Mathematics, 22.06.2019 12:50 jaxondbagley

Solve the equation for x. if a solution is extraneous, be sure to identify it in your final answer. square root of the quantity x minus 2 end quantity plus 8 equals x

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Answer from: sophiagardens227

11 only true real solution

6 is extraneous (not a solution to the original equation)

Step-by-step explanation:

sqrt(x-2)+8=x

Get the radical by itself first (the square root thing)

Subtract 8 on both sides

sqrt(x-2)=x-8

Now square both sides

x-2=(x-8)^2

x-2=x^2-16x+64 (multiply the (x-8)^2 out aka multiply (x-8) and (x-8) use foil)

Subtract x and add 2 on both sides

0=x^2-17x+66

Find two numbers that multiply to be 66 and add to be -17

66=6(11) so 66=-6(-11) and -6+(-11)=-17

So the factored form of the equation is

0=(x-6)(x-11)

x=6 or x=11

Check to see if any of those are bad solutions (solutions that actually don't work-aka extraneous)

Check x=6

sqrt(6-2)+8=6

sqrt(4)+8=6

2+8=6

Not right so 6 is extraneous

Check x=11

sqrt(11-2)+8=11

sqrt(9)+8=11

3+8=11

Good

So 11 is the only true real solution

Answer from: 23kaevans

-2 is a real solution to the equation

-5 is extraneous (not a solution to the original)

Step-by-step explanation:

sqrt(x+6)-4=x

Get square root thing by itself

Add 4 on both sides

sqrt(x+6)=x+4

Square both sides

x+6=(x+4)^2

x+6=x^2+8x+16

Subtract x and subtract 6 on both sides

0=x^2+7x+10

What two integers multiply to be 10 and add to be 7

Hope you said 2 and 5

0=(x+2)(x+5)

x=-2 or x=-5

Check!

x=-2 gives:

sqrt(-2+6)-4=-2

sqrt(4)-4=-2

2-4=-2

-2=-2 true so x=-2 is good

Check!

x=-5 gives:

sqrt(-5+6)-4=-5

sqrt(1)-4=-5

1-4=-5

-3=-5 not true so x=-5 is bad

Answer from: damienlopezram

ANSWER

Extraneous solution: x=6

Real solution: x=11

EXPLANATION

The given expression is

$\sqrt{x - 2} + 8 = x$

Add -8 to both sides:

$\sqrt{x - 2} + 8 + - 8= x + - 8$

$\implies\sqrt{x - 2} = x - 8$

Square both sides.

$\implies(\sqrt{x - 2} )^{2} =( x - 8)^{2}$

$x - 2=( x - 8)^{2}$

We expand the to get

$x - 2 = {x}^{2} - 16x + 64$

Write in standard quadratic form.

${x}^{2} - 16x - x + 64 + 2 = 0$

${x}^{2} - 17x + 66 = 0$

Factor to get:

(x - 6)(x - 11) = 0

$x = 6 \: or \: \: x = 11$

We check for extraneous solutions by substituting each value of x into the original equation.

When x=6

$\sqrt{6 - 2} + 8 = 6$

$\sqrt{4} + 8 =6$

$2 + 8 = 10 \ne8$

Hence x=6 is an extraneous solution.

When x=11

$\sqrt{11- 2} + 8 = 11$

$\sqrt{9} + 8 = 11$

3 + 8 = 11

This statement is true.

Hence x=11 is the only solution.

Answer from: kinglightskin2k

Step-by-step explanation:

$\sqrt{x-3} +5=x\\\sqrt{x-3} =x-5\\squaring ~both~sides\\x-3=x^2-10x+25\\x^2-10x-x+25+3=0\\x^2-11x+28=0\\x^2-7x-4x+28=0\\x(x-7)-4(x-7)=0\\(x-7)(x-4)=0\\x=7,4$

put x=7 in the given equation

$\sqrt{7-3} +5=7\\\sqrt{4} +5=7\\2+5=7\\7=7$

which is true .

∴ x=7 is a solution of the given eq.

now put x=4 in the given eq.

$\sqrt{4-3} +5=7\\1+5=7\\6=7\\$

which is not true.

∴x=4 is an extraneous solution.

Answer from: starxx05235

Step-by-step explanation:

√x+6 -4=x

Move -4 to the R.H.S

√x+6=x+4

Take square on both sides

(√x+6)²=(x+4)²

solve (x+4)² by whole square formula

x+6=x²+8x+16

Now shift all the terms at one side.

x²+8x+16-x-6=0

x²+7x+10=0

Now solve the equation by factorization method.

x²+5x+2x+10=0

x(x+5)+2(x+5)=0

(x+5)(x+2)=0

x+5=0

x= -5

x+2=0

x=-2

We obtained two possible solutions. Now let's check if any of them is extraneous.

