solve the equation:  x^2 = 3x+40

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

Posted on

x^2 = 3x+40

First we will move all terms to the left side:

==> x^2 - 3x - 40 = 0

Now we can either factor or use the formula:

To fctor:

x^2 - 3x - 40 = (x-8)(x+5)= 0

==> x1= 8

== x2= -5

OR:

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

x1= [3 + sqrt(9 +160)[/2 = [ 3+ 13)/2 = 8

x2= [ 3- 13]/2 = -10/2 = -5

==> x= { -5, 8}

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve the equation x^2=3x+40.

Rearrange the equation as x^2-3x-40 = 0

We split the middle term -3x = -8x and 5x to factor the by grouping:

x^2-8x+5x -40 = 0

x(x-8) +5(x- 8)) = 0

(x-8)(x+5) = 0

x-8 = 0 or x+5 = 0

x-8 = 0 gives x= 8

x+5 = 0 gives x = -5.

Therefore x= 8 or x= -5 are the solutions of the given equation.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll subtract 3x + 40 both sides:

x^2 - 3x-40 = 0

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

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

We'll identify a,b,c:

a=1

b=-3

c=-40

We'll calculate delta:

delta = b^2-4ac

delta = 9+160

delta = 169

Since delta is positive, we'll have 2 distinct real roots of the equation x^2 - 3x-40 = 0.

x1=[3+sqrt(9+160)]/2

x1 = (3+13)/2

x1 = 8

x2 =  -5

The real roots are: {-5 ; 8}.

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We have to solve the equation: x^2 = 3x + 40

subtract 3x + 40 from both sides

=> x^2 - 3x - 40 =0

=> x^2 - 8x + 5x - 40 =0

=> x(x-8) + 5(x-8) = 0

=> (x+5)(x-8) =0

=> x + 5 = 0  or x -8 =0

=> x = -5 or  = 8

Therefore x can take the values -5 and 8.