(√2y+7)+3=y Where 2y +7 is all under the same sr sign?

Answer 1

(√2y+7)+3=y
(√2y+7) = y -3
2y + 7 = y^2 -6y +9
y^2-8y +2 = 0
y = [8 +/- sqrt(64-4*1*2)]/2
y = [8 +/- sqrt(56)]/2
y = 4+/- sqrt(14)
y = 4 + sqrt(14), and
y = 4 - sqrt(14)

Answer 2

First, isolate the square root:

√(2y+7)=y-3

Now square both sides:

2y+7=y²-6y+9

Render this into standard form:

0=y²-8y+2

Complete the square by adding 14:

14=y²-8y+16

Factor:

14=(y-4)²

Take the square roots:

y-4=±√14

Add 4:

y=4±√14

Note that it is not necessarily the case that both of these solutions are correct, since one of the transformations we did (squaring both sides) was not invertible. Therefore, we test:

y=4-√14 -- This cannot be a solution, because y-3=1-√14 is negative, but in the original problem, y-3 would have to equal the _positive_ square root of 2y+7.

y=4+√14 -- In this case, y-3, which is 1+√14, is positive, so if its square is equal to 2y+7, then it is indeed the positive square root of 2y+7. So we check: (1+√14)²=15+2√14, and 2y+7 = 15+2√14, so this holds. Therefore, this is a solution.

Thus your unique solution is y=4+√14

Answer 3

sqrt(2y+7) =y-3 The number under the sqrt must be >=0

2y+7>=0 y>=-7/2 If you want to sqare both sides they must have the same sign. As sqrt>=0 y-3>=0 y>=3

2y+7 = y^2 -6y +9 y^2-8y +2=0 y= (( 8+- sqrt(64-8))/2=

4+1/2sqrt(56) The minus sign gives you a value less than 3

which is intrduced through squaring

Answer 4

Rewrite as sqrt(2y +7) = y -3. Square both sides:
2y + 7 = Y^2 -6y + 9. Collect terms: 0 = y^2 -8y +2. Solve by quadratic formula.

Answer 5

2y + 7 = y² - 6y + 9

y² - 8y + 2 = 0

y = (8 ± √ 56) / 2

y = (8 ± 2√14) / 2

y = 4 ± √14

Answer 6

Subtract 3 from both sides:
(√2y+7)=y-3

Square both sides:
(√2y+7)^2=(y-3)^2
2y+7=y^2-6y+9

Subtract 2y and 7 from both sides:
y^2-8y+2=0

Since this is not factorable, you need to use the quadratic formula:
x=-b±√b^2-4ac/2a
a=1,b=-8,c=2
x=-(-8)±√(-8)^2-4(1*2)/2(1)
x=8±√64-8/2
x=8±√56/2
x=8±2√14/2
x=4±√14

Answer 7

sqrt(2y + 7) + 3 = y
sqrt(2y + 7) = y - 3
2y + 7 = (y - 3)^2
2y + 7 = (y - 3)(y - 3)
2y + 7 = y^2 - 6y + 9
y^2 - 8y + 2 = 0

y = (-b ± sqrt(b^2 - 4ac))/(2a)

y = (-(-8) ± sqrt((-8)^2 - 4(1)(2)))/(2(1))
y = (8 ± sqrt(64 - 8))/2
y = (8 ± sqrt(56))/2
y = (8 ± sqrt(4 * 14))/2
y = (8 ± 2sqrt(14))/2
y = 4 ± sqrt(14)

ANS : y = 4 - sqrt(14) or 4 + sqrt(14)

Answer 8

y^4-4y^3-28y^2-8y+4=0