Asked by Jillian
4. REVIEW PROBLEM: Solve).
√3-2x + √3x +13 =5
the term (3-2x) is under a radical as well as the term (3x+13). Please show all steps!! i keep on getting x=9 but plugging it in it is wrong!
√3-2x + √3x +13 =5
the term (3-2x) is under a radical as well as the term (3x+13). Please show all steps!! i keep on getting x=9 but plugging it in it is wrong!
Answers
Answered by
Steve
yes, x=9 will certainly not work.
√(3-2x) + √(3x+13) = 5
√(3-2x) = 5 - √(3x+13)
3-2x = 25 - 10√(3x+13) + 3x+13
35+5x = 10√(3x+13)
x+7 = 2√(3x+13)
x^2+14x+49 = 12x+52
x^2+2x-3 = 0
(x+3)(x-1) = 0
x = 1 or -3
check
x=1: √1 + √16 = 5
x=-3: √9 + √4 = 5
yep.
√(3-2x) + √(3x+13) = 5
√(3-2x) = 5 - √(3x+13)
3-2x = 25 - 10√(3x+13) + 3x+13
35+5x = 10√(3x+13)
x+7 = 2√(3x+13)
x^2+14x+49 = 12x+52
x^2+2x-3 = 0
(x+3)(x-1) = 0
x = 1 or -3
check
x=1: √1 + √16 = 5
x=-3: √9 + √4 = 5
yep.
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