Solve the equation . Show all work, and verify the solution(s).

$\sqrt{2x-5}-4=-x,x>5\div2$√2x−5−4=−x,x>5÷2

3 years ago

Answered By Khevna D

Sqrt(2x-5)= - x+4

2x-5=(-x+4)^{2}

2x-5=(-x+4)(-x+4)

2x-5= x^{2} - 4x - 4x +16

X^{2 }-8x +16 - 2x+5=0

X^{2 }-10x+21=0

(x-7)(x-3)=0

Therefore x= 3 and 7 hence, they are greater than 5/2

3 years ago

Answered By Albert S

The restriction on x is the non-permisable values (NPV); a negative under the radical sign is not allowed. Set the term under the radical sign equal to zero

(2x-5)>=0

2x-5+5>=0+5

2x>=5

2x/2>=5/2

x>=5/2

To solve this we don't like the radical sign. To get rid of it we get it by itself by adding 4 to each side. Once it is by itself on the RHS we square it. Whatever we do one side we have to do to the other.

sqrt(2x-5) - 4 = -x

sqrt (2x-5) - 4 + 4 = -x +4

sqrt(2x-5) = (-x+4)

(sqrt(2x-5))^{2}=(-x+4)^{2}

(2x-5)=(-x+4)(-x+4)

2x-5 = x^{2}-4x-4x+16

2x-5 = x^{2}-8x+16

switch sides just to make it easier

x^{2}-8x+16 = 2x -5

move all the terms to the RHS

x^{2}-8x-2x+16 = 2x -2x -5

x^{2}-10x+16 = -5

x^{2}-10x+16+5=-5+5

x^{2}-10x+21=0

Think of two numbers that mutlitply to give 21 and add to give -10; the numbers are -3 and -7; these are the factors

For the expression x>5 $\div2$÷2 , it is just letting us know that the solutions to 'x' needs to be greater than 5/2, or 2.5, so you can use that information to check your answer when you've worked through the problem.

Working through the problem,

$\sqrt{2x-5}-4=-x$√2x−5−4=−x

since you can't take the square root of a negative number, we want to isolate the term under the square root so that we can get rid of the negative number, and better simplify the equation. To have the square root term by itself on the left side, we can remove the '-4' from the left side of the equation by adding 4 and cancelling it out. what we do to the left side we have to do to the right side. this means that we put '+4' on the right side of the equation. now the equation will look like:

$\sqrt{2x-5}=-x+4$√2x−5=−x+4

The next thing we want to do is get rid of the square root sign. We can do that by squaring both sides of the equation.

2x-5=(-x+4) $^2$^{2}

Now with the square root eliminated we can expand the parenthises with the squared term

2x-5=(-x+4)(-x+4) then we can distribute the terms in the parenthises to create a polynomial expression

2x-5= $x^2-4x-4x+16$x^{2}−4x−4x+16 we can also combine like terms and create a simplified equation equal to 0. starting with -4x-4x simplifying to -8x

$2x-5=x^2-8x+16$2x−5=x^{2}−8x+16

We are getting closer to creating a polynomial equal to 0. we can keep combining like terms by getting the 'x' terms to the right side. here we will minus 2x from both sides.

-5= $x^2-10x+16$x^{2}−10x+16

Then the integers can be moved to the right side by adding 5 to both sides of the equation. this will allow the equation to be equal to 0

0= $x^2-10x+21$x^{2}−10x+21

$x^2-10x+21=0$x^{2}−10x+21=0

the values in this trinomial make it a perfect square trinomial. the solution for a perfect square trinomial equation can be found by figuring out what two integers can be added to = the middle value of the polynomial (-10x) and multiply to = the last value (21). In this case the values would be -7 and -3. (x-3)(x-7)=0

since the values in parenthises will be equal to zero anyways, they can be isolated individually to solve for x. 3 can be added to both sides to eliminate -3 from the left side and solve for x.

