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Big Ideas Math Algebra 2, 2014

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Big Ideas Math Algebra 2, 2014 View details

4. Solving Radical Equations and Inequalities

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Exercise 20 Page 266

Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.

x=1/2

Practice makes perfect

Solving a radical equation usually involves three main steps.

Isolate the radical on one side of the equation.
Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.
Solve the resulting equation. Remember to check your results!

Now we can analyze the given radical equation. sqrt(3-8x^2)=2x Notice that in this equation there is an isolated radical with index equal to 4 on the left-hand side. Then, we will raise each side of the equation to the power of 4.

sqrt(3-8x^2)=2x

LHS^4=RHS^4

(sqrt(3-8x^2))^4=(2x)^4

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Simplify

RootPowToNumber

sqrt(a^n)=a

3-8x^2=(2x)^4

(a b)^m=a^m b^m

3-8x^2=2^4x^4

Calculate power

3-8x^2=16x^4

LHS+8x^2=RHS+8x^2

3=16x^4+8x^2

LHS-3=RHS-3

0=16x^4+8x^2-3

Rearrange equation

16x^4+8x^2-3=0

We obtained a polynomial equation. To solve it, we will define another variable. If we let z=x^2, we can rewrite the last equation in terms of the z-variable. 16x^4+8x^2-3=0 ⇔ 16z^2+8z-3=0 Note that the above equation in terms of z is a quadratic equation. There are many ways to solve a quadratic equation. We will use the Quadratic Formula to solve the equation. Let's determine a, b, and c. 16z^2+8z-3=0 ⇔ 16z^2+ 8z+( - 3)=0 We see above that a= 16, b= 8, and c= - 3. Let's substitute these values into the Quadratic Formula to solve the equation.

z=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

z=- 8±sqrt(8^2-4( 16)( - 3))/2( 16)

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Simplify right-hand side

Calculate power

z=- 8±sqrt(64-4(16)(- 3))/2(16)

Multiply

z=- 8±sqrt(64+192)/32

Subtract terms

z=- 8±sqrt(256)/32

Calculate root

z=- 8± 16/32

Now we can calculate the first root using the positive sign and the second root using the negative sign.

z=- 8± 16/32
z=- 8+ 16/32	z=- 8- 16/32
z=1/4	z=-3/4

We found that the solutions for 16z^2+8z-3=0 are z= 14 and z=- 34. This means that x^2= 14 and x^2=- 34. Since x^2 is non-negative, the equation x^2=- 34 has no solutions. Let's solve x^2= 14.

x^2=1/4

sqrt(LHS)=sqrt(RHS)

x=±1/2

Checking the Solutions

Next, we will check the solutions by substituting 12 and - 12 for x into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution. Let's first check the 12.

sqrt(3-8x^2)=2x

x= 1/2

sqrt(3-8( 1/2)^2)? =2( 1/2)

▼

Simplify

Calculate power

sqrt(3-8(1/4))? =2(1/2)

MoveLeftFacToNum

a*b/c= a* b/c

sqrt(3-8/4)? =2/2

Calculate quotient

sqrt(3-2)? =1

Subtract terms

sqrt(1)? =1

Calculate quotient

1=1 ✓

Because our substitution produced a true statement, we know that our answer, x= 12, is correct. Let's check now - 12.

sqrt(3-8x^2)=2x

x= -1/2

sqrt(3-8( -1/2)^2)? =2( -1/2)

▼

Simplify

Calculate power

sqrt(3-8(1/4))? =2(-1/2)

MoveLeftFacToNum

a*b/c= a* b/c

sqrt(3-8/4)? =-2/2

Calculate quotient

sqrt(3-2)? =-1

Subtract terms

sqrt(1)? =-1

Calculate quotient

1≠ -1 *

Because our substitution produced a false statement, we know that our answer, x= 12, is not correct.

Subchapter links

Communicate Your Answer p.261

Monitoring Progress p.262-265

Exercises p.266-268