Elementary Algebra 2e · free preview

10.1 Solve Quadratic Equations Using the Square Root Property

Learning objectives: Solve quadratic equations of the form $a{x}^{2}=k$ using the Square Root Property; Solve quadratic equations of the form $a{(x-h)}^{2}=k$ using the Square Root Property

1. Learning objectives

Learning objectives

Learning objectives

  • Solve quadratic equations of the form ax2=ka{x}^{2}=k using the Square Root Property
  • Solve quadratic equations of the form a(x−h)2=ka{(x-h)}^{2}=k using the Square Root Property

Quadratic equations are equations of the form ax2+bx+c=0a{x}^{2}+bx+c=0, where a≠0a\ne{}0. They differ from linear equations by including a term with the variable raised to the second power. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will use three other methods to solve quadratic equations.

2. Solve Quadratic Equations of the Form *ax*${}^{2}$ = *k* Using the Square Root Property

Solve Quadratic Equations of the Form ax2{}^{2} = k Using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x2=9{x}^{2}=9.

x2=9Puttheequationinstandardform.x2−9=0Factortheleftside.(x−3)(x+3)=0UsetheZeroProductProperty.(x−3)=0, (x+3)=0Solveeachequation.x=3, x=−3Combinethetwosolutionsinto ± form.x=± 3(Thesolutionisread ‘x isequaltopositiveornegative3.′)\begin{array}{cccrlr} & & & {x}^{2} & = & 9 \\ \mathrm{Put the equation in standard form.} & & & {x}^{2}-9 & = & 0 \\ \mathrm{Factor the left side.} & & & (x-3)(x+3) & = & 0 \\ \mathrm{Use the Zero Product Property.} & & & (x-3)=0,\,(x+3) & = & 0 \\ \mathrm{Solve each equation.} & & & x=3,\,x & = & -3 \\ \mathrm{Combine the two solutions into}\,\pm\,\mathrm{form.} & & & x & = & \pm\,3 \\ \mathrm{(The solution is read}\,\mathrm{`}x\,\mathrm{is equal to positive or negative 3.')} & & \\ \end{array}

We can easily use factoring to find the solutions of similar equations, like x2=16{x}^{2}=16 and x2=25{x}^{2}=25, because 16 and 25 are perfect squares. But what happens when we have an equation like x2=7{x}^{2}=7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

These equations are all of the form x2=k{x}^{2}=k.We defined the square root of a number in this way:

If n2=m, then n isasquarerootof m.\mathrm{If}\,{n}^{2}=m,\,\mathrm{then}\,n\,\mathrm{is a square root of}\,m.

This leads to the Square Root Property.

If x2=k{x}^{2}=k, and k≥0k\ge{}0, then x=k or x=−kx=\sqrt{k}\,\mathrm{or}\,x=-\sqrt{k}.

Notice that the Square Root Property gives two solutions to an equation of the form x2=k{x}^{2}=k: the principal square root of kk and its opposite. We could also write the solution as x=± kx=\pm\,\sqrt{k}.

Now, we will solve the equation x2=9{x}^{2}=9 again, this time using the Square Root Property.

x2=9UsetheSquareRootProperty.x=± 9Simplifytheradical.x=± 3Rewritetoshowthetwosolutions.x=3,x=−3\begin{array}{cccrll} & & & {x}^{2} & = & 9 \\ \mathrm{Use the Square Root Property.} & & & x & = & \pm\,\sqrt{9} \\ \mathrm{Simplify the radical.} & & & x & = & \pm\,3 \\ \mathrm{Rewrite to show the two solutions.} & & & x=3,x & = & -3 \\ \end{array}

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x2=7{x}^{2}=7.

UsetheSquareRootProperty.x2=7x=± 7Rewritetoshowtwosolutions.x=7, x=−7Wecannotsimplify 7, soweleavetheanswerasaradical.\begin{array}{lccclccc} \begin{array}{l} \mathrm{Use the Square Root Property.} \\ \end{array} & & & \begin{array}{rcl} {x}^{2} & = & 7 \\ x & = & \pm\,\sqrt{7} \\ \end{array} \\ \mathrm{Rewrite to show two solutions.} & & & x=\sqrt{7},\,x=-\sqrt{7} \\ \mathrm{We cannot simplify}\,\sqrt{7},\,\mathrm{so we leave the answer as a radical.} & & \\ \end{array}

3. Example 10.1

Example 10.1

Example 10.1

Solve: x2=169{x}^{2}=169.

