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8. Solving Quadratic Equations
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United States of AmericaFL
Grade 912

8.03 Solving quadratic equations using square roots

Lesson
Practice
Lesson

Concept summary

We can solve quadratic equations of the form a\left(x-h\right)^2=k, for positive values of a and k, using square roots. This method is particularly useful if k or \dfrac{k}{a} is a square number, as we will get rational solutions.

Remember that when we take the square root of a number k, we get two solutions \pm \sqrt{k}.

Worked examples

Example 1

Solve the following equations by using square roots:

a

x^2=9

Approach

This is a straight forward equation, the solutions are the square roots of 9.

Solution

x=\pm 3

b

\left(x-2\right)^2-100=0

Approach

In order to use square roots to solve, the squared expression must be isolated. In this example we want to isolate the term \left(x-2\right)^2.

Solution

\displaystyle \left(x-2\right)^2-100\displaystyle =\displaystyle 0
\displaystyle \left(x-2\right)^2\displaystyle =\displaystyle 100Add 100 to both sides
\displaystyle x-2\displaystyle =\displaystyle \pm 10Square root both sides

This leaves us with two equations x-2=10 and x-2=-10. Add 2 to solve both equations and we find that the solutions are x=-8,\, x=12.

c

\left(3x-8\right)^2=7

Approach

We can see that 7 is not a square number so our answer will involve radicals. Since the squared expression is already isolated, we are ready to solve by taking square roots.

Solution

\displaystyle \left(3x-8\right)^2\displaystyle =\displaystyle 7
\displaystyle 3x-8\displaystyle =\displaystyle \pm \sqrt{7}Square root
\displaystyle 3x\displaystyle =\displaystyle 8\pm \sqrt{7}Add 8
\displaystyle x\displaystyle =\displaystyle \frac{8\pm \sqrt{7}}{3}Divide by 3

The solutions are x=\dfrac{8}{3} - \dfrac{\sqrt{7}}{3},\, x=\dfrac{8}{3} +\dfrac {\sqrt{7}}{3}

Example 2

A square field has perpendicular lines drawn across it dividing it into 36 equal sized smaller squares. If the total area of the field is 200 square feet, determine the approximate side length of one of the smaller squares.

Approach

We know that there are 36 smaller squares in total on a larger square grid, so there must be 6 by 6 smaller squares on the grid. If we let the side of a smaller square be x, then the side of the larger square can be 6x. This gives us the quadratic equation \left(6x\right)^2=200. We can then solve this equation by taking square roots.

Solution

\displaystyle (6x)^2\displaystyle =\displaystyle 200
\displaystyle 6x\displaystyle =\displaystyle \pm \sqrt{200}Square root both sides
\displaystyle x\displaystyle =\displaystyle \frac{\pm \sqrt{200}}{6}Divide by 6

Simplifying the expression gives us the solutions x\approx\pm2.36 . We can exclude the negative solution as the length of the square must be positive. So the smaller square has a side length of 2.36 feet.

Reflection

In most real life applications, we will exclude the negative solution as it will not fit in the range of possible solutions for that context.

Outcomes

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.

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