How to factor trinomials to the third power

Chapter 9, “Quadratics” from Beginning and Intermediate Algebra by Tyler Wallace is available under a Creative Commons Attribution 3.0 Unported license. © 2010.

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9.1

Quadratics - Solving with Radicals

Objective: Solve equations with radicals and check for extraneous solu- tions.

Here we look at equations that have roots in the problem. As you might expect, to clear a root we can raise both sides to an exponent. So to clear a square root we can rise both sides to the second power. To clear a cubed root we can raise both sides to a third power. There is one catch to solving a problem with roots in it, sometimes we end up with solutions that do not actually work in the equation. This will only happen if the index on the root is even, and it will not happen all the time. So for these problems it will be required that we check our answer in the original problem. If a value does not work it is called an extraneous solution and not included in the final solution.

When solving a radical problem with an even index: check answers!

Example 442.

7x + 2 √

= 4 Even index!Wewill have to check answers

( 7x+ 2 √

)2 =42 Square both sides, simplify exponents

7x + 2= 16 Solve

− 2 − 2 Subtract 2 fromboth sides 7x = 14 Divide both sides by 7

7 7

x = 2 Need to check answer in original problem

7(2)+ 2 √

= 4 Multiply

14+ 2 √

= 4 Add

16 √

= 4 Square root

4= 4 True! It works!

x = 2 Our Solution

Example 443.

x− 13 √

=− 4 Odd index,we don′t need to check answer ( x− 13 √

)3 =(− 4)3 Cube both sides, simplify exponents x− 1 =− 64 Solve

326

+ 1 +1 Add 1 to both sides

x =− 63 Our Solution

Example 444.

3x +64 √

=− 3 Even index!Wewill have to check answers ( 3x +64 √

) = (− 3)4 Rise both sides to fourth power 3x +6 = 81 Solve

− 6 − 6 Subtract 6 fromboth sides 3x = 75 Divide both sides by 3

3 3

x = 25 Need to check answer in original problem

3(25) + 64 √

=− 3 Multiply 75+64

√ =− 3 Add

814 √

=− 3 Take root 3 =− 3 False, extraneous solution

No Solution Our Solution

If the radical is not alone on one side of the equation we will have to solve for the radical before we raise it to an exponent

Example 445.

x + 4x +1 √

= 5 Even index!Wewill have to check solutions

− x −x Isolate radical by subtracting x fromboth sides 4x +1

√ = 5−x Square both sides

( 4x+ 1 √

)2 =(5− x)2 Evaluate exponents, recal (a− b)2 = a2− 2ab + b2 4x +1 = 25− 10x + x2 Re− order terms 4x +1 =x2− 10x + 25 Make equation equal zero

− 4x− 1 − 4x − 1 Subtract 4x and 1 fromboth sides 0 =x2− 14x + 24 Factor

0= (x− 12)(x− 2) Set each factor equal to zero x− 12=0 or x− 2= 0 Solve each equation + 12+ 12 +2+ 2

x = 12 or x = 2 Need to check answers in original problem

(12) + 4(12) + 1 √

= 5 Checkx =5first

327

12+ 48+1 √

= 5 Add

12+ 49 √

= 5 Take root

12+7= 5 Add

19= 5 False, extraneous root

(2) + 4(2) +1 √

= 5 Checkx =2

2+ 8 +1 √

= 5 Add

2+ 9 √

= 5 Take root

2 +3= 5 Add

5= 5 True! Itworks

x = 2 Our Solution

The above example illustrates that as we solve we could end up with an x2 term or a quadratic. In this case we remember to set the equation to zero and solve by factoring. We will have to check both solutions if the index in the problem was even. Sometimes both values work, sometimes only one, and sometimes neither works.

World View Note: The babylonians were the first known culture to solve quadratics in radicals - as early as 2000 BC!

If there is more than one square root in a problem we will clear the roots one at a time. This means we must first isolate one of them before we square both sides.

Example 446.