Solution

UsetheSquareRootProperty.Simplifytheradical. x2=169x=± 169x=± 13Rewritetoshowtwosolutions. x=13, x=−13\begin{array}{lccclccccc} \begin{array}{l} \mathrm{Use the Square Root Property.} \\ \mathrm{Simplify the radical.} \\ \end{array} & & & \,\begin{array}{rll} {x}^{2} & = & 169 \\ x & = & \pm\,\sqrt{169} \\ x & = & \pm\,13 \\ \end{array} \\ \mathrm{Rewrite to show two solutions.} & & & \,x=13,\,x=-13 \\ \end{array}

4. Example 10.2

Example 10.2

Example 10.2

How to Solve a Quadratic Equation of the Form ax2=k{ax}^{2}=k Using the Square Root Property

Solve: x2−48=0{x}^{2}-48=0.

Solve a quadratic equation using the Square Root Property.

  1. Step 1. Isolate the quadratic term and make its coefficient one.
  2. Step 2. Use Square Root Property.
  3. Step 3. Simplify the radical.
  4. Step 4. Check the solutions.

To use the Square Root Property, the coefficient of the variable term must equal 1. In the next example, we must divide both sides of the equation by 5 before using the Square Root Property.

5. Example 10.3

Example 10.3

Example 10.3

Solve: 5m2=805{m}^{2}=80.

Solution

The Square Root Property started by stating, ‘If x2=k{x}^{2}=k, and k≥0k\ge{}0’. What will happen if k<0k<0? This will be the case in the next example.

6. Example 10.4

Example 10.4

Example 10.4

Solve: q2+24=0{q}^{2}+24=0.

Solution

q2+24=0{q}^{2}+24=0
Isolate the quadratic term. q2=−24\,{q}^{2}=-24
Use the Square Root Property. q=± −24\,q=\pm\,\sqrt{-24}
The −24\sqrt{-24} is not a real number.There is no real solution.

Remember, we first isolate the quadratic term and then make the coefficient equal to one.

7. Example 10.5

Example 10.5

Example 10.5

Solve: 23u2+5=17\frac{2}{3}{u}^{2}+5=17.

Solution

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

8. Example 10.6

Example 10.6

Example 10.6

Solve: 2c2−4=452{c}^{2}-4=45.

Solution

Isolate the quadratic term.Divide by 2 to make the coefficient 1.Simplify.Use the Square Root Property.Simplify the radical.Rationalize the denominator.Simplify. 2c2−4=45 2c2=49 2c22=492 c2=492 c=± 492 c=± 492 c=± 49⋅22⋅2 c=± 722\begin{array}{rll} \,2{c}^{2}-4 & = & 45 \\ \,2{c}^{2} & = & 49 \\ \,\frac{2{c}^{2}}{2} & = & \frac{49}{2} \\ \,{c}^{2} & = & \frac{49}{2} \\ \,c & = & \pm\,\sqrt{\frac{49}{2}} \\ \,c & = & \pm\,\frac{\sqrt{49}}{\sqrt{2}} \\ \,c & = & \pm\,\frac{\sqrt{49}\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} \\ \,c & = & \pm\,\frac{7\sqrt{2}}{2} \\ \end{array}
Rewrite to show two solutions.c=722, c=−722c=\frac{7\sqrt{2}}{2},\,c=-\frac{7\sqrt{2}}{2}
Check. We leave the check for you.

9. Solve Quadratic Equations of the Form *a*(*x* − *h*)${}^{2}$ = *k* Using the Square Root Property

Solve Quadratic Equations of the Form a(x − h)2{}^{2} = k Using the Square Root Property

We can use the Square Root Property to solve an equation like (x−3)2=16{(x-3)}^{2}=16, too. We will treat the whole binomial, (x−3)(x-3), as the quadratic term.

10. Example 10.7

Example 10.7

Example 10.7

Solve: (x−3)2=16{(x-3)}^{2}=16.

Solution

11. Example 10.8

Example 10.8

Example 10.8

Solve: (y−7)2=12{(y-7)}^{2}=12.

Solution

Remember, when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

12. Example 10.9

Example 10.9

Example 10.9

Solve: (x−12)2=54.{(x-\frac{1}{2})}^{2}=\frac{5}{4}.

