Radicals and imaginary numbers in Algebra! Help!?

Question

I can't figure these problems out. =/ Any help is good, but if you could explain what you did, that'd be great.

*note: assume ∞ is equal to a square root sign. I couldn't find on on my keyboard. >.< so i.e. ∞(3) is the sqrt of 3. etc.

1. Simplify: the cubed root of 40

2. Simplify: 10/2∞(3)-∞(7)

3. Solve: x^2 + 6x + 2 = 0

4. Solve: (x^2/2)+1=x/5

5. Simplify: (-3+i∞(5))/(-i∞(5) -3)

6. Simplify: (1 + i)^3

7. Solve: 36x^-4 - 5x^-2 - 1 = 0

8. Simplify: 7^(3∞(2)) divided by 49.

9. Solve: 3^-(x+5) = 9^(4x)

Any questions, please ask. I know math can be hard to type/read on a computer, because of the lack of symbol options.

Answer 1

Go to START, Programs, Accessories, System tools and Character Map to find some of your symbols like sq root sign, etc.  I just write the words out &quot;sqrt&quot; for square root sometimes.

I&#039;ll help you a little....

1.) cube root 40 = cube root (2*2*2*5)= 2*cubrt(5)

2.) if you mean [10/(2√3-√7)] then we can simplify by multiplying top and bottom by the conjugate of dnominator:

(2√3+√7) to get

10(2√3+√7)/[(2√3-√7)(2√3+√7)]

= 10(2√3+√7)/[4(3)+2(√21)-2(√21)-7]

=  10(2√3+√7)/(12-7) = 10(2√3+√7)/5

= 2(2√3+√7)  simple answer

3.) This is a quadratic equation: x^2 + 6x + 2 = 0

Use quadratic formula  x=[-b+-sqrt(b^2-4ac)]/(2a) with a=1, b=6, and c=2

x= [-6+√(36-4(1)(2))]/2   or x= [-6-√(36-4(1)(2))]/2

x=  [-6+√(28)]/2     or    x=   [-6-√(28)]/2

x= -3+√(28)        or     x= -3-√(28)  

x= -3+2√7           or     x= -3-2√7

4.) Rearrange to the quadratic equation form:

Common denominator =10  so we have:

5x^2+10 = 2x or  5x^2-2x+10=0 quadratic form

Use quadratic equation to solve with a=5, b=-2, c=10

Do the discriminant check b^2-4ac to make sure it is not negative. It is... so your answer will be imaginery with i=√-1

x={-(-2)+sqrt[(-2)^2-4(5)(10)]}/2(5)  or

x={-(-2)-sqrt[(-2)^2-4(5)(10)]}/2(5)

Again x= {2+sqrt[4-200]}/10  or   x= {2-sqrt[4-200]}/10



           x= {2+sqrt[-196]}/10      or     x= {2-sqrt[-196]}/10

        

           x= {2+14√[-1]}/10      or     x= {2-14√[-1]}/10

           x=(2+14i)/10              or       x=(2-14i)/10  

        

or         x=(1+7i)/5                   or         x= (1-7i)/5

5.) Oh you know about i = √-1. Good...so remember i^2 = -1

(-3+i√5)(i√5-3) = (-3+i√5)^2= 9-6i√5+5(i^2)

=9-6i√5-5 = (4-6i√5) or 2(2-3i√5) answer

6.)(1+i)^3 = (1+i)(1+i)(1+i)= (1+2i+i^2)(1+i) = (1+2i-1)(1+i)

= (2i)(1+i) = 2i+2(i^2)= (2i-2) or 2(i-1) answer

7.) Simplify each term bby multiplying all by x^4 to get:

36-5x^2-x^4=0

Change problem around so x^4 is positive by multiplying everything by -1 then we have:

x^4+5(x^2)-36=0

Factor to get: (x^2+?)(x^2+)  Looking at last term and middle, find factors of 36 that will give 5 when added or subtracted.

(x^2+9)(x^2-4)=0

Now you have two equations to solve x^2+9=0 and x^2-4=0

x^2=-9 or x=√-9 = 3√-1=3i

and x^2=4 or x=+2 and x=-2

Collect your answers

8.) [7^(3√2)]/49, remember 49=7^2. So we can rewrite

[7^(3√2)]/(7^2),

remember 1/(x^a) =x^(-a) and (x^a)(x^b)= x^(a+b)

=  7^(3√2-2) simple answer

9.) 3^-(x+5) = 9^(4x); change 9=3^2 now you have:

        3^-(x+5) = 3^2(4x) or 3^-(x+5) = 3^(8x)

Since both have a base 3 we can drop it or log base 3 both to get:

-(x+5) = 8x

-x-5= 8x

9x=-5

x=-5/9 answer

done