I need help in Algebra! Please answer!?

Question

Please solve all three if possible but if not try to solve one or two
how do u solve the followings:
1) the square root of "a" squared + 2a - 10 = a
2) square root of "b" squared - 7b + 6 = b - 4
3)square root of 3y squared + 5y + 6 = y +3

o and the - sign is subtraction

Answer 1

1) sqrt(a^2 + 2a - 10) = a

Square both sides,

a^2 + 2a - 10 = a^2

Subtract a^2 both sides,

2a - 10 = 0
2a = 10
a = 5

2) sqrt(b^2 - 7b + 6) = b - 4

Square both sides,

b^2 - 7b + 6 = (b - 4)^2
b^2 - 7b + 6 = b^2 - 8b + 16
-7b + 6 = -8b + 16
b = 10

3) sqrt(3y^2 + 5y + 6) = y + 3

Square both sides,

3y^2 + 5y + 6 = (y + 3)^2
3y^2 + 5y + 6 = y^2 + 6y + 9

Move everything to the left hand side,

2y^2 - y - 3 = 0

Factor.

(2y - 3)(y + 1) = 0

Solve for y.

2y - 3 = 0
y + 1 = 0

y = { 3/2, -1 }

However, we need to test these values into the original equation to ensure they both work.

If y = 3/2:

LHS = sqrt(3y^2 + 5y + 6)
LHS = sqrt(3(3/2)^2 + 5(3/2) + 6)
LHS = sqrt(3(9/4) + 15/2 + 6)
LHS = sqrt(27/4 + 15/2 + 6)
LHS = sqrt(27/4 + 30/4 + 24/4)
LHS = sqrt(81/4) = 9/2

RHS = y + 3
RHS = (3/2) + 3
RHS = (3/2) + (6/2)
RHS = 9/2

LHS = RHS, so y = 3/2 is indeed a solution.

Test y = -1:

LHS = sqrt(3y^2 + 5y + 6)
LHS = sqrt(3(-1)^2 + 5(-1) + 6)
LHS = sqrt(3 - 5 + 6)
LHS = sqrt(4)
LHS = 2

RHS = y + 3
RHS = (-1) + 3
RHS = 2

LHS = RHS, so y = -1 is indeed a solution.

Therefore, y = { 3/2, -1 }

Answer 2

1. The square root of a^2 can be simply transformed to a.

therefore: a+2a-10=a
3a-10=a
3a-a=10
2a=10
answer a=5

2. the square root of b^2 can be simply transformed to b.

therefore: b-7b+6=b-4
6b+6=b-4
6b-b=-4-6
5b=-10
b=-2

Answer 3

1. sq. rt. A^2 + 2a -10 = a
a+2a-10=a
3a -10 = a
2a-10=0
2a=10
a=5

2. sq.rt. b^2 -7b + b = b - 4
b-7b+b=b-d
-5b = b -4
-4b = -4
b=1

3. sq.rt. 3y^2 +5y + 6 = y+ 3
3y + 5y +6 = y + 3
8y + 6 = y + 3
7y + 6 = 3
7y = -3

Answer 4

I checked the work, and as long as in your question you mean the square root of everything on the left side, in other words, the square root of (a^2+2a-10), and not just the square root of (a^2), then "Puggy"'s answers are correct. This goes for questions 2 and 3 also. Good luck in your class!

Answer 5

It is unclear from your questions what is under the square root sign.
Is question (1): sqrt(a^2) + 2a - 10 = a?
If so, then that simplifies to:
a + 2a - 10 = a
2a = 10
a = 5.

Answer 6

(a^2 + 2a - 10)^0.5 = a
a^2 + 2a - 10 = a^2
2a = 10
a = 5

(b^2 - 7b + 6)^0.5 = b - 4
b^2 - 7b + 6 = b^2 - 8b +16
b = 10

(y^2 + 5y + 6)^0.5 = y + 3
3y^2 + 5y + 6 = y^2 + 6y + 9
2y^2 - y - 3 = 0
(2y + 3)(y - 1) = 0
y = -2/3, 1

Answer 7

Anytime you have a squared term, the best way to go is to set the equation equal to zero and either 1) factor and set each factor equal to zero, 2) complete the square, or 3) use the quadratic formula, whatever you're learning now.

Good luck!

Answer 8

If we give you the answers we arnt helping you..but if we tell you how to and let you do it...then were helping....here is the first step....make it equal to zero....anybody wanna enligten her ont eh 2nd step...