### Solving Other Non-linear Equations

let's just have some review!

let's solve for:

**x**by completing the square, factoring, and using the quadratic formula. Take note...we must get the same answers for all of this.

^{2}-2x-24Completing the square:

x

^{2}-2x-24=0

x

^{2}-2x=24

x

^{2}+1=24+1

square root of (x-1)(x-1)=square root of 25

x-1=±square root of 25

x-1=5

x=6,-4

Factoring:

x

^{2}-2x-24= 0

(x-6) (x+4)

x= 6,-4

Quadratic formula:

Solve for: y= x^{4} -2x^{2} -15

Let p= x^{2}

y= p^{2} -2p-15

p^{2}-2p-15=0

p^{2} -2p=15

(p-1)^{2} =15 +1

square root of (p-1)^{2} =square root of 16

p-1=± 4

p= -3, 5

substitute p back to x^{2}

x^{2}= square root of -3

x= ± i square root of 3

x= ± square root of 5

Solve for: y= x^{8}-1

Factor x^{8}-1

(x^{4}-1) (x^{4}+1)=0

Factor (x^{4}-1)

(x^{2}-1) (x^{2}+1) (x^{4}+1)=0

(x-1)(x+1)(x^{2}+1)(x^{4}+1)

x-1=0 x+1=0 x^{2}+1=0

x=1 x=-1 x^{2}=-1 x^{4}+1=0

x=± square root of -1 x^{4}=-1

x=± i let p=x^{2}

p^{2}=-1

p=± i

substitute p back to x^{2}

square root of x^{2}=square root of ± i

x=±square root of i

therefore: x= ±1, ± i, ± square root of i

Solve for: y=x^{3}-8

0=x^{3}-8

cube root of x^{3}=cube root of 8

x=2

**by: fevie javier**

^{}

## 1 Comments:

Who's the next scribe then?

9:50 PM

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