(x+a)(x+b)=2x^2+20x+48
if (x+a)(x+b) = 0 , sum of roots = -a - b, product of roots = ab
if 2x^2 + 20x + 48 = 0 , sum of roots = -20/2 = -10, product of roots = 48/2 = 24
-a - b = -10
a+b = 10
b = 10-b
ab = 24
a(10-b) = 24
10a - a^2 - 24 = 0
a^2 - 10a + 24 = 0
(a-6)(a-4) = 0
a = 6 or a = 4
if a=6, then b = 4
if a = 4, then b = 6
in either case a^2 + b^2 = 16 + 36 = 52
given (x+a)(x+b)=2x^2+20x+48, what is the value of a^2+b^2?
Please help. Thank you!
4 answers
Thank you so much Mr. Reiny!
Using Reiny's fine introduction, we can proceed as follow:
(a+b)^2 = a^2 + 2ab + b^2 = a^2+b^2 + 2ab
(-10)^2 = a^2+b^2 + 2*24
100 = a^2+b^2 + 48
a^2+b^2 = 52
(a+b)^2 = a^2 + 2ab + b^2 = a^2+b^2 + 2ab
(-10)^2 = a^2+b^2 + 2*24
100 = a^2+b^2 + 48
a^2+b^2 = 52
Thank you so much oobleck!