Problem Solving

[Math Revolution GMAT math practice question]

If x^2+y^2=28 and xy=11, what is the value of (x+y)^2?

A. 28
B. 39
C. 50
D. 61
E. 72

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x^2+y^2=28 and xy=11, what is the value of (x+y)^2?

A. 28
B. 39
C. 50
D. 61
E. 72

$$? = {\left( {x + y} \right)^2}$$
$$\left\{ \matrix{
{x^2} + {y^2} = 28 \hfill \cr
2xy = 2 \cdot 11 \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\,? = 28 + 2 \cdot 11 = 50$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x^2+y^2=28 and xy=11, what is the value of (x+y)^2?

A. 28
B. 39
C. 50
D. 61
E. 72

Every test-taker should know the following quadratic identities:
(x+y)² = x² + y² + 2xy
(x-y)² = x² + y² - 2xy
(x+y)(x-y)² = x² - y²

Substituting x²+y²=28 and xy=11 into (x+y)² = x² + y² + 2xy, we get:
(x+y)² = 28 + 2*11 = 50.

The correct answer is C.

An alternate approach is to BALLPARK.
x and y must have a product of 11, while x²+y² = 28.
If x=3 and y=4, then xy=12 and x²+y² = 3²+4² = 25.
The values in blue are reasonably close to the values in red.
If x=3 and y=4, then (x+y)² = (3+4)² = 49.
Thus, the correct answer must be close to 49.
Only C is viable.

=>

(x+y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 + 2xy = 28 + 2*11 = 28 + 22 = 50.


Therefore, the answer is C.
Answer: C