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If (2x^2+3x+5)^1/2 +(2x^2+3x+20)^1/2 = 15, then fractional value of x is:

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madmoni06 | Student | Honors

Posted January 14, 2013 at 3:12 PM via web

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If (2x^2+3x+5)^1/2 +(2x^2+3x+20)^1/2 = 15, then fractional value of x is:

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 14, 2013 at 4:34 PM (Answer #1)

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The fractional value of x has to be determined given that `(2x^2+3x+5)^(1/2) +(2x^2+3x+20)^(1/2) = 15`

 `(2x^2+3x+5)^(1/2) +(2x^2+3x+20)^(1/2) = 15`

Let `y = 2x^2 + 3x + 5`

=> `sqrt y + sqrt(y + 15) = 225`

Take the square of both the sides

=> `y + y + 15 + 2*sqrt(y^2 + 15y) = 225`

=> `2y + 2*sqrt(y^2 + 15y) = 210`

=> `sqrt(y^2 + 15y) = 105 - y`

=> `y^2 + 15y = y^2 + 11025 - 210y`

=> `225y = 11025`

=> `y = 49`

As `2x^2 + 3x + 5 = y`

=> `2x^2 + 3x + 5 = 49`

=> `2x^2 + 3x - 44 = 0`

=> `2x^2 + 11x - 8x - 44 = 0`

=> `x(2x + 11) - 4(2x + 11) = 0`

=> `(x - 4)(2x + 11) = 0`

=> x = 4 and x = `-11/2`

The fractional value of x that satisfies the equation is `x = -11/2`

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