The original equation is √x+6 -4=0

when x = -5 we get,

√-5+6 -4= -5

√1 -4= -5

1-4 = -5

-3 ≠ -5

So x = -5 is not a solution. It is an extraneous solution.

when x= -2 we get,

√-2+6 -4= -2

√4 -4 = -2

2 - 4= -2

-2 = -2

So x= -2 is a solution

Answer from: jose55221

The solution is x=11

Step-by-step explanation:

we have

$\sqrt{x - 2} + 8 = x$

Solve for x

Subtract 8 both sides

$\sqrt{x - 2} + 8-8 = x-8$

$\sqrt{x - 2}= x-8$

Squared both sides

$(\sqrt{x - 2})^2= (x-8)^2$

(x - 2)= (x-8)^2

Expand the right side

(x - 2)= (x^2-16x+64)

x^2-17x+66=0

Solve the quadratic equation by graphing

The solutions are x=6, x=11

see the attached figure

Verify the solutions

For x=6

substitute the value of x in the original equation

$\sqrt{6 - 2} + 8 = 6$

$\sqrt{4} + 8 = 6$

2 + 8 = 6

$10\neq 6$

therefore

x=6 -----> is an extraneous solution

For x=11

substitute the value of x in the original equation

$\sqrt{11 - 2} + 8 = 11$

$\sqrt{9} + 8 = 11$

3 + 8 = 11

11=11 ---> is true

therefore

The solution is

x=11

show all work to solve the equation for x. if a solution is extraneous, be sure to identify it in yo

Answer from: luhhbitilia566

The solution of the equation is x = 7 only

Step-by-step explanation:

* Lets solve this equation and then check the answer

∵ √(x - 3) + 5 = x

- At first subtract 5 from both sides

∴ √(x - 3) = x - 5

- To cancel the square root, square the two sides

∵ [√(x - 3)]² = (x - 5)²

- cancel the square root of the left hand side and open the bracket

power 2 in the right hand side

∴ x - 3 = (x)² + 2(-5)x + (-5)²

∴ x - 3 = x² - 10x + 25

- Add 3 and subtract x from both sides

∴ 0 = x² - 11x + 28

- Its quadratic equation lets factorize it into two brackets

∵ x × x = x²

∵ -4 × -7 = 28

∵ -4x + -7x = -11x

∴ x² - 11x + 28 factorized to (x - 7)(x - 4)

- Equate the two factors by 0

∵ x² - 11x + 28 = 0

∴ (x - 7)(x - 4) = 0

- Let each factor = 0

∴ x - 7 = 0 ⇒ add 7 to both sides

∴ x = 7

x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

* Lets check these value of x if both are solution to the given equation

- Substitute the value of x in the given equation

∵ √(x - 3) + 5 = x

- If x = 7

∴ √(7 - 3) + 5 = 7

∴ √4 + 5 = 7

∴ 2 + 5 = 7

∴ 7 = 7

- The left hand side = the right hand side

∴ x = 7 is a solution of the given equation

- If x = 4

∴ √(4 - 3) + 5 = 4

∴ √1 + 5 = 4

∴ 1 + 5 = 4

∴ 6 ≠ 4

- The left hand side ≠ the right hand side

∴ x = 4 is not a solution of the given equation

Answer from: alex43079

The solutions to the equation $\sqrt{x+6}-4=x$ are x = -5 or x = -2.

Further Explanation:

We first isolate the square root. To do this, we cancel the 4 by adding it to each side:

$\sqrt{x+6}-4+4=x+4\\\\\sqrt{x+6}=x+4$

Next we want to cancel the square root sign. To do this, we square both sides of the equation:

$(\sqrt{x+6})^2=(x+4)^2\\\\x+6=(x+4)(x+4)\\\\x+6=x(x)+4(x)+4(x)+4(4)\\\\x+6=x^2+4x+4x+16\\\\x+6=x^2+8x+16$

Now we want to gather all of the variables on one side of the equation. The simplest way to do this will be to subtract x from each side:

$x+6-x=x^2+8x+16-x\\\\6=x^2+7x+16$

To solve quadratic equations (those with an x² term), we want ot set them equal to 0. To do this, we subtract 6 from each side:

$6-6=x^2+7x+16-6\\\\0=x^2+7x+10$

We can now factor the right hand side of the equation. To do this, we find factors of the constant, 10, that sum to the x term, 7. Looking at all of the factors of 10, we see that 5 and 2 have a sum of 7; this gives us

0=(x+5)(x+2)

Using the zero product property, we know that in order to have a product of 0, one or more of the factors must be equal to 0. This tells us that we have

$x+5=0\text{ or }x+2=0$

To solve the first equation, we subtract 5 from each side:

$x+5-5=0-5\\\\x=-5$

To solve the second equation, we subtract 2 from each side:

$x+2-2=0-2\\\\x=-2$

This means the solutions are x = -5 or x = -2.

Learn More

Learn more about quadratic equations at:

Learn more about factoring at:

Learn more about radical equations at:

Keywords: radical equation, identify extraneous solution, square root equation

Answer from: seanholmes91405

There is one extraneous solution at x = 4

The only solution is x=7

Step-by-step explanation:

Given:

The equation to solve is given as:

$\sqrt{x-3}+5=x$

In order to solve, we will first isolate the radical part.

Adding -5 on both the sides, we get:

$\sqrt{x-3}+5-5=x-5\\\sqrt{x-3}=x-5$

Now, squaring both the sides, we get:

$(\sqrt{x-3})^2=(x-5)^2\\\\x-3=x^2+25-10x\\\\x^2-10x-x+25+3=0\\\\x^2-11x+28=0\\\\(x-4)(x-7)=0\\\\x=4\ or\ x=7$

Now, let us check whether the calculated values satisfy the equations or not.

Plug in 4 and 7 for 'x' and check the value of the equation.

At x=4 ,

$\sqrt{4-3}+5=4\\\sqrt1+5=4\\1+5=4\\6=4$

But, 6 is not equal to 4 and hence 4 is not a solution to the given equation. It is an extraneous solution for the given equation.

Now, at x=7

$\sqrt{7-3}+5=7\\\sqrt4+5=7\\2+5=7\\7=7(True)$

Therefore, x=7 is a solution to the equation.

Hence, there is one extraneous solution at x = 4

The only solution is x=7

Answer from: Ryzuko

x = 3

Step-by-step explanation:

This is hard to explain so i'll show you the work

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