+0

0
336
3

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

#1
+2328
+2

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. $$y=-15-x$$

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.

$$x(-15-x)=-54$$

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
#2
+1

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
#3
+2328
0

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. $$x=\frac{15\pm\sqrt{225-252}}{2}$$ $$x=\frac{15\pm\sqrt{-27}}{2}$$ 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