Solution

Use the Square Root Property.Rewrite the radical as a fraction of square roots.Simplify the radical.Solve for x.(x−12)2=54x−12=± 54x−12=± 54x−12=± 52x=12±52\begin{array}{rll} {(x-\frac{1}{2})}^{2} & = & \frac{5}{4} \\ x-\frac{1}{2} & = & \pm\,\sqrt{\frac{5}{4}} \\ x-\frac{1}{2} & = & \pm\,\frac{\sqrt{5}}{\sqrt{4}} \\ x-\frac{1}{2} & = & \pm\,\frac{\sqrt{5}}{2} \\ x & = & \frac{1}{2}\pm\frac{\sqrt{5}}{2} \\ \end{array}
Rewrite to show two solutions.x=12+52, x=12−52x=\frac{1}{2}+\frac{\sqrt{5}}{2},\,x=\frac{1}{2}-\frac{\sqrt{5}}{2}
Check. We leave the check for you.

We will start the solution to the next example by isolating the binomial.

13. Example 10.10

Example 10.10

Example 10.10

Solve: (x−2)2+3=30{(x-2)}^{2}+3=30.

Solution

 (x−2)2+3=30\,\begin{array}{rll} {(x-2)}^{2}+3 & = & 30 \\ \end{array}
Isolate the binomial term. (x−2)2=27\,\begin{array}{rll} {(x-2)}^{2} & = & 27 \\ \end{array}
Use the Square Root Property. x−2=± 27\,\begin{array}{rll} x-2 & = & \pm\,\sqrt{27} \\ \end{array}
Simplify the radical. x−2=± 33\,\begin{array}{rll} x-2 & = & \pm\,3\sqrt{3} \\ \end{array}
Solve for x. x=2±33\,\begin{array}{rll} x & = & 2\pm{}3\sqrt{3} \\ \end{array}
Rewrite to show two solutions. x=2+33, x=2−33\,x=2+3\sqrt{3},\,x=2-3\sqrt{3}
Check. We leave the check for you.

14. Example 10.11

Example 10.11

Example 10.11

Solve: (3v−7)2=−12{(3v-7)}^{2}=-12.

Solution

Use the Square Root Property.(3v−7)2=−123v−7=± −12\begin{array}{rll} {(3v-7)}^{2} & = & -12 \\ 3v-7 & = & \pm\,\sqrt{-12} \\ \end{array}
The −12\sqrt{-12} is not a real number.There is no real solution.

The left sides of the equations in the next two examples do not seem to be of the form a(x−h)2a{(x-h)}^{2}. But they are perfect square trinomials, so we will factor to put them in the form we need.

15. Example 10.12

Example 10.12

Example 10.12

Solve: p2−10p+25=18{p}^{2}-10p+25=18.

Solution

The left side of the equation is a perfect square trinomial. We will factor it first.

Factor the perfect square trinomial.Use the Square Root Property.Simplify the radical.Solve for p.p2−10p+25=18(p−5)2=18p−5=± 18p−5=± 32p=5±32\begin{array}{rll} {p}^{2}-10p+25 & = & 18 \\ {(p-5)}^{2} & = & 18 \\ p-5 & = & \pm\,\sqrt{18} \\ p-5 & = & \pm\,3\sqrt{2} \\ p & = & 5\pm{}3\sqrt{2} \\ \end{array}
Rewrite to show two solutions.p=5+32, p=5−32p=5+3\sqrt{2,}\,p=5-3\sqrt{2}
Check. We leave the check for you.

16. Example 10.13

Example 10.13

Example 10.13

Solve: 4n2+4n+1=164{n}^{2}+4n+1=16.

Solution

Again, we notice the left side of the equation is a perfect square trinomial. We will factor it first.

17. Section 10.1 Exercises

Section 10.1 Exercises

Section 10.1 Exercises

Solve Quadratic Equations of the form ax2=ka{x}^{2}=k Using the Square Root Property

In the following exercises, solve the following quadratic equations.

Solve Quadratic Equations of the Form a(x−h)2=ka{(x-h)}^{2}=k Using the Square Root Property

In the following exercises, solve the following quadratic equations.

Mixed Practice

In the following exercises, solve using the Square Root Property.

