Help with solving radical equations?

Question

26. 2√(x+7)=x-1 or 2√(x+7)+10=x-1

answer is 9 which expression is correct?

29. √(3x+3)+√(x+2)=5

30. √(x)+6=√(x+72)

Answer 1

26.

 2√(x+7)+10=x-1

 √(4x+28)+10=x-1

 subtract 10 from each side

√(4x+28) = x-11

square both sides

4x + 28 = (x -11)²

4x + 28 = x² - 22x +121

subtract 4x from each side

28 = x² - 26x +121

subtract 28 from each side

x² - 26x + 93 = 0

Separate variables and constants aside by subtracting 93 from each side.

 x² - 26x = - 93

 To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression

 Here x coefficient = -26. so, (half the x coefficient)² = (-26/2)²= 169 

Add 169 to each side

x² - 26x  +169= - 93 +169

(x - 13)²= 124

Take square root both sides

x - 13= √124

x = 13 ± √124

x = 13 + √124 or x = 13 - √124

Plugin the values none of the vaue is working. so no solution

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Let's try with expression  2√(x+7)=x-1, then answer is 9.

 2√(x+7)=x-1

 √(4x+28)=x-1

square both sides

4x + 28 = (x -1)²

4x + 28 = x² - 2x +1

subtract 4x from each side

28 = x² - 6x + 1

subtract 28 from each side

x² - 6x + 27 = 0

x² - 9x +3x - 27 = 0

x ( x -9 ) + 3 ( x -3 ) = 0

(x - 9) ( x + 3) =0

x = 9 or x = -3

 

Comments

2√(x+7) = x-1 is the correct expression.

Answer 2

30.

√x+6=√(x+72)

square both sides

x + 12√x + 36 = x +72

cancel common terms

12√x + 36 =  +72

subtract 36 from each side

12√x =  36

divide each side by 12

√x = 3

square both sides

x = 9

Answer 3

29) The equation is √(3x + 3) + √(x + 2) = 5

√(3x + 3) + √(x + 2) = 5

Squaring on both sides we get,

[√(3x + 3) + √(x + 2)]² = 5²

[√(3x + 3)]² + [√(x + 2)]² + 2*√(3x + 3)*√(x + 2) = 5²

(3x + 3) + (x + 2) + 2*√[(3x + 3)*(x + 2)] = 5²

3x + 3 + x + 2 + 2*√[(3x + 3)*(x + 2)] = 25

4x + 5 + 2√[(3x + 3)*(x + 2)] = 25

2√[(3x + 3)*(x + 2)] = 25 - 4x - 5

2√[(3x + 3)*(x + 2)] = 20 - 4x

Squaring on both sides we get,

{2√[(3x + 3)*(x + 2)]}² = (20 - 4x)²

4[(3x + 3)*(x + 2)] = 20² + (4x)² - 2*20*4x

4[3x² + 6x + 3x + 6] = 20² + (4x)² - 2*20*4x

4[3x² + 9x + 6] = 20² + (4x)² - 2*20*4x

4[3x² + 9x + 6] = 400 + 16x² - 160x

4[3x² + 9x + 6] = 4[100 + 4x² - 40x]

3x² + 9x + 6 = 100 + 4x² - 40x

3x² + 9x + 6 - 100 - 4x² + 40x = 0

-x² + 49x - 94 = 0

x² - 49x + 94 = 0

x² - 47x - 2x + 94 = 0

x(x - 47) - 2(x - 47) = 0

(x - 47)(x - 2) = 0

x - 47 = 0 or x - 2 = 0

x = 47 or x = 2

Check:

If x = 47 then  √(3*47 + 3) + √(47 + 2) = √(141+3) + √49 = 12 + 7 = 19 (False)

If x = 2 then  √(3*2 + 3) + √(2 + 2) = √9 + √4 = 3 + 2 = 5 (True)

Therefore x = 2 is the solution.

Answer 4

26. The radical equation is 2√(x + 7) = x - 1, and solve for x.

Write the number 2 as √4.

√4√(x + 7) = x - 1

√[ 4(x + 7) ] = x - 1

√(4x + 28) ] = x - 1

Square each side.

4x + 28 = (x - 1)²

4x + 28 = x² - 2x + 1

Subtract (4x + 28) from each side.

4x + 28 - (4x + 28) = x² - 2x + 1 - (4x + 28)

0 = x² - 6x - 27

0 = x² - 9x + 3x - 24

0 = x(x - 9) + 3(x - 9)

0 = (x + 3)(x - 9)

Apply zero product property : If ab = 0 then both a and b zero (or) a = 0 or b = 0.

x + 3 = 0 and x - 9 = 0

x = - 3 and x = 9.

It is very important that check the solution. Sometimes you will obtain a number that does not satisfy the original equation. Such a number is called an extraneous solution.

Subtitue the value of x = - 3 in the original radical equation, 2√(x + 7) = x - 1.

2√[ ( - 3) + 7 ] ≟ (- 3) - 1

2√[4] ≟ - 4

2(2) ≟ - 4

4 ≟ - 4

The above statement is false, the value of x = - 3 is not a solution of the original equation.

Substitute the value of x = 9 in the original radical equation, 2√(x + 7) = x - 1.

2√[ (9) + 7 ] ≟ (9) - 1

2√(16) ≟ 8

2(4) ≟ 8

8 = 8

The above statement is true, the value of x = 9 is a solution of the original equation.