3x− 8 √

− x√ =0 Even index!Wewill have to check answers + x √

+ x √

Isolate first root by adding x √

to both sides

3x− 8 √

= x √

Square both sides

( 3x− 8 √

)2 = ( x √

)2 Evaluate exponents

3x− 8= x Solve − 3x − 3x Subtract 3x fromboth sides − 8=− 2x Divide both sides by− 2 − 2 − 2

4= x Need to check answer in original

3(4)− 8 √

− 4 √

=0 Multiply

12− 8 √

− 4 √

=0 Subtract

4 √

− 4 √

=0 Take roots

328

2− 2=0 Subtract 0=0 True! It works

x =4 Our Solution

When there is more than one square root in the problem, after isolating one root and squaring both sides we may still have a root remaining in the problem. In this case we will again isolate the term with the second root and square both sides. When isolating, we will isolate the term with the square root. This means the square root can be multiplied by a number after isolating.

Example 447.

2x + 1 √

− x√ = 1 Even index!Wewill have to check answers + x √

+ x √

Isolate first root by adding x √

to both sides

2x +1 √

= x √

+ 1 Square both sides

( 2x +1 √

)2 = ( x √

+1)2 Evaluate exponents, recall (a + b)2 = a2 +2ab+ b2

2x +1= x +2 x √

+ 1 Isolate the termwith the root

−x− 1−x − 1 Subtract x and 1 fromboth sides x =2 x

√ Square both sides

(x)2 =(2 x √

)2 Evaluate exponents

x2 =4x Make equation equal zero

− 4x− 4x Subtract x fromboth sides x2− 4x= 0 Factor

x(x− 4)= 0 Set each factor equal to zero x =0 or x− 4= 0 Solve

+4+ 4 Add 4 to both sides of second equation

x = 0 or x= 4 Need to check answers in original

2(0)+ 1 √

− (0) √

= 1 Checkx =0first

1 √

− 0 √

= 1 Take roots

1− 0= 1 Subtract 1= 1 True! Itworks

2(4)+ 1 √

− (4) √

= 1 Checkx =4

8+ 1 √

− 4 √

= 1 Add

9 √

− 4 √

= 1 Take roots

3− 2= 1 Subtract 1= 1 True! Itworks

329

x =0 or 4 Our Solution

Example 448.

3x + 9 √

− x + 4 √

=− 1 Even index!Wewill have to check answers + x +4 √

+ x + 4 √

Isolate the first root by adding x + 4 √

3x + 9 √

= x +4 √

− 1 Square both sides ( 3x + 9 √

)2 =( x + 4 √

− 1)2 Evaluate exponents 3x + 9=x + 4− 2 x +4

√ + 1 Combine like terms

3x +9 =x + 5− 2 x + 4 √

Isolate the termwith radical

−x− 5−x− 5 Subtractx and 5 fromboth sides 2x +4=− 2 x + 4

√ Square both sides

(2x + 4)2 =(− 2 x +4 √

)2 Evaluate exponents

4x2 + 16x + 16=4(x + 4) Distribute

4x2 + 16x + 16=4x + 16 Make equation equal zero

− 4x− 16− 4x− 16 Subtract 4x and 16 fromboth sides 4x2 + 12x = 0 Factor

4x(x + 3)= 0 Set each factor equal to zero

4x = 0 or x +3= 0 Solve

4 4 − 3− 3 x =0 or x =− 3 Check solutions in original

3(0)+ 9 √

− (0)+ 4 √

=− 1 Checkx= 0first 9

√ − 4 √

=− 1 Take roots 3− 2=− 1 Subtract

1=− 1 False, extraneous solution

3(− 3) +9 √

− (− 3)+ 4 √

=− 1 Checkx=− 3 − 9+9

√ − (− 3)+ 4 √

=− 1 Add 0

√ − 1 √

=− 1 Take roots 0− 1=− 1 Subtract − 1=− 1 True! Itworks

x =− 3 Our Solution

330

9.1 Practice - Solving with Radicals

Solve.

1) 2x + 3 √

− 3= 0

3) 6x− 5 √

− x =0

5) 3+ x= 6x + 13 √

7) 3− 3x √

− 1 =2x

9) 4x + 5 √

− x + 4 √

=2

11) 2x +4 √

− x + 3 √

=1

13) 2x +6 √

− x + 4 √

=1

15) 6− 2x √

− 2x +3 √

= 3

2) 5x +1 √

− 4= 0

4) x + 2 √

− x√ = 2

6) x− 1= 7−x √

8) 2x +2 √

=3 + 2x− 1 √

10) 3x +4 √

− x + 2 √

=2

12) 7x +2 √

− 3x + 6 √

=6

14) 4x− 3 √

− 3x +1 √

= 1

16) 2− 3x √

− 3x +7 √

= 3

331

9.2

Quadratics - Solving with Exponents

Objective: Solve equations with exponents using the odd root property and the even root property.

