) Let's find the value of ab when a? + b2 = 52, a - b = 2
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Answer:a+b=6………………………………..(1)
a-b=4…………………………………(2)
Adding equation (1) and (2),we get,
(a+b)+(a-b)=10
=> a+b+a-b=10
=> 2a=10
=> 2a/2=10/2
=> a=5
Putting a=5 in equation (1),we get,
5+b=6
=> b=6–5
=> b=1
Now,
a^2+b^2
=(5)^2+(1)^2
=25+1
=26
2nd Method:
a+b=6
On squaring both sides,
(a+b)^2=(6)^2
=> a^2+2ab+b^2=36…………………….(1)
a-b=4
On squaring both sides,
(a-b)^2=(4)^2
=a^2–2ab+b^2=16……………..………….(2)
Adding equation (1) and (2), we get,
(a^2+2ab+b^2)+(a^2–2ab+b^2)=36+16
=> a^2+2ab+b^2+a^2–2ab+b^2=52
=> a^2+a^2+b^2+b^2+2ab-2ab=52
=> 2a^2+2b^2=52
=> 2(a^2+b^2)=52
=> a^2+b^2=52/2
=>a^2+b^2=26
Hence, a^2+b^2=26
Step-by-step explanation:
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