# Please solve and show work.  2x^2 + 3x + 12 = 0

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may also try to solve the problem by completing the square with respect to the following formula `(a+b)^2 = a^2+2ab+b^2` .

`2x^2 + 3x + b^2 = (sqrt2*x)^2 + 2*sqrt2*x*b + b^2`

Equating like parts yields:

`3x = 2*sqrt2*x*b => b = (3x)/(2*sqrt2*x) => b = 3sqrt2/4`

The expansion of binomial is completed such that:

`2x^2 + 3x + (3sqrt2/4)^2 = (sqrt2 x + 3sqrt2/4)^2`

`2x^2 + 3x + 12 = (sqrt2 x + 3sqrt2/4)^2 - (3sqrt2/4)^2 + 12`

Since `2x^2 + 3x + 12 = 0` , then `(sqrt2 x + 3sqrt2/4)^2 - (3sqrt2/4)^2 + 12 = 0`  such that:

`(sqrt2 x + 3sqrt2/4)^2 - (3sqrt2/4)^2 + 12 = 0`

`(sqrt2 x + 3sqrt2/4)^2 = 18/16 - 12`

`(sqrt2 x + 3sqrt2/4)^2 = (18 - 192)/16`

`(sqrt2 x + 3sqrt2/4)^2 = -174/16 = -87/8`

`sqrt2 x + 3sqrt2/4 = +-sqrt(-87/8)`

Notice that the equation has not real roots since +-sqrt(-87/8) !in R, but it has complex roots.

Hence, using complex number theory, `sqrt(-1) = i`  such that:

`sqrt2 x + 3sqrt2/4 = +-i*sqrt(87/8)`

`sqrt2*x = -3sqrt2/4+-i*sqrt(87/8)`

`x_(1,2)= (-3sqrt2/4+-i*sqrt(87/8))/sqrt2`

Hence, evaluating the solutions to the given quadratic equation yields that there exists two complex conjugate roots `x_(1,2) = (-3sqrt2/4+-i*sqrt(87/8))/sqrt2.`

najm1947 | Elementary School Teacher | (Level 1) Valedictorian

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We have to solve 2x^2 + 3x + 12 = 0

This is in the form ax^2+bx+c=0 where a=2, b=3 and c=12 and its roots are as:

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

Hence the roots are:

x = [-3+sqrt{(3)^2-4*2*(12)}]/(2*2) and

x = [-3-sqrt{(3)^2-4*2*(12)}]/(2*2)

=> x = [-3+sqrt{9-96}]/4 , [3-sqrt{9-96}]/4

=> x = [-3+sqrt{-87}]/4 , [-3-sqrt{-87}]/4

=> x = -3/4+sqrt(-87)/4 , -3/4-sqrt(-87)/4

There appears to be some problem is the description of the equation as it yields imaginary roots instead of real ones. Any way the solution of the equation is

x = -3/4+sqrt(-87)/4 , -3/4-sqrt(-87)/4