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One more please! 9x − 61 √ x + 80 = 0?

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Find all the real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.)
asked Feb 3, 2014 in ALGEBRA 1 by andrew Scholar

1 Answer

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Given 9x-61√x+80 = 0

Add 61√x to each side.

9x-61√x+61√x+80 = 61√x

9x+80 = 61√x

Apply squring on each side.

(9x+80)^2 = (61√x)^2

81x^2+6400+1440x = 3721x

81x^2+6400+1440x-3721x = 0

81x^2-2281x+6400 = 0

Roots are x = [-b±√(b^2-4ac)]/2a

x = [2281±√(5202961-2073600)]/162

x = [2281±√3129361]/162

x = [2281±√1769]/162

x = (2281+1769)/162 and x = (2281-1769)/162

x = 25 and x = 3.160

Real solutions of 9x-61√x+80 are x = 25 , and x = 3.160

x = {25,3.160}

answered Feb 3, 2014 by david Expert

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