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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
10. Roots and Zeros
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Exercise 23 Page 77

Apply the difference of two squares.

Roots: x=- 52, x= 52, x=- 52i, x= 52i
Number and Type of Roots: two real roots, two imaginary roots

Practice makes perfect

We want to solve the given equation. To do so, we will apply a factoring technique called the difference of two squares. a^2-b^2 ⇔ (a+b)(a-b) To do so, we will start by writing both terms as perfect squares.

16x^4-625=0
WritePow

Write as a power

4^2x^4-25^2=0
SplitIntoFactors

Split into factors

4^2x^(2*2)-25^2=0
ProdInExponent

a^(m* n)=(a^m)^n

4^2(x^2)^2-25^2=0
ProdPow

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

(4x^2)^2-25^2=0
FacDiffSquares

a^2-b^2=(a+b)(a-b)

( 4x^2+25 )( 4x^2-25 )=0

We have written the left-hand side of the equation as the product of two factors. To solve the equation, we will apply the Zero Product Property.

( 4x^2+25 )( 4x^2-25 )=0
ZeroProdProp

Use the Zero Product Property

lc4x^2+25=0 & (I) 4x^2-25=0 & (II)
â–¼
Solve for x
SubEqn

(I): LHS-25=RHS-25

l4x^2=-25 4x^2-25=0
AddEqn

(II): LHS+25=RHS+25

l4x^2=-25 4x^2=25

(I), (II): .LHS /4.=.RHS /4.

lx^2=-25/4 x^2=25/4
MoveNegNumToFrac

(I): Put minus sign in front of fraction

lx^2=-25/4 x^2=25/4

(I), (II): sqrt(LHS)=sqrt(RHS)

lx=±sqrt(-25/4) x=±sqrt(25/4)
SqrtNegToSqrtI

(I): sqrt(- a)= sqrt(a)* i

lx=±sqrt(25/4)* i x=±sqrt(25/4)

(I), (II): sqrt(a/b)=sqrt(a)/sqrt(b)

lx=± sqrt(25)/sqrt(4)* i x=±sqrt(25)/sqrt(4)

(I), (II): Calculate root

lx=± 5/2* i x=±5/2
Multiply

(I): Multiply

lx=± 5/2i x=±5/2

We found four roots of the given equation. Roots 5/2i, - 5/2i, 5/2, - 5/2 We can see that there are four roots. Two of these roots are real and two of them are imaginary.

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Exercises 2 (Page 77)
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