Find x:If sqrt( x^2 - 5x + 4 )=sqrt( x^2 + 3x -2)  find x.

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to use the property of square root function such that:

`sqrt(x^2 - 5x + 4) = sqrt(x^2 + 3x - 2) => x^2 - 5x + 4 = x^2 + 3x -2`

Reducing duplicate terms yields:

`-5x + 4 = 3x - 2 => -5x - 3x = -4 - 2 => -8x = -6 => x = 6/8 => x = 3/4`

Testing the value `x = 3/4` in equation yields:

`sqrt((3/4)^2 - 5*3/4 + 4) = sqrt((3/4)^2 + 3*3/4 - 2)`

`sqrt(9/16 - 15/4 + 4) = sqrt(9/16 + 9/4 - 2)`

`sqrt((9 - 60 + 64)/16) = sqrt((9 + 36 - 32)/16)`

`sqrt(13/16) = sqrt(13/16)`

Hence, evaluating the solution to the equation, under the given conditions, yields `x = 3/4.`

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll impose the constraint of existence of square roots:

x^2 - 5x + 4>=0

We'll re-write the inequality:

(x-1)(x-4)>=0

The intervals of admissible values are (-infinte ,1)U(4 , +infinite).

x^2 + 3x -2>=0

x1 = [-3+sqrt(9+8)]/2

x1 = (-3+sqrt17)/2

x2 = (-3+sqrt17)/2

The intervals of admissible values are (-infinte ,(-3+sqrt17)/2)U((-3+sqrt17)/2 , +infinite).

The solution of the equation has to belong to the intervals of admissible values:

(-infinte ,(-3+sqrt17)/2)U(4 , +infinite).

Now, we can solve the equation by raising to square both sides, to get rid of the square roots:

( x^2 - 5x + 4 ) = ( x^2 + 3x -2)

We'll eliminate and combine like terms:

-5x - 3x + 4 + 2 = 0

-8x + 6 = 0

-8x = -6

x = 6/8

x = 3/4

We notice that the value for x doesn't belong to the interval of admissible values for x, so the equation has no solution.

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question