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Solve by using the quadratic formula 3x^2+5=-2x

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Hello from MrBillDoesMath!

Answer: -1/3 + i* (1/3) * sqrt(14) and -1/3 - i* (1/3) * sqrt(14)

Discussion :

Recall the solutions of the quadratic equation "ax^2 + bx + c = 0" are given by the formula

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


Rewriting the given equation in this for gives

3x^2 + 2x + 5 = 0


so a = 3, b = 2, and c = 5. Substituting the values in the formula gives


x = (- 2 + sqrt ( 2^2 - 4*3*5) ) / 2a #1

and

x = (- 2 - sqrt ( 2^2 - 4*3*5) ) / 2a #2


Simplifying, #1 = ( -2 + sqrt(4-60) )/ (2 * 3) =

-2/(2*3) + (sqrt(-56)) / (2*3) =

-1/3 + i * (sqrt (56)) /6 =

-1/3 + i * (sqrt(4 *14)) /6 =

-1/3 + i * (sqrt(4)/6) * sqrt(14) =

-1/3 + i * (2/6) * sqrt(14) =

-1/3 + i* (1/3) * sqrt(14)


Similarly, the solution of # 2= -1/3 - i* (1/3) * sqrt(14). The only difference between the two solutions is the minus sign boldfaced above.


Regards, MrB




User Whatang
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