Find the solution for the following quadratic equations:   a) 3x^2 + 9x - 27 = 0     b) x^2 + 2x - 15 = 0

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

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a) 3x^2 + 9x - 27 = 0 

First let us divide by 3 to simplify the equation:

==> x^2 + 3x - 9) = 0

Now we will use the formula to solve.

We know that:

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

Let us substitute.

x1 = ( -3 + sqrt(9 -4*-9) / 2 = ( -3 + sqrt(45)/2

==> x1 = (-3+3sqrt5)/2

==>x2= (-3-3sqrt5)/2

   

 b) x^2 + 2x - 15 = 0

To solve the equation we will use the factoring method.

First, we will factor the equation.

==> ( x+5) ( x-3) = 0

==> x1= -5

==> x2= 3

==> x values are { -5, 3}

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the roots of the quadratic, we'll use the quadratic formula:

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

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

Where a,b,c are the coefficients of the quadratic.

We'll identify a,b,c:

a = 3

b = 9

c = -27

We'll substitute them into the formula:

x1 = [-9+sqrt(81 + 324)]/6

x1 = (-9+sqrt405)/6

x1 = 9(-1 + sqrt5)/6

x1 = 3(-1+sqrt5)/2

x2 = 3(-1-sqrt5)/2

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the solution to a) 3x^2 + 9x - 27 = 0     b) x^2 + 2x - 15 = 0.

a)

3x^2+9x-27 = 0.

We divide by 3 to make the equation in smallest terms. This won't affect the solution.

(3x^2+9x+27)/3 = 0.

x^2+3x-9 = 0.

We add (3/2)^2 and subtract (3/2)^2.

x^2+3x+(3/2)^2 - (3/2)^2 -9 = 0.

(x-3/2)^2  = 9+(3/2)^2  = 45/4.

(x-3/2)^2 = 45/4.

(x-3/2) = sqrt(45/4) or x-3/2 = -sqrt45/4.

x = 3/2 +(3/2)sqrt5 or x = 3/2 - (3/2)sqrt5.

b)

x^2+2x-15 = 0.

We complete  x^2+2x as a perfect square and rewrite the equation as:

(x^2+2x +1) - 16 = 0

(x+1)^2 = 16.

x+1 = sqrt16 = 4 or -4.

 So x= 4-1, or x= -4-1.

Therefore x= 3 or x= -5.

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