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Question 30472: I need help with the equation:
sq root of x-9 + sq root of x-21 = sq root of x+11 Solve for x
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! sq root of x-9 + sq root of x-21 = sq root of x+11 Solve for x
sq root of x-9 + sq root of x-21 = sq root of x+11 ----(1)
Squaring both the sides
[sqrt of (x-9)]^2 + [sqrt of (x-21)]^2 + 2X[sqrt of (x-9)]X[sqrt of (x-21)] = [sq root of (x+11)]^2
(x-9)+(x-21)+2X[sqrt of (x-9)]X[sqrt of (x-21)]= (x+11)
2X[sqrt of (x-9)]X[sqrt of (x-21)]= (x+11)-(x-9)-(x-21)
=(x-x-x)+(11+9+21)
=(-x)+(41)
=(41-x)
Therefore 2X[sqrt of (x-9)]X[sqrt of (x-21)]= (41-x)
Squaring both the sides
{2X[sqrt of (x-9)]X[sqrt of (x-21)]}^2 = [-(x+1)]^2
4(x-9)(x-21) = (41-x)^2
4[x(x-21)-9(x-21)] = 1681-82x+x^2
4(x^2-21x-9x+189)= 1681-82x+x^2
(4x^2-x^2)+(-84x-36x+82x)+(756-1681)=0 (grouping like terms)
3x^2-38x-925=0 ----(*)
Now 3X925= 3X5X185= 3X5X5X37= (3X5X5)X(37)
The product is (3x^2)X(925)=[-(3X5X5)x]X[(37)x]and -38x= (-75x)+(37x)
3x^2+[(-75x)+(37x)]-925=0
(3x^2-75x)+(37x-925) = 0
3x(x-25)+37(x-25) = 0
3xp+37p=0 where p= (x-25)
p(3x+37) = 0
That is (x-25)(3x+37)=0
(x-25)= 0 gives x = 25
(3x+37)=0 gives x = -37/3
But x=-37/3 gives
sq root of x-9 = sq root of(-37/3-9) = sq root of(a negative quantity) which is not defined
Let us verify for x = 25 in
sq root of x-9 + sq root of x-21 = sq root of x+11 ----(1)
LHS = sq root of (25-9) + sq root of (25-21)
= sq root of 16+ sq root of 4
= 4 + 2
= 6
= sq root of x+11 for x = 25
=RHS
Therefore our value x = 25 is correct
Answer: x =25
Remark:The technique used is
[sqrt(a)+sqrt(b)] = [sqrt(c)]
squaring
(a+b)+ 2(sqrta)(sqrtb) =c
2(sqrta)(sqrtb) = [c-(a+b)]=k say
2(sqrta)(sqrtb) =k
squaring
4ab = k^2
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