Jack

10-10-2013, 11:04 AM

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

(A) 9

(B) 3

(C) \frac{21}{2}

(D) \frac{23}{2}

Thanks

View Full Version : expression sat math

Jack

10-10-2013, 11:04 AM

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

(A) 9

(B) 3

(C) \frac{21}{2}

(D) \frac{23}{2}

Thanks

miranda

10-10-2013, 11:06 AM

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).

(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, 11:06 AM

thanks miranda!

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