Another type of equation we can solve is one with exponents. As you might expect we can clear exponents by using roots. This is done with very few unex- pected results when the exponent is odd. We solve these problems very straight forward using the odd root property

OddRootProperty: if an = b, then a = b n √

whenn is odd

Example 449.

x5 = 32 Use odd root property

x5 5 √

= 325 √

Simplify roots

x =2 Our Solution

However, when the exponent is even we will have two results from taking an even root of both sides. One will be positive and one will be negative. This is because both 32 = 9 and (− 3)2 = 9. so when solving x2 = 9 we will have two solutions, one positive and one negative: x = 3 and− 3

EvenRoot Property: if an = b, thena = ± bn √

whenn is even

Example 450.

x4 = 16 Use even root property (± )

332

x4 4 √

=± 164 √

Simplify roots

x =± 2 Our Solution

World View Note: In 1545, French Mathematicain Gerolamo Cardano pub- lished his book The Great Art, or the Rules of Algebra which included the solu- tion of an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only three dimensions!

Example 451.

(2x + 4)2 = 36 Use even root property (± ) (2x +4)2

√

=± 36 √

Simplify roots

2x + 4=± 6 To avoid sign errors we need two equations 2x + 4= 6 or 2x + 4=− 6 One equation for+ , one equation for− − 4− 4 − 4 − 4 Subtract 4 fromboth sides

2x =2 or 2x =− 10 Divide both sides by 2 2 2 2 2

x =1 or x =− 5 Our Solutions

In the previous example we needed two equations to simplify because when we took the root, our solutions were two rational numbers, 6 and− 6. If the roots did not simplify to rational numbers we can keep the ± in the equation.

Example 452.

(6x− 9)2 = 45 Use even root property (± ) (6x− 9)2

√

=± 45 √

Simplify roots

6x− 9=± 3 5 √

Use one equation because root did not simplify to rational

+ 9 + 9 Add 9 to both sides

6x = 9± 3 5 √

Divide both sides by 6

6 6

x = 9± 3 5

√

6 Simplify, divide each termby 3

x = 3± 5

√

2 Our Solution

333

When solving with exponents, it is important to first isolate the part with the exponent before taking any roots.

Example 453.

(x +4)3− 6= 119 Isolate part with exponent + 6 +6

(x +4)3 = 125 Use odd root property

(x +4)33 √

= 125 √

Simplify roots

x + 4=5 Solve

− 4− 4 Subtract 4 fromboth sides x =1 Our Solution

Example 454.

(6x +1)2 +6 = 10 Isolate partwith exponent

− 6− 6 Subtract 6 fromboth sides (6x + 1)2 =4 Use even root property (± )

(6x +1)2 √

=± 4 √

Simplify roots

6x +1 =± 2 To avoid sign errors,weneed two equations 6x +1 =2 or 6x +1 =− 2 Solve each equation

− 1− 1 − 1 − 1 Subtract 1 fromboth sides 6x = 1 or 6x =− 3 Divide both sides by 6 6 6 6 6

x = 1

6 or x =− 1

2 Our Solution

When our exponents are a fraction we will need to first convert the fractional

exponent into a radical expression to solve. Recall that a m

n = ( an √

) m

. Once we

have done this we can clear the exponent using either the even ( ± ) or odd root property. Then we can clear the radical by raising both sides to an exponent (remember to check answers if the index is even).

Example 455.

(4x +1) 2

5 = 9 Rewrite as a radical expression

( 4x + 15 √

)2 = 9 Clear exponent first with even root property (± ) ( 4x +15 √

)2 √

=± 9 √

Simplify roots

334

4x +15 √

=± 3 Clear radical by raising both sides to 5th power ( 4x + 15 √

)5 = (± 3)5 Simplify exponents 4x +1=± 243 Solve, need 2 equations!

4x +1= 243 or 4x +1=− 243 − 1 − 1 − 1 − 1 Subtract 1 fromboth sides

4x = 242 or 4x=− 244 Divide both sides by 4 4 4 4 4

x = 121

2 ,− 61 Our Solution

Example 456.

