Question 109513
Solve each of the equations. Be sure to check your solutions.
Sqrt(y) + 7 = 5
sqrt(y) = -2
Square both sides to get:
y = 4
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Checking y = 4 in sqrt(y) + 7 = 5
sqrt(4) + 7 = 5
2 + 7 = 5
False:
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There is no solution for sqrt(y)+7 = 5
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I get -2 but not sure what the square root of -2 is Is there one?
No.  Sqrt(4) is 2, not -2.
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Solve each of the equations. Be sure to check your solutions.
Sqrt (3x+ 1) = 4 
3x+1= 16
3x = 15
x = 5
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Checking x= 5
sqrt(3*5+1) = 4
sqrt(16) = 4
4=4
True
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I get one is one the square root of one? 
Yes
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Solve each of the equations. Be sure to check your solutions. 
sqrt 2x + 1 + x = 7 
sqrt(2x+1) + x = 7
sqrt(2x+1) = 7-x
2x+1 = 49-14x+x^2
x^2-16x-49=0
x = [16 +- sqrt(16^2-4*-49)]/2
x = [16 +- sqrt(452)]/2
x = [16 +- 2sqrt(113)]/2
x = 8 + sqrt(113) or x = 8 -sqrt(113)
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Cheers,
Stan H.





I got x=2 and am not sure what the square root of two is. 
Help on this one please.
Solve for a 
c = sqrt a^2 + b^2 
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