What 2 numbers multiply to 30 but add to 16
The Correct Answer and Explanation is:
Let’s break this down step-by-step. You are looking for two numbers that multiply to 30 and add to 16.
To solve this, let’s denote the two numbers as x and y.
- Multiplication condition:
x×y=30x \times y = 30x×y=30 - Addition condition:
x+y=16x + y = 16x+y=16
Step 1: Express one variable in terms of the other
From the second equation x+y=16x + y = 16x+y=16, solve for one of the variables. Let’s solve for xxx:x=16−yx = 16 – yx=16−y
Step 2: Substitute into the multiplication equation
Now, substitute x=16−yx = 16 – yx=16−y into the first equation x×y=30x \times y = 30x×y=30:(16−y)×y=30(16 – y) \times y = 30(16−y)×y=30
Step 3: Expand and simplify
Now expand the equation:16y−y2=3016y – y^2 = 3016y−y2=30
Rearrange this into a standard quadratic form:y2−16y+30=0y^2 – 16y + 30 = 0y2−16y+30=0
Step 4: Solve the quadratic equation
Now, we can solve this quadratic equation using the quadratic formula:y=−(−16)±(−16)2−4(1)(30)2(1)y = \frac{-(-16) \pm \sqrt{(-16)^2 – 4(1)(30)}}{2(1)}y=2(1)−(−16)±(−16)2−4(1)(30)
Simplify the terms inside the square root:y=16±256−1202y = \frac{16 \pm \sqrt{256 – 120}}{2}y=216±256−120y=16±1362y = \frac{16 \pm \sqrt{136}}{2}y=216±136y=16±11.662y = \frac{16 \pm 11.66}{2}y=216±11.66
So, the two possible solutions for yyy are:y=16+11.662≈13.83y = \frac{16 + 11.66}{2} \approx 13.83y=216+11.66≈13.83
ory=16−11.662≈2.17y = \frac{16 – 11.66}{2} \approx 2.17y=216−11.66≈2.17
Step 5: Find corresponding xxx values
Using the equation x=16−yx = 16 – yx=16−y, we can now find the corresponding xxx values:
For y≈13.83y \approx 13.83y≈13.83,x≈16−13.83=2.17x \approx 16 – 13.83 = 2.17x≈16−13.83=2.17
For y≈2.17y \approx 2.17y≈2.17,x≈16−2.17=13.83x \approx 16 – 2.17 = 13.83x≈16−2.17=13.83
Conclusion:
So, the two numbers are approximately 2.17 and 13.83. These values satisfy both the conditions:
- Their product is close to 2.17×13.83≈302.17 \times 13.83 \approx 302.17×13.83≈30,
- Their sum is 2.17+13.83=162.17 + 13.83 = 162.17+13.83=16.
These numbers are not exact integers, but they are the closest real solutions to this problem.