(3x− 2) 3

4 = 64 Rewrite as radical expression

( 3x− 24 √

)3 = 64 Clear exponent firstwith odd root property

( 3x− 24 √

)33 √

= 643 √

Simplify roots

3x− 24 √

= 4 Even Index!Check answers.

( 3x− 24 √

)4 =44 Raise both sides to 4th power

3x− 2= 256 Solve + 2 + 2 Add 2 to both sides

3x = 258 Divide both sides by 3

3 3

x = 86 Need to check answer in radical form of problem

( 3(86)− 24 √

)3 = 64 Multiply

( 258− 24 √

)3 = 64 Subtract

( 2564 √

)3 = 64 Evaluate root

43 = 64 Evaluate exponent

64= 64 True! It works

x = 86 Our Solution

With rational exponents it is very helpful to convert to radical form to be able to see if we need a ± because we used the even root property, or to see if we need to check our answer because there was an even root in the problem. When checking we will usually want to check in the radical form as it will be easier to evaluate.

335

9.2 Practice - Solving with Exponents

Solve.

1) x2 = 75

3) x2 + 5= 13

5) 3x2 + 1= 73

7) (x +2)5 =− 243

9) (2x +5)3− 6= 21

11) (x− 1) 2

3 = 16

13) (2−x) 3

2 = 27

15) (2x− 3) 2

3 = 4

17) (x + 1

2 ) − 2

3 =4

19) (x− 1)− 5

2 = 32

21) (3x− 2) 4

5 = 16

23) (4x +2) 3

5 =− 8

2) x3 =− 8

4) 4x3− 2= 106

6) (x− 4)2 = 49

8) (5x +1)4 = 16

10) (2x +1)2 +3 = 21

12) (x− 1) 3

2 = 8

14) (2x +3) 4

3 = 16

16) (x +3) − 1

3 = 4

18) (x− 1)− 5

3 = 32

20) (x +3) 3

2 =− 8

22) (2x +3) 3

2 = 27

24) (3− 2x) 4

3 =− 81

336

9.3

Quadratics - Complete the Square

Objective: Solve quadratic equations by completing the square.

When solving quadratic equations in the past we have used factoring to solve for our variable. This is exactly what is done in the next example.

Example 457.

x2 + 5x +6= 0 Factor

(x + 3)(x +2)= 0 Set each factor equal to zero

x + 3=0 or x +2= 0 Solve each equation

− 3− 3 − 2− 2 x =− 3 or x =− 2 Our Solutions

However, the problem with factoring is all equations cannot be factored. Consider the following equation: x2 − 2x − 7 = 0. The equation cannot be factored, however there are two solutions to this equation, 1 + 2 2

√ and 1 − 2 2

√ . To find these two

solutions we will use a method known as completing the square. When completing the square we will change the quadratic into a perfect square which can easily be solved with the square root property. The next example reviews the square root property.

Example 458.

(x +5)2 = 18 Square root of both sides

(x + 5)2 √

=± 18 √

Simplify each radical

x +5 =± 3 2 √

Subtract 5 fromboth sides

− 5 − 5 x =− 5± 3 2

√ Our Solution

337

To complete the square, or make our problem into the form of the previous example, we will be searching for the third term in a trinomial. If a quadratic is of the form x2 + bx + c, and a perfect square, the third term, c, can be easily

found by the formula (

1

2 · b )2

. This is shown in the following examples, where we

find the number that completes the square and then factor the perfect square.

Example 459.

x2 +8x + c c =

(

1

2 · b )2

and our b =8

(

1

2 · 8 )2

= 42 = 16 The third term to complete the square is 16

x2 + 8x + 16 Our equation as aperfect square, factor

(x +4)2 Our Solution

Example 460.

x2− 7x + c c = (

1

2 · b )2

and our b =7

(

1

2 · 7 )2

=

(

7

2

)2

= 49

4 The third term to complete the square is

49

4

x2− 11x + 49 4

Our equation as aperfect square, factor

(

x− 7 2

)2

Our Solution

Example 461.

x2 + 5

3 x + c c =

(

1

2 · b )2

and our b =8

(

1

2 · 5 3

)2

=

(

5

6

)2

= 25

36 The third term to complete the square is

25

36

338

x2 + 5

3 x+

25

36 Our equation as aperfect square, factor

(

x + 5

6

)2

Our Solution

The process in the previous examples, combined with the even root property, is used to solve quadratic equations by completing the square. The following five steps describe the process used to complete the square, along with an example to demonstrate each step.

