# Algebra

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1. √(9y - 196) + √196 = √49

My work:

I squared all 3 roots and got

9y - 196 + 196 = 49

The - 196 + 196 cancels out and we are left with 9y = 49.
divide 49 by 9. = 49/9. I'm pretty sure this is wrong. Can you show me where I messed up?

2. √x(√9 - 4/3) = √25

3. (4 - 3√2)^2

My work: Distributive property gives me 4^2 - 3^2√2^2
The √2^2 simplifies to 2, 4^2 is 16, 3^2 is 9, So we have 16-9x2 Multiply first, so 16 - 18 = -2.

Is this correct?

4. √(135*b^2*c^3*d) * √(5b^2*d

Okay, so I broke this down like this: √135 * √b^2 * √c^2 * √c * √d * √5 * √b^2 * √d

Both the √b^2's and the √c^2 simplify down to b,b, and c

so now we have b^2c√(135 * 5cd^2) Simplifies to b^2cd√(5^2 * 3^2 * 3) which simplifies to 15b^2cd√3

Is this correct? If not can you tell me where I went wrong?

Thank you!

• Algebra -

If a+b = c you cannot say that a^2+b^2 = c^2

√(9y - 196) + √196 = √49
√(9y - 196) + 14 = 7
√(9y - 196) = -7
But, √N is positive. So, you cannot have √(9y - 196) = -7

√x(√9 - 4/3) = √25
√x(3 - 4/3) = 5
√x(5/3) = 5
√x = 3
x = 9

(4 - 3√2)^2
recall that (a-b)^2 = a^2-2ab+b^2, so we have
16 - 24√2 + 18
44 - 24√2

almost right - you lost a √c
√(135*b^2*c^3*d) * √(5b^2*d)
√(135*5) √(b^2*b^2) √c^3 √(d*d)
15√3 b^2 c√c d
15b^2cd √(3c)

• Algebra -

I have a couple questions about # 1.
Why can't it be a^2 + B^2 = c^2?
I think that's what I'm taught in class. Also, if you square everything, isn't still going to come out the same? If you do something to the entire expression, it doesn't change the value, right? If your conclusion is correct, what would I put for that answer? "No solution"? "No real solution"?

On #3. Would I do 44-24 = 20√2.. or?

• Algebra -

For #1: When I get √(9y - 196) = -7 can I square each side to get 9y - 196 = 49? Then I can add 196 to both sides. It cancels out on the left side and I have 9y = 245. Then I can divide both sides by 9. I would get y = 27.2. Would that be correct?

• Algebra -

You say that I can't use a^2 + b^2 = c^2, but that formula is called the Pythagorean theorem. I was instructed to use that formula. Am I misunderstanding what you're saying?

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