how do i solve for y?

Question

given a = -5, b = 1, c = 2, if
y^2 - a^2 = (c^2 + b)^2

i am sure there are two solutions. i asked this beforre and accidentally closed it. i think one solution is square root of 50

Answer 1

The answer is, in fact, square root of 50. Otherwise known as 5 Radical 2. That would be the reduced form.

If you plug the given in the formula, you should get:
y^2 - (-5)^2 = [(2)^2 + 1]^2

y^2 - 25 = (4+1)^2

y^2 - 25 = 5^2

y^2 = 50

And you square root both sides. Remember that a negative number that is square rooted will give you the same answer as a positive number.

Hope that helped!

Answer 2

I think alienhuman... was on the right track, but I think she made a mistake at the end. Here is my alternative solution:

y^2 - (-5)^2 = [(2)^2 +1]^2
y^2 - 25 = 5^2
y^2 - 25 = 25
y^2 = 50
y = plus or minus sqrt 50
y = plus or minus 5 sqrt 2
y = 7.071067812 or -7.071067812
you will always get two values for y when y is squared in the original formula.

You can test my answers by plugging the value of y back into the original formula:

(7.071067812)^2 - (-5)^2 = ((2^2) + 1)^2
50 - 25 = 25
50 = 25 + 25
50 = 50 -- yep, that's a true value for y

(-7.071067812)^2 -(-5)^2 = 25
50 - 25 = 25
50 = 50 --this is also a true value for y

Answer 3

Solving for y

Given

A = -5
b = 1
c = 2

Insert the above values into the a, b, and c position in the in the equation

-- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The Equation

y² - a² = (c² + b)²
+a² + a²

adding +a² to both sides

y² = √a² + (c² + b)²

y² = √ (- 5)² + [ (2)² + 1)²]

y² = √ 25 + [(4 + 1)²]

y² = √25 + [(5)²]

y² = √25 + [25]

y²= √25 + 25

y² = √50

y =7.071067812

y = 7.07 rounded to two decimal places

Answer 4

y^2=(c^2 + b)^2 +a^2

to find the value of y substitute the given value of a,b,c

y^2= (2^2 + 1)^2 + (-5)^2
y^2= ( 4 + 1 )^2 + 25
y^2= 5^2 + 25
y^2= 25 + 25
y^2= 50
y= squareroot of 50

Answer 5

just substitute the known values into the equation:

y^2 - a^2 = (c^2 + b)^2
y^2 - (-5)^2 = (2^2 + 1)^2

now just expand...

y^2 - 25 = 25
y^2 = 25 + 25
y^2 = 50

therefore, y = + sqrt50 and - sqrt50

always remember that when you get an equation like a^2 = 16,
a has two values, +4 and -4. even the brightest students sometimes miss this out and forget.

have fun with math!

Answer 6

substitute the given equivalents of the variables then solve.

y^2-a^2=(c^2+b)^2
y^2-(-5)^2=(2^2+1)^2
y^2-25=(4+1)^2
y^2-25=5^2
y^2-25=25
y^2=50

then get the square root of both sides

y=square root of 50 or y=square root of -50

(a positive number's square root would include the posive square root of the number and the negative square root of th said number.)

Answer 7

y^2=a^2+(c^2+b)^2. now substituting the values of a,b and c
y^2=(-5)^2+(2^2+1)^2=>25+25=2*25 and so y=(2*25)^1/2
=5 rt 2 (answer)

Answer 8

by substituting the values we get,

y^2 - (-5)^2 = (2^2 + 1)^2
y^2 - 25 = 5^2
y^2 = 25 + 25
y^2 = 50
y = Sqrt 50
or y = +7.07 and -7.07

Answer 9

the two answers are
1) the square root of 50
2) the negative square root of 50

basically, plus or minus the square root of 50

Answer 10

y^2 - a^2 = (c^2 + b)^2

y^2 - (-5)^2 = (2^2 + 1)^2
y^2 - (25) = (4 + 1)^2
y^2 - 25 = 5^2
y^2 - 25 = 25
y^2 = 50
y = sqrt(50)
y = sqrt(25 * 2)
y = 5sqrt(2)

y = -5sqrt(2) or 5sqrt(2)