Solve the equation.

Question

Solve the equation.

 

√¯¯¯x-5 = x - 7

Answer 1

√(x-5) = x - 7

 

We need to square both sides to solve, but the equation we then get may have solutions that don't work in the original equation. So we will have to check for these "extraneous solutions".

 

Square both sides:

 

x - 5 = x^2 - 14x + 49

 

0 = x^2 - 15x + 54

 

0 = (x - 6)(x - 9)

 

x = 6 or x = 9

 

check:

 

√(6-5) =? 6 - 7

√(1) ≠ -1   ==> x = 6 IS NOT a solution; it's EXTRANEOUS.

 

√(9-5) =? 9 - 7

√(4) = 2  √  ==> x = 9 is the only solution.

Answer 2

Hi Keith;

I think this is...

√(x-5)=x-7

Let's square both sides...

[√(x-5)]=(x-7)²

x-5=(x-7)(x-7)

Let's FOIL the two parenthetical equations...

FIRST...(x)(x)=x²

OUTER...(-7)(x)=-7x

INNER...(x)(-7)=-7x

LAST...(-7)(-7)=49

x-5=x²-7x-7x+49

x-5=x²-14x+49

Let's subtract (x-5) from both sides...

(x-5)-(x-5)=x²-14x+49-(x-5)

0=x²-14x+49-x+5

0=x²-15x+54

Let's factor.

For the FOIL...

FIRST must be (x)(x)=x²

OUTER must add-up to -15x.

INNER must be either (27)(2) or (2)(27) or (6)(9) or (9)(6) and both numbers must be negative to have a positive result for the last, and a negative result for OUTER and INNER.

0=(x-9)(x-6)

Let's FOIL...

FIRST...(x)(x)=x²

OUTER...(x)(-6)=-6x

INNER...(-9)(x)=-9x

LAST...(-9)(-6)=+54

0=x²-6x-9x+54

0=x²-15x+54

0=(x-9)(x-6)

Either or both parenthetical equation(s) must equal zero...

0=(x-9)

9=x

0=(x-6)

6=x

 

Answer 3

Beautiful work by Vivian L. of Middletown, CT!

 

Student also needs to know the strategy of solution this type of equation.

 

Step 1:  square both side.   (or arrange into this format if it is given differently)

 

Step 2: rearrange to format a normal quadratic equation and then solve it accordingly.

 

I think recognizing this pattern and squaring it is the key to solving it!