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Jack
10-10-2013, 12:04 PM
If x^2+y^2=12 and xy=-\frac{3}{2}, what is the value of (x+y)^2 ?

(A) 9

(B) 3

(C) \frac{21}{2}

(D) \frac{23}{2}


Thanks

miranda
10-10-2013, 12:06 PM
We know that
(x+y)^2= x^2+2xy+y^2 =12+2(-\frac{3}{2})=12-3=9?

Therefore the correct answer is (A).

Jack
10-10-2013, 12:06 PM
thanks miranda!