Expert: Josh - 2/27/2008

Question
I am really struggling with these few. Is there anyway you would be able to check my work and help me with these three? √(x) + √(x + 8) = 2
√(x - 3) + √(x + 5) = 2√y
2√(x - 8) + √(x - 5) = √(x + 7)
Is there anyway you would be able to check my work for these? √(2x-7)= -3 Answer: 8
√(8x-5) - √(7x-1) = 0 Answer: 4 x =
√(x² - 2x – 4) Answer: -2
√(3x – 2) = x – 2 Answer: 1 or 6
√(-x) – 6 = x Answer: -9 or -4
³√(2x +5) = -3 Answer: -16
√(x + 4) – 1 = √(x - 1) Answer: 5

Answer
Hi Sarah,

I'll do a couple to show you the way. You should then be able to attempt the remaining questions.

The main difficulty here is that the algebraic expressions contain the square root (sqrt) of "x-a", where "a" is some number. To solve for "x", we need to square both sides of the equation, after appropriate rearrangment of terms. This requires that you obtain the quadratic expansion on one side of the equation.

Consider squaring "a+b". (a+b)^2 = (a+b)*(a+b) = a^2+ 2ab +b^2.

In Question 1, we can rearrange the equation as follows:
sqrt(x)+sqrt(x+8)=2
sqrt(x)-2=-sqrt(x+8) ...Next, square both sides.
[sqrt(x)-2]^2 = [-sqrt(x+8)]^2 ..Hint: use quadratic expansion on LHS
x-4sqrt(x)+4=(x+8) ...Now simplify. Observe the two "x"'s cancel
-4sqrt(x)=4 ...divide both sides by -2
sqrt(x)=-1
This has no real-valued solution. i.e., you cannot take the square root of a real number to obtain a negative quantity, unless you work with complex numbers. Working with complex numbers, we first define i=sqrt(-1). If x=i^4, then, sqrt(x) = sqrt(i^2*i^2) = i^2 = -1 and it works out.

Q2. Not sure what this question actually asks for. However, you can draw a graph for ¡Ô(x - 3) + ¡Ô(x + 5) = 2¡Ôy.

Q3. Try using the technique introduced in Q1.
2¡Ô(x - 8) + ¡Ô(x - 5) = ¡Ô(x + 7)
4*(x-8)+4sqrt((x-8)(x-5))+(x-5)=(x+7)
4x-32 + 4sqrt((x-8)(x-5)) = 12 ...this is good, we can isolate the square root term by moving "4x-32" to the RHS, before we square both sides once again.
4sqrt((x-8)(x-5)) = 44-4x ...next, factorize
4sqrt((x-8)(x-5)) = 4(11-x) ...cancel common factor (the 4)
sqrt((x-8)(x-5)) = 11-x ...now square both sides
(x-8)(x-5) = (11-x)^2 ...can you take over from here? you will end up solving a quadratic equation.

Work space for you to have a go:







Complete solution (don't look):
x^2-13x+40 = 121-22x+x^2
9x = 81
x = 9

Q4: You tell me...what should we do to "sqrt(2x-7)= -3"?
Again, the idea is to solve for x. We cannot have x inside a square root. So, we square both sides and proceed from there. Doing this, we get
2x-7 = 9
2x = 16
x = 8.

Good luck with the rest:)