ \ \quad {\color{#0000FF}{13x}}\ &&+ 3 \\ 4x^2 &+ \ \ \ {\color{#0000FF}{12x +x}} &&+ 3 \end{aligned} Notice that the expression hasn't been changed. Rather, it has been rewritten.

4

Factor out the GCF in the first two terms

Now, the expression has four terms, which can be grouped into the first two terms and the last two terms. Then, the GCF of each group can be factored out. Here, begin with the first group of terms, 4x24x^2 and 12x.12x. Notice how each can be written as a product of its factors. 4x2 + 12x4xx + 43x\begin{aligned} 4x^2 \ &+ \ 12x \\ 4 \cdot x \cdot x \ &+ \ 4 \cdot 3 \cdot x \end{aligned} The GCF is 4x.4x. Therefore, it's possible to factor out 4x.4x. 4x2+12x+x+34x(x+3)+x+3\begin{aligned} 4x^2 +12x \textcolor{lightgray}{+}\textcolor{lightgray}{x} \textcolor{lightgray}{+} \textcolor{lightgray}{3} \\ 4x(x+3)\textcolor{lightgray}{+} \textcolor{lightgray}{x}\textcolor{lightgray}{+}\textcolor{lightgray}{3} \end{aligned}

5

Factor out the GCF in the last two terms

Next, repeat the same process with the last two terms. In this case, xx and 33 do not have any common factors, but it's always possible to write expressions as a product of 11 and itself. 4x(x+3)+x+34x(x+3)+1(x+3)\begin{aligned} \textcolor{lightgray}{4x(x+3)} & \textcolor{lightgray}{+} x+3 \\ \textcolor{lightgray}{4x(x+3)} & \textcolor{lightgray}{+} 1(x+3) \end{aligned}

6

Factor out the common factor

If all previous steps have been performed correctly, there should now be two terms with a common factor, which can be seen as a repeated parentheses. Factoring (x+3)(x+3) out of both terms gives 4x(x+3)+1(x+3)(x+3)(4x+1).\begin{aligned} {\color{#0000FF}{4x}}(x+3) + & {\color{#0000FF}{1}} (x+3) \\ (x+3) ({\color{#0000FF}{4}}&{\color{#0000FF}{x}}+{\color{#0000FF}{1}}). \end{aligned} This means that 4x2+13x+34x^2+13x+3 can be written in factored form as (x+3)(4x+1).(x+3)(4x+1).

Exercise

Solve the following equation by factoring. x22x3=0 x^2-2x-3=0

Solution
We'll focus on the left-hand side of the equation before solving for x.x. Notice that 11 and -3\text{-} 3 are the factors that multiply to equal -3\text{-} 3 and add to -2.\text{-} 2.
x22x3=0x^2 - 2x-3 = 0
x2+x3x3=0x^2+x - 3x-3 = 0
x(x+1)3x3=0x(x+1) - 3x-3 = 0
x(x+1)3(x+1)=0x(x+1) - 3(x+1) = 0
(x+1)(x3)=0(x+1)(x-3) = 0
Notice that, in factored form, the numbers in the parentheses with xx are the numbers from the chosen factor pair. This is because the coefficient of x2x^2 is 1.1. We could have written the expression in factored form immediately after finding the factor pair. Lastly, we can use the zero product property to solve the equation. x+1=0x=-1x3=0x=-3\begin{aligned} x+1=0 \quad &\leftrightarrow \quad x=\text{-} 1\\ x-3=0 \quad &\leftrightarrow \quad x= \phantom{\text{-}} 3 \end{aligned} Thus, the solutions to the equation x22x3=0x^2-2x-3=0 are x=-1x=\text{-} 1 and x=3.x=3.
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