Find all real solutions to the quadratic equation  x2 + 2 = x + 5

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x^2 + 2 = x+ 5

First we will rewrite the equation with the right side equal o:

==> x^2 + 2 - x - 5 = 0

Now we will rearrange:

==> x^2 -x - 3 = 0

Now we will use the formula to solve:

We know that:

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

==> a = 1  b = -1    c = -3

==> x1=  [ 1 + sqrt( 1- 4*1*-3) ]/ 2

             = ( 1+ sqrt(13) / 2

==> x2= ( 1- sqrt13)/ 2

Then the solution is :

 x = { (1+ sqrt13)/2  ,   (1- sqrt13)/ 2 }

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

x^2+x = x+5. To find the real solutions.

We subtract x+5 from both side of the given equation:

x^2+x-x-5 = 0

x^2 - 5 = 0

x^2 = 5.

We take square root of both sides:

x = sqrt5 , or x= -sqrt5. Both are real solutions of x.

Therefore x= sqrt5 and x = -sqrt are the real solutions os x^2+x = x+5.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-arrange the terms of the quadratic by subtracting x + 5 both sides:

x^2 + 2 - x - 5 = 0

We'll combine like terms:

x^2 - x - 3 = 0

We'll apply the quadratic formula:

x1 = [-b+sqrt(b^2 - 4ac)]/2a

x2 = [-b-sqrt(b^2 - 4ac)]/2a

We'll identify a,b,c:

a = 1 , b = -1 , c = -3

We'll substitute the coefficients a,b,c into the formula of the quadratic:

x1 = [1 + sqrt(1 + 12)]/2

x1 = (1+sqrt13)/2

x2 = (1-sqrt13)/2

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