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Solve the equation (5x^2-6x+8)^1/2 - (5x^2-6x-7)^1/2 = 1

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nena1993 | Student, Grade 11 | eNoter

Posted August 16, 2010 at 2:10 AM via web

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Solve the equation (5x^2-6x+8)^1/2 - (5x^2-6x-7)^1/2 = 1

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giorgiana1976 | College Teacher | Valedictorian

Posted August 16, 2010 at 2:14 AM (Answer #1)

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For the beginning, let's note the square roots as:

A = sqrt (5x^2-6x+8) => A^2 = 5x^2-6x+8

B = sqrt (5x^2-6x-7) => B^2 = 5x^2-6x-7

We'll re-write the equation:

A-B = 1 (1)

A^2 - B^2 = 5x^2-6x+8-5x^2+6x+7

We'll eliminate like terms:

A^2 - B^2 =  15

We'll divide A^2 - B^2 by A-B:

 (A^2 - B^2) / (A-B) = 15/1 = 15

But  (A^2 - B^2) / (A-B) = (A-B)(A+B)/(A-B) = A+B

A+B = 15 (2)

We'll add (1)+(2):

A+B+A-B = 16

2A = 16

A = 8

But A = sqrt (5x^2-6x+8) => sqrt (5x^2-6x+8) = 8

We'll square raise both sides:

5x^2-6x+8 = 64

5x^2-6x+8-64 = 0

5x^2-6x-56 = 0

We'll apply the quadratic formula;

x1 = [6+sqrt(36+1120)]/10

x1 = (6+34)/10

x1 = 4

x2 = (6-34)/10

x2 = -28/10

x2 = -14/5

The roots of the equation are: {-14/5 ; 4}.

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jeyaram | Student , Undergraduate | Valedictorian

Posted August 16, 2010 at 2:31 PM (Answer #2)

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(5x^2-6x+8)^1/2 - (5x^2-6x-7)^1/2 = 1

 

put y=5x^2-6x+8

so   y^1/2 - (y-15)^1/2=1

(y-15)^1/2=y^1/2-1

y-15=y-2y^1/2+1

y^1/2=8

y=64

so            5x^2-6x+8=64

5x^2-6x-56=0

5x^2-20x+14x-56 =0

5x(x-4)+14(x-4)=0

(5x+14)(x-4)=0

(5x+14)=0   or   (x-4)=0

x=(-14/5) or   x=4

 

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