A Nice Algebra Problem | Find the values of a and b | 3 Methods

[@OlgaPlaysMatch-3Game]

It was hard, but by inpsection trial and error, I found that {a, b} = {5, 7}. It’s not a=5 and b=7, but to say that a = {5, 7} and b = {5, 7}. Reason: *Addition has the commutative property.*

[@aminehl6025]

It is exactly the way he put it. solutions for a and b are (a,b) = {5,7} , {7,5} where whenever a is 5, then b is 7 and vice versa.

[@idvbane8580]

Very gd explain 🥰🥰🥰

[@MathBooster]

Thank you 🙂

[@devondevon4366]

5 and 7 a^2 + b^2 + 2ab = 144 74 + 2ab = 144 2ab = 70 (a-b)^2 = n a^2 + b^2 - 2ab = n^2 74 -70 = n^2 4 =n^2 n = 2 Henc a+ b =12 a-b =2 2a =14 a = 7 and b =5

[@OlgaPlaysMatch-3Game]

Your method is extremely confusing. Why do you use n, and where have you solved for b?

[@devondevon4366]

@@OlgaPlaysMatch-3Game Since the values for a^2 + b^2 and 2ab are known, the square root of n^2 would give the value for (a-b) since (a-b)^2 = a^2 + b ^2 - 2ab. Hene (a-b)^2 = 74 - 70 = 4 (a-b) = sqrt 4 = 2 So n=2 is the value for a-b or the difference between "a" and "b". The reason for doing this is that since the value for a + b = 12, KNOWING the value for a-b = 2 will enable us to find "a' and "b" since you can add both, thus eliminating the 'b' value; hence 2a =14; hence a =7; hence "b" =5 since the difference between "a" and "b" is 2.

[@vijayannair2316]

So nice

[@MathBooster]

Thank you 🙂

[@OlgaPlaysMatch-3Game]

What’s nice in mathematics? *Factorization.* Factorize 3n² + 17n + 10.

[@piman9280]

Again, one can quickly see by inspection that {(a, b)} = {(5, 7), (7,5)}. It's good to show the algebraic method, but it would be better to do so where the solution isn't so obvious.

[@idvbane8580]

🥰🥰🥰sir me 1 , math is gd🥰🥰