Math Help Needed?

Question

It says to Simplify. Give exact answers. Do not use your calculater and round. The problem is 4 sqroot 150. The 4 is on the outside of the square root of 150.

Solve equation by finding square roots. Simplify answer.
2a^2 - 5 = 35

Use quadratic formula to solve equation. Simplify all radicals and fractions. I'm cunfused on this one. x^2 = 5x + 1. First I know you put it in standard form, so wouldnt that be -x^2 + 5x + 1 ?

Also if you wanna explain how to do these next ones, please do so. Evaluate and round answers to two decimals.
5 + - 3 sqroot 3
____________
2

6 + - 2 sqroot 3
____________
5

Answer 1

When looking at square roots you look for factors that are perfect squares because sqrt(ab)=sqrt(a)sqrt(b)

4sqrt(150) = 4sqrt(6*25) = 4*sqrt(6)sqrt(25) = 4sqrt(6)5

20sqrt(6)
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2a^2 - 5 = 35

Here you just simplify

2a^2=35+5=40

a^2=40/2=20

a=sqrt(20) = sqrt(4*5) = sqrt(4)(sqrt(5) = +/-2sqrt(5)

Make sure you understand where the answer must be positive or negative or could be either.
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It's usually helpful if you keep the x^2 term positive but it really doesn't matter if you use the quadratic formula.

-x^2+5x+1=0

That means a=-1, b=+5 and c=+1

the formula is x = [-b+/-sqrt(b^2-4ac)] / (2a)

x = {-5+/-sqrt[25-4(-1)(1)]} / (2*-1)
x = [-5+/-sqrt(25+4)] / -2

x=[5-sqrt(29)]/2 or [5+sqrt(29)]/2

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well if you want to do these you need to use a calculator (or at least a square root table)

sqrt(3)=1.732

so just multiplying it out you get

[5+3sqrt(3)]/2 = 5.10
[5-3sqrt(3)]/2 = -0.10

[6+2sqrt(3)]/5 = 1.89
[6-2sqrt(3)]/5 = 0.51

Answer 2

4*SQRT(150) = 4*SQRT(6*25) = 4*SQRT(25)*SQRT(6)
SQRT(25) = 5
4*5*SQRT(6) = 20*SQRT(6)

The square of 2.4 is 5.76
The square of 2.5 is 6.25

Therefore, the square root of 6 is very close to 2.45

4*SQRT(150) = approx. 20*2.45 = 49
You cannot give an exact value because SQRT(6) is not a rational number.

In some cases, you have to indicated that the answer is "plus or minus" a value.

For example, the square root of 4 is +2 or -2 (because -2 times -2 equals +4). However, if you are using the sqaure root as a function, then you must pick only one (usually the positive answer) because a function can only have one answer.

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2a^2 - 5 = 35
2a^2 = 40
a^2 = 20
a = SQRT(20) = SQRT(4)*SQRT(5) = 2*SQRT(5)

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Yes.

However, it is more common to use a positive factor for the higher degree:
x^2 - 5x -1 = 0
corresponding to a x^2 + b x + c = 0
where (in this case) a=1 b=-5 c=-1

I suspect that you are expected to use 'the formula'

x = [ -b +/- SQRT( b^2 - 4*a*c)] / (2*a)
x = [ 5 +/- SQRT(25 + 4)] / 2
x = (5/2) + SQRT(29/4) and
x = (5/2) - SQRT(29/4)
(there are two answers; +/- means plus or minus)

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(1/2)*( 5 +/- 3*SQRT(3))
For this one, I happen to remember that SQRT(3) = 1.73205080756... (and it is the tangent of 60 degrees)

You will get two answers:
2.5 + 1.5*(1.732) = 5.1 (or 5.098), and
2.5 - 1.5*(1.732) = -0.1 (or -0.098).

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1.2 * 0.4*(1.732), and
1.2 - 0.4*(1.732)
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Answer 3

I'm asssuming you're OK with the first two.

On the third, you goofed on your standard form.
x^2 = 5x + 1
x^2 - 1 = 5x + 1 - 1
x^2 - 1 = 5x
x^2 - 1 - 5x = 5x - 5x
x^2 - 1 - 5x = 0
x^2 -5x -1 = 0 THAT is the correct standard form. Now you can use the quadratic equation successfully.

Answer 4

4 sqrt(150) = 4 sqrt(6*25) = 4*5sqrt(6) = 20sqrt(6)

2aa-5 = 35
2aa=35+5=40
aa = 20
a = sqrt(20) = sqrt(4*5) = 2sqrt(5)

xx = 5x+1
xx-5x=1
xx-5x-1=0
x = (-b+-sqrt(bb-4ac))/2a = (5 +- sqrt(5*5-4(1)(-1))/2
= (5 +- sqrt(25+4))/2 = (5 +- sqrt(29))/2

(5 +- 3(1.73))/2 = (5 +- 5.19)/2 = 10.19/2 or -.19/2 - 5.09 or -.09

Answer 5

find x....HERE it is hahahahaha