Use a quadratic equation to find the zeroes of a functions wherein their sum adds to -15 and their product is -54.

 Oct 14, 2017

This one requires some thinking. We know that a quadratic equation has 2 zeroes. One is located at \((x,0)\), and the other is located at \((y,0)\)


Now, what do we know about the relationship of these two numbers? Well, we know that the sum of both zeroes is -15.


\(x+y=-15\) Let's solve for y here.


Since y=-15-x, we know that a zero is located at \((-15-x,0)\). We also know that, when multiplied, both  zeroes yield -54. Knowing this information, we can generate a quadratic equation.




This quadratic equation has everything that we are looking for. Now, solve for the zeroes of this quadratic.


\(x(-15-x)=-54\) Distribute the x into the binomial in the parentheses.
\(-15x-x^2=-54\) Add 54 to both sides.
\(-x^2-15x+54=0\) Divide by -1. This is not strictly necessary, but it does make the x^2-term positive, which could make methods of solving easier.
\(x^2+15x-54=0\) Factor this by finding two numbers that multiply to -54 and add to 15. These happen to be -3 and 18.
\((x-3)(x+18)=0\) By the zero product theorem, set both products equal to zero.
\(x-3=0\) \(x+18=0\)


Add the constant in both equations and move it move to the right hand side.
\(x_1=3\) \(x_2=-18\)




These zeroes definitely fit the criteria.

 Oct 14, 2017

You're the greatest. It's well-explained, too.


However, I tried the next one, and I still cannot figure out what the answer is. The sum is 15 and the product is 63.

Guest Oct 14, 2017
edited by Guest  Oct 14, 2017

Let's go through the process again, shall we?


Let x represent one of the numbers. Then 15-x is the other number. Knowing this, the equation becomes \(x(15-x)=63\).


\(x(15-x)=63\) Distribute.
\(15x-x^2=63\) Subtract 63 from both sides.
\(-x^2+15x-63=0\) I like to divide by -1 out of habit.
\(x^2-15x+63=0\) In this case, this quadratic cannot be factored. I'll use the quadratic formula here.
\(x=\frac{-(-15)\pm\sqrt{(-15)^2-4(1)(63)}}{2(1)}\) Let's do some simplification here.
\(x=\frac{15\pm\sqrt{-27}}{2}\) I see an issue here, do you? This is probably why you were unable to come up with a solution. The square root of -27 results in a nonreal answer. 


This means that no quadratic equation with real solutions can have property aforementioned, which is a sum of 15 and a product of 63.

TheXSquaredFactor  Oct 14, 2017

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