Topics
8. Solving Quadratic Equations
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United States of AmericaFL
Grade 912

8.06 Solving quadratic equations using appropriate methods

Lesson
Practice
Lesson

Concept summary

We have several methods we can use to solve quadratic equations. To determine which method is the most suitable we need to look at the form of the quadratic equation. Some are easily solved by factoring for example, or by taking the square root.

If we are unable to solve the quadratic easily using one of these methods, the quadratic formula is often the best approach since it can be used to solve any quadratic equation once it's written in standard form. If we have access to technology, drawing the graph of the corresponding quadratic function can help us find exact solutions or approximate a solution if it is not an integer value.

Worked examples

Example 1

For the following quadratic equations, find the solution using an efficient method. Justify which method you used.

a

x^2-7x+12=0

Approach

The leading coefficient of x^2 is 1, so we can check if this can be easily factored. The prime factors of 12 are \pm 1,\, \pm 2,\, \pm3,\,\pm4,\, \pm6,\, \pm 12, and we want to find two factors that have a product of 12 and sum to -7. As the product is positive but the sum is negative we know both factors must be negative.

Solution

The two factors that have a product of 12 and a sum of -7 are -3 and -4. We can write the equation in factored form as \left(x-4\right)\left(x-3\right)=0, which gives us two solutions x=3,\, x=4.

Reflection

In general, if the coefficients are small, and especially if a=1, it is worth checking to see if we can easily factor the equation to solve.

b

x^2-6=21

Approach

Here we have b=0, and can easily isolate the x^2, which means we can solve this by using square roots.

Solution

\displaystyle x^2-6\displaystyle =\displaystyle 21
\displaystyle x^2\displaystyle =\displaystyle 27Add 6
\displaystyle x\displaystyle =\displaystyle \pm \sqrt{27}Square root

x=\pm \sqrt{27} giving us two solutions x \approx 5.196, x \approx -5.196

Reflection

In general, if we can easily rearrange the equation into the form \left(x-h\right)^2=k for some positive value of k then solving using square roots is a suitable method.

Example 2

A rectangular enclosure is to be constructed from 100 meters of wooden fencing. The area of the enclosure is given by A = 50 x - x^{2}, where x is the length of one side of the rectangle. If the area is 525 m^2, determine the side lengths.

Approach

We can set up and solve a quadratic equation, 50x-x^{2}=525. Since the values are large we will try solving this problem with the quadratic formula. The two solutions will be the side lengths of the enclosure.

Solution

Rearranging the equation into standard form we get x^2-50x+525=0.

We can solve this using the quadratic equation:

1\displaystyle x\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}Quadratic formula
2\displaystyle x\displaystyle =\displaystyle \dfrac{-\left(-50\right) \pm \sqrt{\left(-50\right)^2-4\left(1\right)\left(525\right)}}{2\left(1\right)}Substitute values for a,b,c
3\displaystyle x\displaystyle =\displaystyle \dfrac{50 \pm \sqrt{400}}{2}Simplify
4\displaystyle x\displaystyle =\displaystyle \dfrac{50 \pm 20}{2}Evaluate the square root

This leaves us with two values, x=\dfrac{50+20}{2} and x=\dfrac{50-20}{2}. Evaluating each expression for x we get x=35 and x=15.

Reflection

We can confirm our answer is correct by checking the conditions of the problem. We had 100 meters of fencing and 2\left(35+15\right)=100. We needed the area to be 525 m^2 and 35\left(15\right)=525 as required.

Since there are two rational solutions, the quadratic equation was also factorable:

x^2-50x+525=\left(x-35\right)\left(x-15\right), but these factors are not immediately obvious.

Outcomes

MA.912.AR.1.1

Identify and interpret parts of an equation or expression that represent a quantity in terms of a mathematical or real-world context, including viewing one or more of its parts as a single entity.

MA.912.AR.1.2

Rearrange equations or formulas to isolate a quantity of interest.

MA.912.AR.1.7

Rewrite a polynomial expression as a product of polynomials over the real number system.

MA.912.AR.2.1

Given a real-world context, write and solve one-variable multi-step linear equations.

MA.912.AR.3.1

Given a mathematical or real-world context, write and solve one-variable quadratic equations over the real number system.

MA.912.AR.3.8

Solve and graph mathematical and real-world problems that are modeled with quadratic functions. Interpret key features and determine constraints in terms of the context.

MA.912.NSO.1.4

Apply previous understanding of operations with rational numbers to add, subtract, multiply and divide numerical radicals.

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