(x-3)=0 x=3

the same can be done with (x-7)=0. to solve for x, 7 can be added to both sides to isolate 'x' and eliminate -7 from the left side of the equation

3 years ago

## Answered By Khevna D

Sqrt(2x-5)= - x+4

2x-5=(-x+4)

^{2}2x-5=(-x+4)(-x+4)

2x-5= x

^{2}- 4x - 4x +16X

^{2 }-8x +16 - 2x+5=0X

^{2 }-10x+21=0(x-7)(x-3)=0

Therefore x= 3 and 7 hence, they are greater than 5/2

3 years ago

## Answered By Albert S

The restriction on x is the non-permisable values (NPV); a negative under the radical sign is not allowed. Set the term under the radical sign equal to zero

(2x-5)>=0

2x-5+5>=0+5

2x>=5

2x/2>=5/2

x>=5/2

To solve this we don't like the radical sign. To get rid of it we get it by itself by adding 4 to each side. Once it is by itself on the RHS we square it. Whatever we do one side we have to do to the other.

sqrt(2x-5) - 4 = -x

sqrt (2x-5) - 4 + 4 = -x +4

sqrt(2x-5) = (-x+4)

(sqrt(2x-5))

^{2}=(-x+4)^{2}(2x-5)=(-x+4)(-x+4)

2x-5 = x

^{2}-4x-4x+162x-5 = x

^{2}-8x+16switch sides just to make it easier

x

^{2}-8x+16 = 2x -5move all the terms to the RHS

x

^{2}-8x-2x+16 = 2x -2x -5x

^{2}-10x+16 = -5x

^{2}-10x+16+5=-5+5x

^{2}-10x+21=0Think of two numbers that mutlitply to give 21 and add to give -10; the numbers are -3 and -7; these are the factors

(x-3)(x-7)=0

set each term to zero

x-3=0 x-7=0

x=3 x=7

3 years ago

## Answered By Kareem B

$\sqrt{2x-5}-4=-x,x>5\div2$√2x−5−4=−x

$\sqrt{2x-5}=-x+4$√2x−5=−x+4

$\left(\sqrt{2x-5}\right)^2=\left(-x+4\right)^2$(√2x−5)

^{2}=(−x+4)^{2}$2x-5=\left(-x+4\right)\left(-x+4\right)$2x−5=(−x+4)(−x+4)

$2x-5=x^2-8x+16$2x−5=x

^{2}−8x+16$0=x^2-8x-2x+16+5$0=x

^{2}−8x−2x+16+5$x^2-10x+21=0$x

^{2}−10x+21=0$\left(x-7\right)\left(x-3\right)=0$(x−7)(x−3)=0

$x-7=0$x−7=0

$x-3=0$x−3=0

$x=7$x=7

$x=3$x=3

2 years ago

## Answered By Mat B

$\sqrt{2x-5}-4=-x,x>5\div2$√2x−5−4=−x,x>5÷2

For the expression x>5 $\div2$÷2 , it is just letting us know that the solutions to 'x' needs to be greater than 5/2, or 2.5, so you can use that information to check your answer when you've worked through the problem.

Working through the problem,

$\sqrt{2x-5}-4=-x$√2x−5−4=−x

since you can't take the square root of a negative number, we want to isolate the term under the square root so that we can get rid of the negative number, and better simplify the equation. To have the square root term by itself on the left side, we can remove the '-4' from the left side of the equation by adding 4 and cancelling it out. what we do to the left side we have to do to the right side. this means that we put '+4' on the right side of the equation. now the equation will look like:

$\sqrt{2x-5}=-x+4$√2x−5=−x+4

The next thing we want to do is get rid of the square root sign. We can do that by squaring both sides of the equation.

2x-5=(-x+4) $^2$

^{2}Now with the square root eliminated we can expand the parenthises with the squared term

2x-5=(-x+4)(-x+4) then we can distribute the terms in the parenthises to create a polynomial expression

2x-5= $x^2-4x-4x+16$x

^{2}−4x−4x+16 we can also combine like terms and create a simplified equation equal to 0. starting with -4x-4x simplifying to -8x$2x-5=x^2-8x+16$2x−5=x

^{2}−8x+16We are getting closer to creating a polynomial equal to 0. we can keep combining like terms by getting the 'x' terms to the right side. here we will minus 2x from both sides.

-5= $x^2-10x+16$x

^{2}−10x+16Then the integers can be moved to the right side by adding 5 to both sides of the equation. this will allow the equation to be equal to 0

0= $x^2-10x+21$x

^{2}−10x+21$x^2-10x+21=0$x

^{2}−10x+21=0the values in this trinomial make it a perfect square trinomial. the solution for a perfect square trinomial equation can be found by figuring out what two integers can be added to = the middle value of the polynomial (-10x) and multiply to = the last value (21). In this case the values would be -7 and -3. (x-3)(x-7)=0

since the values in parenthises will be equal to zero anyways, they can be isolated individually to solve for x. 3 can be added to both sides to eliminate -3 from the left side and solve for x.

(x-3)=0 x=3

the same can be done with (x-7)=0. to solve for x, 7 can be added to both sides to isolate 'x' and eliminate -7 from the left side of the equation

(x-7)=0 x=7