What are the solutions of equations square root (x^2-8x+31)-squareroot (x^2-8x+24)=1?

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beckden | High School Teacher | (Level 1) Educator

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While it is not necessary to impose restrictions on the answers, it is possible because of the loss of information when squaring a variable to lose the sign of the result, a simple example is sqrt(x)=-3.  You can see that there is no answer to this problem but squaring both sides gives x^2 = 9 and has solutions -3 and +3.  You need ot check all answers to determine if they are extraneous or not.  In this case it does not matter but it is possible to get answers that do not solve the squared equation but do not solve the original equation.

√(x²-8x+31)-√(x²-8x+24)=1
(√(x²-8x+31)-√(x²-8x+24))²= 1
-2√(x²-8x+24)√(x²-8x+31)-16x+2x²+ 55=1
-2√(x²-8x+24)√(x²-8x+31)=16x-2x²-54
√(x²-8x+24)√(x²-8x+31)=-8x+x²+27
(√(x²-8x+24)√(x²-8x+31))²= x⁴-16x³+119x²-440x+744
(-8x+x²+27)²= x⁴-16x³+118x²-432x+729
x⁴-16x³+119x²-440x+744=x⁴-16x³+118x²-432x+729
x²-8x+15=0
(x-3)(x-5)=0
x=3, x=5
√(3²-8(3)+31)-√(3²-8(3)+24)= 1   x=3 is a solution
√(5²-8(5)+31)-√(5²-8(5)+24)= 1   x=5 is a solution

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james2011 | Student, Grade 9 | (Level 1) Honors

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from

(1)    sqrt(x^2 -8x + 31) - sqrt(x^2 - +24) = 1,

multiple sqrt(x^2 -8x + 31) + sqrt(x^2 -8x +24) on both sides

(x^2 - 8x + 31) - (x^2 -8x + 24) = sqrt(x^2 -8x + 31) + sqrt(x^2 -8x +24)

simplify it and put it into right order

(2) sqrt(x^2 -8x + 31) + sqrt(x^2 -8x +24) = 7

add (1) and (2)

 

2* sqrt(x^2 -8x + 31) = 8

simplify and square both sids

x^2 - 8x + 31 = 16

we have

x^2 -8x + 15 = 0

(x - 3) (x -5 ) = 0

x = 3 or x = 5

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Let [sqrt(x^2 - 8x + 31) - sqrt(x^2 - 8x + 24)] = 1 (1)

We'll multiply the given expression by it's conjugate:

[sqrt(x^2 - 8x + 31) - sqrt(x^2 - 8x + 24)][sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)] = [sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)]

The product from the left returns the difference of two squares:

x^2 - 8x + 31 - x^2 + 8x - 24 = [sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)]

We'll eliminate like terms:

7 = [sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)] (2)

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

[sqrt(x^2 - 8x + 31) - sqrt(x^2 - 8x + 24)] + [sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)] = 1 + 7

We'll eliminate like terms:

2sqrt(x^2 - 8x + 31) = 8

We'll divide by 2:

sqrt(x^2 - 8x + 31) = 4

We'll raise to square both sides:

(x^2 - 8x + 31) = 16

We'll subtract 16 both sides:

x^2 - 8x + 31 - 16 = 0

x^2 - 8x + 15 = 0

We'll apply quadratic formula:

x1 = [8+sqrt(64 - 60)]/2

x1 = (8+2)/2

x1 = 5

x2 = (8-2)/2

x2 = 3

Since there is no need to impose constraints of existence of square roots, because both radicands are positive for any value of x, therefore the solutions of the equation are: x1 = 5 and x2 = 3.

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