Solved on Sep 29, 2023

Find the values of $x$ where $y=x+\sqrt{x+6}$ and $y=6$ .

STEP 1

Assumptions1. We have two equations $y=x+\sqrt{x+6}$ and $y=6$ . We are to find all values of $x$ that satisfy both equations

STEP 2

Since both equations are equal to $y$ , we can set them equal to each other. This gives us a new equation to solve for $x$ .
$x + \sqrt{x+6} =6$

STEP 3

We want to isolate the square root term, so we subtract $x$ from both sides of the equation.
$\sqrt{x+6} =6 - x$

STEP 4

To get rid of the square root, we square both sides of the equation. Remember that $(a-b)^2 = a^2 -2ab + b^2$ .
$(\sqrt{x+6})^2 = (6 - x)^2$

STEP 5

implify both sides of the equation.
$x + =36 -12x + x^2$

STEP 6

Rearrange the equation to form a quadratic equation. A quadratic equation is in the form $ax^2 + bx + c =0$ .
$x^2 -13x +30 =0$

STEP 7

Factor the quadratic equation. We are looking for two numbers that multiply to $30$ and add to $-13$ .
$(x -10)(x -3) =0$

STEP 8

Set each factor equal to zero and solve for $x$ .
$x -10 =0 \quad \text{and} \quad x -3 =0$

STEP 9

olving the equations gives us two potential solutions for $x$ .
$x = \quad \text{and} \quad x =3$

STEP 10

We need to check these solutions in the original equation $y=x+\sqrt{x+6}$ to make sure they don't result in the square root of a negative number, as this would be undefined in the real number system.
For $x =10$ , we have $y =10 + \sqrt{10+6} =10 + \sqrt{16} =10 +4 =14$ , which is not equal to $6$ .
For $x =3$ , we have $y =3 + \sqrt{3+6} =3 + \sqrt{9} =3 +3 =6$ , which is equal to $6$ .
So, the only solution is $x =3$ .

Was this helpful?

Find the values of $x$ where $y=x+\sqrt{x+6}$ and $y=6$ .

STEP 1

Assumptions1. We have two equations $y=x+\sqrt{x+6}$ and $y=6$ . We are to find all values of $x$ that satisfy both equations

STEP 2

Since both equations are equal to $y$ , we can set them equal to each other. This gives us a new equation to solve for $x$ .
$x + \sqrt{x+6} =6$

STEP 3

We want to isolate the square root term, so we subtract $x$ from both sides of the equation.
$\sqrt{x+6} =6 - x$

STEP 4

To get rid of the square root, we square both sides of the equation. Remember that $(a-b)^2 = a^2 -2ab + b^2$ .
$(\sqrt{x+6})^2 = (6 - x)^2$

STEP 5

implify both sides of the equation.
$x + =36 -12x + x^2$

STEP 6

Rearrange the equation to form a quadratic equation. A quadratic equation is in the form $ax^2 + bx + c =0$ .
$x^2 -13x +30 =0$

STEP 7

Factor the quadratic equation. We are looking for two numbers that multiply to $30$ and add to $-13$ .
$(x -10)(x -3) =0$

STEP 8

Set each factor equal to zero and solve for $x$ .
$x -10 =0 \quad \text{and} \quad x -3 =0$

STEP 9

olving the equations gives us two potential solutions for $x$ .
$x = \quad \text{and} \quad x =3$

STEP 10

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the values of xxx where y=x+x+6y=x+\sqrt{x+6}y=x+x+6​ and y=6y=6y=6.

STEP 1

Assumptions1. We have two equations y=x+x+6y=x+\sqrt{x+6}y=x+x+6​ and y=6y=6y=6 . We are to find all values of xxx that satisfy both equations

STEP 2

Since both equations are equal to yyy, we can set them equal to each other. This gives us a new equation to solve for xxx.x+x+6=6x + \sqrt{x+6} =6x+x+6​=6

STEP 3

We want to isolate the square root term, so we subtract xxx from both sides of the equation.x+6=6−x\sqrt{x+6} =6 - xx+6​=6−x

STEP 4

To get rid of the square root, we square both sides of the equation. Remember that (a−b)2=a2−2ab+b2(a-b)^2 = a^2 -2ab + b^2(a−b)2=a2−2ab+b2.(x+6)2=(6−x)2(\sqrt{x+6})^2 = (6 - x)^2(x+6​)2=(6−x)2

STEP 5

implify both sides of the equation.x+=36−12x+x2x + =36 -12x + x^2x+=36−12x+x2

STEP 6

Rearrange the equation to form a quadratic equation. A quadratic equation is in the form ax2+bx+c=0ax^2 + bx + c =0ax2+bx+c=0.x2−13x+30=0x^2 -13x +30 =0x2−13x+30=0

STEP 7

Factor the quadratic equation. We are looking for two numbers that multiply to 303030 and add to −13-13−13.(x−10)(x−3)=0(x -10)(x -3) =0(x−10)(x−3)=0

STEP 8

Set each factor equal to zero and solve for xxx.x−10=0andx−3=0x -10 =0 \quad \text{and} \quad x -3 =0x−10=0andx−3=0

STEP 9

olving the equations gives us two potential solutions for xxx.x=andx=3x = \quad \text{and} \quad x =3x=andx=3

STEP 10

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the values of $x$ where $y=x+\sqrt{x+6}$ and $y=6$ .

Assumptions1. We have two equations $y=x+\sqrt{x+6}$ and $y=6$ . We are to find all values of $x$ that satisfy both equations

Since both equations are equal to $y$ , we can set them equal to each other. This gives us a new equation to solve for $x$ .
$x + \sqrt{x+6} =6$

We want to isolate the square root term, so we subtract $x$ from both sides of the equation.
$\sqrt{x+6} =6 - x$

To get rid of the square root, we square both sides of the equation. Remember that $(a-b)^2 = a^2 -2ab + b^2$ .
$(\sqrt{x+6})^2 = (6 - x)^2$

implify both sides of the equation.
$x + =36 -12x + x^2$

Rearrange the equation to form a quadratic equation. A quadratic equation is in the form $ax^2 + bx + c =0$ .
$x^2 -13x +30 =0$

Factor the quadratic equation. We are looking for two numbers that multiply to $30$ and add to $-13$ .
$(x -10)(x -3) =0$

Set each factor equal to zero and solve for $x$ .
$x -10 =0 \quad \text{and} \quad x -3 =0$

olving the equations gives us two potential solutions for $x$ .
$x = \quad \text{and} \quad x =3$