SOLUTION: the square root of 4y+16 - the square root of y-5 equals 6
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Question 762329: the square root of 4y+16 - the square root of y-5 equals 6
Answer by smiths(2) (Show Source): You can put this solution on YOUR website!
√(4y + 12) - √(y - 6) = 6
Square both sides:
[√(4y + 12) - √(y - 6)]² = 6²
4y + 12 + (y - 6) - 2√(4y + 12)√(y - 6) = 36 . . By perfect square trinomial, expand [√(4y + 12) - √(y - 6)]².
4y + 12 + y - 6 - 2√(4y + 12)√(y - 6) = 36
5y + 6 - 2√(4y + 12)√(y - 6) = 36
Then, bring 5y + 6 to the right:
-2√(4y + 12)√(y - 6) = 36 - (5y + 6)
-2√(4y + 12)√(y - 6) = 36 - 5y - 6
-2√(4y + 12)√(y - 6) = 30 - 5y
Square both sides again!
(-2√(4y + 12)√(y - 6))² = (30 - 5y)²
4(4y + 12)(y - 6) = 900 - 300y + 25y²
4(4y² - 12y - 72) = 900 - 300y + 25y²
16y² - 48y - 288 = 900 - 300y + 25y²
Bring 16y² - 48y - 288 to the right.
0 = 900 - 300y + 25y² - (16y² - 48y - 288)
0 = 9y² - 252y + 1188
By factoring, we obtain:
0 = 9(y² - 28y + 132)
0 = 9(y - 22)(y - 6)
Finally, by zero-product identity:
y - 22 = 0 and y - 6 = 0
y = {22,6}
it may help u ......!!!
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