Square Roots?

Question

The square root of 2x-7 minus the square root of x minus 3 is equal to -2

Can someone show me the steps. Online class,and it's not explaning it.

Answer 1

√(2x-7) - √x - 3 = -2
First, add 3 to both sides:
√(2x-7) - √x = 1
Square both sides:
(2x-7) -2√x√(2x-7) + x = 1
Subtract 3x-7 and simplify:
-2√(2x²-7x) = -3x+8
Multiply by -1/2:
√(2x²-7x) = 3x/2-4
Square both sides again:
2x²-7x = 9x²/4 - 12x + 16
Subtract 2x²-7x:
0 = x²/4 - 5x + 16
Multiply by 4:
x² - 20x + 64 = 0
Add 36:
x² - 20x + 100 = 36
Factor:
(x-10)² = 36
Take the square root:
x-10=±6
Add 10:
x=10±6

Now test both possible solutions:

x=4:

√1 - √4 - 3 = -4 ≠ -2

x=16:

√25 - √16 - 3 = -2

So x=16

Answer 2

Original equation:
√(2x-7) - √(x-3) = -2, add √(x-3) from both sides
√(2x-7) = √(x-3) - 2, square both sides
2x-7 = x-3 -2√(x-3) - 2√(x-3) + 4, combine like terms
2x-7= x- 4√(x-3) + 1, take x and the integers to the left side
x - 8 = -4√(x-3), square both sides
(x-8)^2 = 16(x-3), distribute (multiply all out)
x^2-16x+64=16x-48, set equation equal to zero
x^2 -32x+112=0, factor
(x-28)(x-4)=0, set each term equal to zero and solve
x-28=0 or x-4=0
x=28 or x=4.
Plug each of these back in to the original equation.
When you do this, you do not get a square root of a negative number so both answers work and therefore are correct answers for this equation.

Answer 3

Restating your question, for syntax sake:

sqrt(2x-7) - sqrt(x - 3) = -2

First, bring one of the square roots over to the right hand side. Then you get:

sqrt(2x-7) = sqrt(x-3) - 2

Square both sides, and note that squaring a square root eliminates the square root. So we get:

2x - 7 = (sqrt(x-3) - 2)^2

Let's expand the right hand side so we know we have to deal with FOIL.

2x - 7 = (sqrt(x-3) - 2) (sqrt(x-3) - 2)

Foiling out the right hand side, we get

2x - 7 = (x - 3) - 2*sqrt(x-3) - 2*sqrt(x-3) + 4

Combining the two square roots into one, we get

2x - 7 = (x - 3) - 4*sqrt(x-3) + 4

Now, we just isolate the term with the square root; shifting everything but the square root term to the left hand side, we get

x - 8 = 4*sqrt(x-3)

Square both sides, to get

(x-8)^2 = 16 * (x-3)

Expanding the left hand side and the right hand side, we get

x^2 - 16x + 64 = 16x - 48

And then shifting everything to the left hand side, we get

x^2 - 32x + 112 = 0

You have to use the quadratic formula at this point, and test your solutions back into the original equation, because the solution may or may not work.

Answer 4

√(2x-7)-√(x-3)=-2 add √(x-3) to each side
√(2x-7)=√(x-3)-2 square both sides
2x-7=x-3+4-4√(x-3) subtract x+1 from each side
x-8=-4√(x-3) square both sides
x^2-16x+64=16(x-3)=16x-48 subtract 16x-48 from both sides
x^2-32x+96=0
x=(32+/-√(1024-384))/2
x=16+/-(1/2)√640=16+/-4√10
x=16+4√10
x=16-4√10

Answer 5

sqrt(2x-7) -sqrt(x-3)= -2
sqrt(2x-7) = sqrt(x-3) -2
Now square both sides of equation, getting:
2x-7 =x-3 -4Sqrt(x-3) +4
2x-7-x+3-4 =-4sqrt(x-3)
x-8= -4sqrt(x-3)
Now square both sides again getting:
x^2 -16x + 64 =16x-48
x^2-32x +112=0
(x-4)(x-28) =0
x-4=0 so x=4, and
x-28=0 so x = 28

Answer 6

√(2x - 7) - √(x - 3) = 0
√(2x - 7) = √(x - 3)
Squaring both sides,
2x - 7 = x - 3
2x = 4
x = 2

Answer 7

If you get all your answers here then you might pay the price later. Many schools () give all the new freshman a basic math test. If you score low enough you have to take remedial algebra.

Answer 8

squaring
(2x-7)+(x-3)-2[(2x-7)(x-3)]^1/2=4
3x-10-2(2x^2-13x+21)^1/2=4
-2(2x^2-13x+21)^1/2=-3x+14
2(2x^2-13x+21)^1/2=3x-14
squaring
4(2x^2-13x+21)=9x^2+196-84x
8x^2-52x+84=9x^2-84x+196
x^2-32x+112=0
solve for x
(x-28)(x-4)=0
x=28 or 4