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If (a+b)^2=64 and ab = 3, a^2+b^2

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Mariamikaylaviz
[tex](a+b)^2=64 \\\\ a^2+2ab+b^2=64 \\\\ a^2+2*3+b^2=64 \\\\ a^2+6+b^2=64 \\\\ a^2+b^2=64-6 \\\\ \boxed{\boxed{a^2+b^2=58}}[/tex]
Jers15viz
(a+b)²=64
a²+2ab+b2=64
a²+2(3)+b²=64
a²+b²=58

Since ab is 3, then twice the value of ab is 6. Therefore just substitute and then subtract it both sides.

Hope this helps =)