What two numbers multiply to negative 64 and add up to 16
The Correct Answer and Explanation is:
To solve for two numbers that multiply to -64 and add up to 16, let’s denote these numbers as xxx and yyy. According to the problem:x×y=−64x \times y = -64x×y=−64x+y=16x + y = 16x+y=16
To solve this system, we can use substitution or trial and error. However, let’s first approach this by setting up a quadratic equation.
If we consider the two numbers as the roots of a quadratic equation, then the sum and product of the roots can be used to form the quadratic equation. The general form of a quadratic equation is:t2−(x+y)t+x×y=0t^2 – (x + y)t + x \times y = 0t2−(x+y)t+x×y=0
Substituting the values for the sum and product from the problem:t2−16t−64=0t^2 – 16t – 64 = 0t2−16t−64=0
Now we need to solve this quadratic equation. To do this, we use the quadratic formula:t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}t=2a−b±b2−4ac
For the equation t2−16t−64=0t^2 – 16t – 64 = 0t2−16t−64=0, the coefficients are:
- a=1a = 1a=1
- b=−16b = -16b=−16
- c=−64c = -64c=−64
Substitute these values into the quadratic formula:t=−(−16)±(−16)2−4(1)(−64)2(1)t = \frac{-(-16) \pm \sqrt{(-16)^2 – 4(1)(-64)}}{2(1)}t=2(1)−(−16)±(−16)2−4(1)(−64)t=16±256+2562t = \frac{16 \pm \sqrt{256 + 256}}{2}t=216±256+256t=16±5122t = \frac{16 \pm \sqrt{512}}{2}t=216±512t=16±22.6272t = \frac{16 \pm 22.627}{2}t=216±22.627
Now, solve for the two possible values of ttt:t1=16+22.6272=38.6272≈19.31t_1 = \frac{16 + 22.627}{2} = \frac{38.627}{2} \approx 19.31t1=216+22.627=238.627≈19.31t2=16−22.6272=−6.6272≈−3.31t_2 = \frac{16 – 22.627}{2} = \frac{-6.627}{2} \approx -3.31t2=216−22.627=2−6.627≈−3.31
Thus, the two numbers are approximately 19.31 and -3.31. When multiplied, they give approximately -64, and when added together, they give approximately 16.
However, exact integer values would require adjusting the approach slightly. Would you like help with that?
