SOLUTION: Could you please help? Solve the following equations. Leave radicals in radical form ( do not calculate their approximate values.) a. {{{sqrt x+6-4=x}}} b. {{{(5)sqrt(3x+1)

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Question 61078: Could you please help?
Solve the following equations. Leave radicals in radical form ( do not calculate their approximate values.)
a.
b. (the 5's should be at the tip of the radical sign)
c. (x+6)^2/5+2=3
Thank you for your help!!
Answer by uma(370) (Show Source):
You can put this solution on YOUR website!
(a) sqrt(x+6) - 4 = x
Addinfg 4 to either side we get,
sqrt (x+6) - 4 + 4 = x + 4
==> sqrt(x + 6) = x + 4
Squaring both the sides we have,
x + 6 = (x + 4)^2
==> x + 6 = x^2 + 8x + 16 [(a+b)^2 = a^2 + 2ab + b^2]
==> x = x^2 + 8x + 16 - 6
==> x = x^2 + 8x + 10
==> 0 = x^2 + 8x + 10 - x [adding -x to both the sides]
==> 0 = x^2 + 7x + 10
==> 0 = x^2 + 5x + 2x + 10 [Splitting the middle term]
==> 0 = x(x+5) + 2(x+5) [Taking out the common factor]
==> 0 = (x+5)(x+2)
==> x + 5 = 0 or x+2 = 0
==> x = -5 or x = -2

(b) (5)sqrt(3x+1)=(5)sqrt(x^2-2x-5)
Dividing both the sides by 5 we get,
sqrt(3x+1)= sqrt(x^2-2x-5)
Squaring both the sides,
3x + 1 = x^2 - 2x - 5
==> 3x + 1 - 3x = x^2 - 2x - 5 - 3x [ adding - 3x to both the sides]
==> 1 = x^2 - 5x - 5
==> 0 = x^2 - 5x - 5 - 1 [adding -1 to both the sides]
==> 0 = x^2 - 5x - 6
==> 0 = x^2 - 6x + x - 6
==> 0 = x(x-6) + 1(x-6)
==> 0 = (x-6)(x+1)
==> x - 6 = 0 or x + 1 = 0
==> x = 6 or x = -1
(c)[(x+6)^2]/5 + 2 = 3
==> (x+6)^2 /5 = 3-2 [Adding -2 to both the sides]
==> (x+6)^2/5 = 1
Multiplying both the sides by 5 we have,
(x+6)^2 = 1*5
==> (x+6)^2 = 5
==> x^2 + 12x + 36 = 5
==> x^2 + 12x + 36 - 5 = 0 [add -5 to both the sides]
==> x^2 + 12x + 31 = 0
This equation does not have exact values for x.
Please check the equation.

Good Luck!!!