Radical expressions?

Question

square root of 2y+7(+4)=y
I cannot seem to get an answer that checks out right.

Answer 1

I will denote the square root of a by sqrt(a). In the original equation sqrt(2y+7) + 4 = y, subtract 4 from both sides to get:

sqrt(2y+7) = y-4.

Now take the squares of both sides. This yields:

2y+7 = (y-4)^2, which can be expanded into:

2y + 7 = y^2 - 8*y +16, which you can finally rearrange into

y^2 - 10*y + 9 = 0, which is a quadratic equation. Applying the quadratic formula, you get the following two solutions:

y_1 = 1 and
y_2 = 9.

You can check which of the roots is the correct one by plugging them back into the original equation.

The first one gives sqrt(2+7) + 4 = 1, which is incorrect (implying that y=1 is a false root we arrived at by squaring both sides), and the second one yields sqrt(18+7) + 4 = 9, which is correct, so the only solution is y=9.

Answer 2

Well, when you're working with radical equations, it's possible to do everything right and still get nothing but false answers. Let's see what happens.

If I read you correctly, you're trying to solve

4 + √(2y + 7) = y

Start by subtracting the 4 from both sides:
√(2y + 7) = y - 4

Squaring both sides:
2y + 7 = (y - 4)²
2y + 7 = y² -8y + 16

Simplifying:
0 = y² -10y + 9

This is factorable:
(y -9)(y-1) = 0

So either y = 9 or y = 1. You do need to check both answers, however. I'll let you proceed from this point, but I'll tell you that, in this particular problem, at least one of the two answers won't work.

Hope that helps!

Answer 3

move the 4 to the other side
sqrt(2y+7) = y-4
square both sides...
2y + 7 = y^2 - 8y + 16
simplify
0 = y^2 - 10 y +11
complete the square
0 = y^2 - 10 y + 25 -14
0= (y-5)^2 - (sq rt 14)^2

since A^2 - B^2 = (A-B) (A+B)
then
(y-5-rt14) (y-5+rt14) = 0

so y = 5 plus or minus sqrt 14

Answer 4

rt2y-y=-28
y(rt-1)=-28
y=-28/(rt2-1)
multiplying th nr and dr by rt2+1
-28(rt+1)/2-1=-28(rt2+1)

Answer 5

i dont know