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Section 9.1 Solving with Radicals

Objective: Solve equations with radicals and check for extraneous solutions.

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 9.1.1.

7x+2=4 Even index! We will have to check answers. (7x+2)2=42 Square both sides, simplify exponents 7x+2=16 Solve โˆ’2 โˆ’2 Subtract 2 from both sides 7x7=147 Divide both sides by 7x=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 9.1.2.

xโˆ’13=โˆ’4 Odd index, we don't need to check answer. (xโˆ’13)3=(โˆ’4)3 Cube both sides, simplify exponents xโˆ’1=โˆ’64 Solve +1+1 Add 1 to both sides x=โˆ’63 Our Solution 

Example 9.1.3.

3x+64=โˆ’3 Even index! We will have to check answers (3x+64)4=(โˆ’3)4 Rise both sides to fourth power 3x+6=81 Solve โˆ’6 โˆ’6 Subtract 6 from both sides 3x3=753 Divide both sides by 3x=25 Need to check answer in original problem 3(25)+64=โˆ’3 Multiply 75+64=โˆ’3 Add 814=โˆ’3 Take root 3=โˆ’3 False, extraneous solution No Solution Our Solution 
In this example one can also notice at the very beginning that in 3x+64=โˆ’3 the left hand side is always nonnegative, no mater what the value of x is, hence it will never be equal to the righ hand side โˆ’3.

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 9.1.4.

x+4x+1=5 Even index! We will have to check solutions โˆ’xโˆ’x Isolate radical by subtracting x from both sides 4x+1=5โˆ’x Squarebothsides (4x+1)2=(5โˆ’x)2 Evaluate exponents, recall (aโˆ’b)2=a2โˆ’2ab+b24x+1=25โˆ’10x+x2 Reorder terms 4x+1=x2โˆ’10x+25 Make equation equal zero โˆ’4xโˆ’1โˆ’4xโˆ’1 Subtract 4x and 1 from both sides 0=x2โˆ’14x+24 Factor 0=(xโˆ’12)(xโˆ’2) Set each factor equal to zero xโˆ’12=0 or xโˆ’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 Check x=5 first 12+48+1=5 Add 12+49=5 Take root 12+7=5 Add 19=5 False, extraneous root  (2)+4(2)+1=5 Check x=22+8+1=5 Add 2+9=5 Take root 2+3=5 Add 5=5 True! It works  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 9.1.5.

3xโˆ’8โˆ’x=0 Even index! We will have to check answers +x +x Isolate first root by adding x to both sides 3xโˆ’8=x Square both sides (3xโˆ’8)2=(x)2 Evaluate exponents 3xโˆ’8=x Solve โˆ’3xโˆ’3x Subtract 3x from both sides โˆ’8=โˆ’2x Divide both sides by โˆ’2โˆ’8โˆ’2=โˆ’2xโˆ’2 4=x Need to check answer in original 3(4)โˆ’8โˆ’4=0 Multiply 12โˆ’8โˆ’4=0 Subtract 4โˆ’4=0 Take roots 2โˆ’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 9.1.6.

2x+1โˆ’x=1 Even index! We will 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+b22x+1=x+2x+1 Isolate the term with the root โˆ’xโˆ’1โˆ’xโˆ’1 Subtract x and 1 from both sides x=2x Square both sides (x)2=(2x)2 Evaluate exponents x2=4x Make equation equal zero โˆ’4x โˆ’4x Subtract x from both sides x2โˆ’4x=0 Factor x(xโˆ’4)=0 Set each factor equal to zero x=0 or xโˆ’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)+1โˆ’(0)=1 Check x=0 first 1โˆ’0=1 Take roots 1โˆ’0=1 Subtract 1=1 True! It works  2(4)+1โˆ’(4)=1 Check x=48+1โˆ’4=1 Add 9โˆ’4=1Take roots3โˆ’2=1 Subtract1=1True! It works x=0 or 4 Our Solution 

Example 9.1.7.

3x+9โˆ’x+4=โˆ’1 Even index! We will have to check answers +x+4 +x+4 Isolate the first root by adding x+43x+9=x+4โˆ’1 Square both sides (3x+9)2=(x+4โˆ’1)2 Evaluate exponents 3x+9=x+4โˆ’2x+4+1 Combine like terms 3x+9=x+5โˆ’2x+4 Isolate the term with radical โˆ’xโˆ’5 โˆ’xโˆ’5 Subtract x and 5 from both sides 2x+4=โˆ’2x+4 Square both sides (2x+4)2=(โˆ’2x+4)2 Evaluate exponents 4x2+16x+16=4(x+4) Distribute 4x2+16x+16=4x+16 Make equation equal zero โˆ’4xโˆ’16 โˆ’4xโˆ’16 Subtract 4x and 16 from both sides 4x2+12x=0 Factor 4x(x+3)=0 Set each factor equal to zero 4x=0 or x+3=0 Solve รท4 รท4โˆ’3 โˆ’3 x=0 or x=โˆ’3 Check solutions in original  3(0)+9โˆ’(0)+4=โˆ’1 Check x=0 first 9โˆ’4=โˆ’1 Take roots 3โˆ’2=โˆ’1 Subtract 1=โˆ’1 False, extraneous solution  3(โˆ’3)+9โˆ’(โˆ’3)+4=โˆ’1 Check x=โˆ’3โˆ’9+9โˆ’โˆ’3+4=โˆ’1 Add 0โˆ’1=โˆ’1 Take roots 0โˆ’1=โˆ’1 Subtract โˆ’1=โˆ’1 True! It works  x=โˆ’3 Our Solution 

Exercises Exercises - Solving with Radicals

Exercise Group.

Solve.