Solve for x. (A) `x^2-4=x` (B) `x^2-5=x`  

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lemjay | High School Teacher | (Level 3) Senior Educator

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(A) `x^2-4=x`

To solve for x, express the equation in quadratic form `ax^2+bx+c=0` . To do so, subtract both sides by x in order to make the right side zero.

   `x^2`  `-`  `4`  `=`  `x`

            `-x`    `-x`

  `-----------`

   `x^2-x-4=0`          

Then, use the quadratic formula to solve for x.

`x=[-b+-sqrt(b^2-4ac)]/(2a)=[-(-1)+-sqrt((-1)^2-4*1(-4))]/(2*1)=[1+-sqrt17]/2`

Hence, the values of x are:

`x_1 = (1+sqrt17)/2`                  and                 `x_2=(1-sqrt17)/2`

(b) `x^2 - 5 =x`

(Solution is the same with A.)

Express the equation in quadratic form `ax^2+bx+c=0` .

   `x^2`  `-`  `5`  `=`  `x`  

           `-x`     `-x`

  `-----------`

   `x^2-x-5=0`

Use the quadratic formula to solve for x.

`x=[-b+-sqrt(b^2-4ac)]/(2a) = [-(-1)+-sqrt((-1)^2-4*1(-5))]/(2*1) = [1+-sqrt21]/2`

Hence, the values of x are:

`x_1=(1+sqrt21)/2`                  and                `x_2=(1-sqrt21)/2`

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