Problem 3x2 + 18x− 6=0

1. Separate constant term from variables + 6+6

3x2 + 18x = 6

2. Divide each term by a 3

3 x2 +

18

3 x =

6

3

x2 +6x =2

3. Find value to complete the square: (

1

2 · b )2 ( 1

2 · 6 )2

= 32 =9

4. Add to both sides of equation x2 +6x =2

+9 +9 x2 +6x + 9= 11

5. Factor (x +3)2 = 11

Solve by even root property

(x +3)2 √

=± 11 √

x+ 3=± 11 √

− 3 − 3 x=− 3± 11

√

World View Note: The Chinese in 200 BC were the first known culture group to use a method similar to completing the square, but their method was only used to calculate positive roots.

The advantage of this method is it can be used to solve any quadratic equation. The following examples show how completing the square can give us rational solu- tions, irrational solutions, and even complex solutions.

Example 462.

2x2 + 20x+ 48= 0 Separate constant term fromvaraibles

339

− 48− 48 Subtract 24 2x2 + 20x =− 48 Divide by a or 2

2 2 2

x2 + 10x =− 24 Find number to complete the square: (

1

2 · b )2

(

1

2 · 10 )2

= 52 = 25 Add 25 to both sides of the equation

x2 + 10x =− 24 + 25 + 25

x2 + 10x+ 25= 1 Factor

(x +5)2 = 1 Solvewith even root property

(x +5)2 √

=± 1 √

Simplify roots

x + 5=± 1 Subtract 5 fromboth sides − 5− 5

x =− 5± 1 Evaluate x =− 4 or − 6 Our Solution

Example 463.

x2− 3x− 2=0 Separate constant fromvariables + 2+2 Add 2 to both sides

x2− 3x =2 No a,find number to complete the square (

1

2 · b )2

(

1

2 · 3 )2

=

(

3

2

)2

= 9

4 Add

9

4 to both sides,

2

1

(

4

4

)

+ 9

4 =

8

4 +

9

4 =

17

4 Need commondenominator (4) on right

x2− 3x + 9 4

= 8

4 +

9

4 =

17

4 Factor

(

x− 3 2

)2

= 17

4 Solve using the even root property

(

x− 3 2

)2 √

=± 17 4

√

Simplify roots

x− 3 2

= ± 17 √

2 Add

3

2 to both sides,

340

+ 3

2 +

3

2 we already have a common denominator

x = 3± 17

√

2 Our Solution

Example 464.

3x2 =2x− 7 Separate the constant from the variables − 2x− 2x Subtract 2x fromboth sides 3x2− 2x =− 7 Divide each termby a or 3 3 3 3

x2− 2 3 x =− 7

3 Find the number to complete the square

(

1

2 · b )2

(

1

2 · 2 3

)2

=

(

1

3

)2

= 1

9 Add to both sides,

− 7 3

(

3

3

)

+ 1

9 =

− 21 3

+ 1

9 =

− 20 9

get commondenominator on right

x2− 2 3 x+

1

3 =− 20

9 Factor

(

x− 1 3

)2

=− 20 9

Solve using the even root property

(

x− 1 3

)2 √

=± − 20 9

√

Simplify roots

x− 1 3

= ± 2i 5

√

3 Add

1

3 to both sides,

+ 1

3 +

1

3 Already have commondenominator

x = 1± 2i 5

√

3 Our Solution

As several of the examples have shown, when solving by completing the square we will often need to use fractions and be comfortable finding common denominators and adding fractions together. Once we get comfortable solving by completing the square and using the five steps, any quadratic equation can be easily solved.

341

9.3 Practice - Complete the Square

Find the value that completes the square and then rewrite as a perfect square.

1) x2− 30x +__ 3) m2− 36m +__ 5) x2− 15x + __ 7) y2− y +__

2) a2− 24a +__ 4) x2− 34x +__ 6) r2− 1

9 r +__

8) p2− 17p +__ Solve each equation by completing the square.

