Question

Solving Square Root Equations

Solve the following equations. Be sure to label any extraneous solutions.

Answer 1

when you have to remove the square root you have to power the both sides of the equation.

1)6x= x^2

0 = x^2 - 6x

0= x(x-6)

x=0 or x-6=0

x=6

2)
`√(5x-6)` = x

5x-6=x^2

x^2 -5x+6=0

(x-3)(x-2)=0

x-3=0 or x-2= 0

x=3 or x=2

3)

`√(2x-1)` = x-3

2x-1 = (x-3)^2

2x-1= x^2 - 6x +9

x^2 -8x +10=0

this equation can't solve by factoring method easily. so we have to use the completing square method.

x^2-8x+10=0

x^2-8x= -10

x^2 - 8x+(-4)^2 = -10+ ( -4)^2

(x-4)^2 = 6

`x= 4 +√(6)` or
`x= 4 -√(6)`

4)
`x= 2 + √(2x-11)`

x^2= 4+4
`√(2x-11)` + 2x-11

x^2= 7+ 2x +4
`√(2x-11)`

x^2-2x-7= 4
`√(2x-11)`

(x^2-2x-7)^2= 16 (2x-11)

x^4 +4x^2 +49-4x^3 + 28x - 14x^2 = 32x-176

x^4 - 4x^3 - 10x^2 - 4x + 225= 0

x(x^3 -4x^2 - 10x-4)+225=0

this is a power 4 th equation this equation can solve by normal steps.

Answer 2

Explanation:

1) To solve the equation √(6x) = x, we can first square both sides of the equation to eliminate the square root:

(√(6x))^2 = x^2

6x = x^2

Next, we can rearrange the equation into standard quadratic form by subtracting 6x from both sides:

x^2 - 6x = 0

Now, we can factor out an x from the left-hand side of the equation:

x(x - 6) = 0

Setting each factor equal to zero, we find two solutions:

x = 0 or x - 6 = 0

Therefore, the solutions to the equation √(6x) = x are x = 0 and x = 6.

2)To solve √(5x-6) = x, we can square both sides of the equation:

(√(5x-6))^2 = x^2

5x-6 = x^2

Rearranging this quadratic equation to the standard form ax^2 + bx + c = 0, we get:

x^2 - 5x + 6 = 0

This can be factored into:

(x - 2)(x - 3) = 0

Therefore, the solutions to the equation √(5x-6) = x are x = 2 and x = 3.

We should check if these solutions satisfy the original equation:

When x=2: √(5x-6) = √(5(2)-6) = √4 = 2, which satisfies the equation.

When x=3: √(5x-6) = √(5(3)-6) = √9 = 3, which also satisfies the equation.

Therefore, the solutions are x = 2 and x = 3.

3) To solve the equation √(2x-1) = x-3, we can square both sides of the equation to eliminate the square root:

(√(2x-1))^2 = (x-3)^2

Simplifying the left-hand side gives:

2x-1 = (x-3)^2

Expanding the right-hand side gives:

2x-1 = x^2 - 6x + 9

Rearranging the equation gives:

x^2 - 8x + 10 = 0

We can solve this quadratic equation :

x^2 - 8x + 10 = 0

Now we can solve this quadratic equation by using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = -8, and c = 10. Substituting these values into the formula:

x = (-(-8) ± √((-8)^2 - 4(1)(10))) / 2(1)

Simplifying:

x = (8 ± √(64 - 40)) / 2

x = (8 ± √24) / 2

x = 4 ± √6

Therefore, the solutions to the equation √(2x - 1) = x - 3 are x = 4 + √6 and x = 4 - √6.

4) has no real solutions.