Finding two numbers when having their sum and product

Question

I have two numbers, their sum is 41 and their product is 238. What are the numbers?

I got during this far in my calculations:

$a+b=41,\quad ab=238,\quad 238=41-b.$

I appreciate answers or tips to help me complete this. Thanks in advance

Answer 1

Your first two equations are correct. One way to continue is to solve one of them for a variable. For example, you could find $$ a = 41 - b $$ and substitute into the other equation to get $$ (41 - b)b = 238. $$ The equation in $b$ can be rewritten $b^2 - 41b + 238 = 0$, which has solutions $7$ and $34$ (you can use the quadratic formula to find these).

If we use this information in $a + b = 41$, we also get solutions of $7$ and $34$ for $a$. As far as we are concerned in this problem, there is no difference between the solution $a = 7$ and $b = 34$ and the solution $a = 34$ and $b = 7$. We may simply say: "The two numbers with sum $41$ and product $238$ are $7$ and $34$."

Answer 2

Hint: Write $$238=2\cdot 7\cdot 17$$

What options for $a$ and $b$ does that leave you? Which of them satisfy $a+b=41$?

Answer 3

From where you reached:

$$b=41-a\implies 238=ab=a(41-a)=41a-a^2\implies a^2-41a+238=0\ldots\ldots$$

Answer 4

$$(t-a)(t-b)=t^2-(a+b)t+ab=t^2-41t+238$$ whose roots are: $7,34$

Answer 5

We use the identity $$(a-b)^2=(a+b)^2-4ab.$$ Thus in our case $(a-b)^2=(41)^2-4(238)=729$.

It follows that $a-b=\pm 27$. Using $a+b=41$, we solve the system of two linear equations for $a$ and $b$. We get $a=34$, $b=7$ or $a=7$, $b=34$.