Everyday Math

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently: Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help: This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no-I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

18. Practice 1

  1. Solve: x2=81{x}^{2}=81.
  1. Solve: y2=121{y}^{2}=121.
  1. Solve: x2−50=0{x}^{2}-50=0.
  1. Solve: y2−27=0{y}^{2}-27=0.

19. Practice 2

  1. Solve: 2x2=982{x}^{2}=98.
  1. Solve: 3z2=1083{z}^{2}=108.
  1. Solve: c2+12=0{c}^{2}+12=0.
  1. Solve: d2+81=0{d}^{2}+81=0.

20. Practice 3

  1. Solve: 12x2+4=24\frac{1}{2}{x}^{2}+4=24.
  1. Solve: 34y2−3=18\frac{3}{4}{y}^{2}-3=18.
  1. Solve: 5r2−2=345{r}^{2}-2=34.
  1. Solve: 3t2+6=703{t}^{2}+6=70.

21. Practice 4

  1. Solve: (q+5)2=1{(q+5)}^{2}=1.
  1. Solve: (r−3)2=25{(r-3)}^{2}=25.
  1. Solve: (a−3)2=18{(a-3)}^{2}=18.
  1. Solve: (b+2)2=40{(b+2)}^{2}=40.

22. Practice 5

  1. Solve: (x−13)2=59.{(x-\frac{1}{3})}^{2}=\frac{5}{9}.
  1. Solve: (y−34)2=716.{(y-\frac{3}{4})}^{2}=\frac{7}{16}.
  1. Solve: (a−5)2+4=24{(a-5)}^{2}+4=24.
  1. Solve: (b−3)2−8=24{(b-3)}^{2}-8=24.

23. Practice 6

  1. Solve: (3r+4)2=−8{(3r+4)}^{2}=-8.
  1. Solve: (2t−8)2=−10{(2t-8)}^{2}=-10.
  1. Solve: x2−6x+9=12{x}^{2}-6x+9=12.
  1. Solve: y2+12y+36=32{y}^{2}+12y+36=32.

24. Practice 7

  1. Solve: 9m2−12m+4=259{m}^{2}-12m+4=25.
  1. Solve: 16n2+40n+25=416{n}^{2}+40n+25=4.
  1. a2=49{a}^{2}=49
  1. r2−24=0{r}^{2}-24=0

25. Practice 8

  1. u2−300=0{u}^{2}-300=0
  1. 4m2=364{m}^{2}=36
  1. x2+20=0{x}^{2}+20=0
  1. 25a2+3=11\frac{2}{5}{a}^{2}+3=11

26. Practice 9

  1. 7p2+10=267{p}^{2}+10=26
  1. (x+2)2=9{(x+2)}^{2}=9
  1. (u−6)2=64{(u-6)}^{2}=64
  1. (m−6)2=20{(m-6)}^{2}=20

27. Practice 10

  1. (r−12)2=34{(r-\frac{1}{2})}^{2}=\frac{3}{4}
  1. (a−7)2+5=55{(a-7)}^{2}+5=55
  1. (5c+1)2=−27{(5c+1)}^{2}=-27
  1. m2−4m+4=8{m}^{2}-4m+4=8

28. Practice 11

  1. 25x2−30x+9=3625{x}^{2}-30x+9=36
  1. 2r2=322{r}^{2}=32
  1. (a−4)2=28{(a-4)}^{2}=28
  1. 9w2−24w+16=19{w}^{2}-24w+16=1

29. Practice 12

  1. a2−18=0{a}^{2}-18=0
  1. (p−13)2=79{(p-\frac{1}{3})}^{2}=\frac{7}{9}
  1. m2+12=0{m}^{2}+12=0
  1. u2−14u+49=72{u}^{2}-14u+49=72

30. Practice 13

  1. (m−4)2+3=15{(m-4)}^{2}+3=15
  1. (x+5)2=4{(x+5)}^{2}=4
  1. 6c2+4=296{c}^{2}+4=29
  1. (x−6)2+7=3{(x-6)}^{2}+7=3

31. Practice 14

  1. Paola has enough mulch to cover 48 square feet. She wants to use it to make three square vegetable gardens of equal sizes. Solve the equation 3s2=483{s}^{2}=48 to find ss, the length of each garden side.
  1. Explain why the equation x2+12=8{x}^{2}+12=8 has no solution.

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