If-a-b-5-a-2-b-2-13-the-value-of-a-b-where-a-gt-b-is-

Question Number 112625 by Aina Samuel Temidayo last updated on 09/Sep/20

If a + b = 5, a^{2} + b^{2} = 13, the value of

a - b (where a > b) is

Commented by MJS_new last updated on 09/Sep/20

lol \dots

b = 5 - a

a^{2} + {(5 - a)}^{2} = 13

a^{2} - 5 a + 6 = 0

(a - 3) (a - 2) = 0

a = 2 \lor a = 3 \land b = 5 - a \land a > b \Rightarrow a = 3 \land b = 2

Answered by MJS_new last updated on 09/Sep/20

at first sight : a = 3 \land b = 2

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

Sure but solve it mathematically .

Answered by ajfour last updated on 09/Sep/20

a - b = s > 0

a + b = 5

\Rightarrow {(s + 5)}^{2} + {(5 - s)}^{2} = 52

\Rightarrow 2 s^{2} = 2 \Rightarrow s = a - b = 1

Answered by john santu last updated on 09/Sep/20

{(a + b)}^{2} = 25 \Rightarrow a^{2} + b^{2} + 2 a b = 25

2 a b = 25 - 13 = 12 \dots (i)

a - b = \sqrt{{(a - b)}^{2}} = \sqrt{a^{2} + b^{2} - 2 a b}

= \sqrt{13 - 12} = \sqrt{1} = 1

Answered by 1549442205PVT last updated on 09/Sep/20

{(a - b)}^{2} = 2 (a^{2} + b^{2}) - {(a + b)}^{2} = 2.13 - 25

= 1 \Rightarrow a - b = 1 (since a > b)