9) x2− 16x + 55=0 11) v2− 8v + 45= 0 13) 6x2 + 12x + 63= 0

15) 5k2− 10k + 48=0 17) x2 + 10x− 57=4 19) n2− 16n + 67= 4 21) 2x2 + 4x + 38=− 6 23) 8b2 + 16b− 37=5 25) x2 =− 10x− 29 27) n2 =− 21+ 10n 29) 3k2 +9= 6k

31) 2x2 + 63=8x

33) p2− 8p =− 55 35) 7n2−n +7= 7n +6n2

37) 13b2 + 15b + 44=− 5+7b2 +3b 39) 5x2 + 5x =− 31− 5x 41) v2 + 5v + 28=0

43) 7x2− 6x + 40=0 45) k2− 7k + 50= 3 47) 5x2 + 8x− 40=8 49) m2 =− 15+9m 51) 8r2 + 10r =− 55 53) 5n2− 8n + 60=− 3n +6+ 4n2

55) − 2x2 + 3x− 5=− 4x2

10) n2− 8n− 12= 0 12) b2 + 2b+ 43= 0

14) 3x2− 6x + 47=0 16) 8a2 + 16a− 1= 0 18) p2− 16p− 52= 0 20) m2− 8m− 3 =6 22) 6r2 + 12r − 24=− 6 24) 6n2− 12n− 14=4 26) v2 = 14v + 36

28) a2− 56=− 10a 30) 5x2 =− 26+ 10x 32) 5n2 =− 10n + 15 34) x2 + 8x+ 15= 8

36) n2 +4n = 12

38) − 3r2 + 12r + 49=− 6r2

40) 8n2 + 16n = 64

42) b2 + 7b− 33=0 44) 4x2 + 4x + 25= 0

46) a2− 5a + 25= 3 48) 2p2− p + 56=− 8 50) n2−n =− 41 52) 3x2− 11x =− 18 54) 4b2− 15b + 56=3b2

56) 10v2− 15v = 27+ 4v2− 6v

342

9.4

Quadratics - Quadratic Formula

Objective: Solve quadratic equations by using the quadratic formula.

The general from of a quadratic is ax2 + bx + c = 0. We will now solve this for- mula for x by completing the square

Example 465.

ax2 + bc + c =0 Separate constant fromvariables

− c− c Subtract c fromboth sides ax2 + bx =− c Divide each termby a a a a

x2 + b

a x =

− c a

Find the number that completes the square (

1

2 · b a

)2

=

(

b

2a

)2

= b2

4a2 Add to both sides,

b2

4a2 − c

a

(

4a

4a

)

= b2

4a2 − 4ac

4a2 =

b2− 4ac 4a2

Get common denominator on right

x2 + b

a x +

b2

4a2 =

b2

4a2 − 4ac

4a2 =

b2− 4ac 4a2

Factor

(

x + b

2a

)2

= b2− 4ac

4a2 Solve using the even root property

(

x + b

2a

)2 √

=± b 2− 4ac 4a2

√

Simplify roots

x + b

2a =

± b2− 4ac √

2a Subtract

b

2a fromboth sides

x = − b± b2− 4ac

√

2a Our Solution

This solution is a very important one to us. As we solved a general equation by completing the square, we can use this formula to solve any quadratic equation. Once we identify what a, b, and c are in the quadratic, we can substitute those

343

values into x = − b± b2− 4ac

√

2a and we will get our two solutions. This formula is

known as the quadratic fromula

Quadratic Formula: if ax2 + bx + c = 0 then x = − b ± b2 − 4ac

√

2a

World View Note: Indian mathematician Brahmagupta gave the first explicit formula for solving quadratics in 628. However, at that time mathematics was not done with variables and symbols, so the formula he gave was, “To the absolute number multiplied by four times the square, add the square of the middle term; the square root of the same, less the middle term, being divided by twice the square is the value.” This would translate to

4ac+ b2 √

− b 2a

as the solution to the equation ax2 + bx = c.

We can use the quadratic formula to solve any quadratic, this is shown in the fol- lowing examples.

Example 466.

x2 +3x + 2=0 a= 1, b = 3, c =2, use quadratic formula

x = − 3± 32− 4(1)(2)

√

2(1) Evaluate exponent andmultiplication

x = − 3± 9− 8

√

2 Evaluate subtraction under root

x = − 3± 1

√

2 Evaluate root

x = − 3± 1

2 Evaluate± to get two answers

x= − 2 2