SOLUTION: 4x^= 13x + 12 Am I right by stating that there is no solution to this problem?

Question 113094: 4x^= 13x + 12
Am I right by stating that there is no solution to this problem?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I presume you mean . I'm not certain why you are saying there is no solution. In fact there is always a solution to a quadratic, in fact there are always two solutions. They aren't always to be found in the real number system, but there are always two answers.

I suspect that you tried to use the quadratic formula: , without putting the equation in standard form. That would result in a negative radicand (the part under the radical) and that may have led you to believe that there is no solution. (That, in itself, is an erroneous belief. The only thing is that when you have a negative radicand, the solution is no longer a pair of real numbers, it becomes a pair of complex numbers -- but that is a tale for another time)

Remember, the quadratic formula is the solution to , so in order to process your equation, you needed to put it in that form.

Now we can see that a = 4, b = -13, and c = -12. Putting those values into the quadratic formula yields:

,
,
, but so
or
Hence, or

Now that I have done this with the quadratic formula and discovered that the solution set contains two rational numbers, I know that the trinomial could have been factored.

Hope this helps,
John
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