If x(2x + 3) = 90 and 7y^(−1/2) + 2y^(−1/2) = y^(1/2) for positive real numbers x and y, then what is the value of x² + y²?

Introduction / Context:
This algebra problem combines a quadratic equation in x with an equation involving powers of y. The aim is to first find the individual values of x and y and then compute the sum x² + y². Such composite questions test comfort with quadratic factorisation and manipulation of fractional exponents.

Given Data / Assumptions:

• x(2x + 3) = 90.
• 7y^(−1/2) + 2y^(−1/2) = y^(1/2).
• x and y are positive real numbers.
• We need to find x² + y².

Concept / Approach:
For the x equation, expand and rearrange to form a quadratic equation, then factorise or use the quadratic formula. For the y equation, recognise that y^(−1/2) equals 1 divided by y^(1/2). Thus the equation can be simplified by introducing a new variable t = y^(1/2). Both equations become much easier under these transformations, and we can then compute the required sum of squares.

Step-by-Step Solution:
First handle x: x(2x + 3) = 90.Expand: 2x^2 + 3x = 90.Rearrange: 2x^2 + 3x − 90 = 0.Factorise: (x − 6)(2x + 15) = 0.Thus x = 6 or x = −15/2, but x is positive, so x = 6.Now handle y: 7y^(−1/2) + 2y^(−1/2) = y^(1/2).Combine like terms on the left: 9y^(−1/2) = y^(1/2).Let t = y^(1/2), so y^(−1/2) = 1/t. Then 9(1/t) = t.This gives 9 = t^2, so t = 3 (positive because y is positive).Therefore y^(1/2) = 3 and y = 9.Now compute x² + y² = 6^2 + 9^2 = 36 + 81 = 117.

Verification / Alternative check:
Verify x: substitute x = 6 into x(2x + 3). We get 6(12 + 3) = 6 × 15 = 90, which matches the original equation. Verify y: for y = 9, we have y^(−1/2) = 1 / 3 and y^(1/2) = 3. Then 7y^(−1/2) + 2y^(−1/2) = 7(1/3) + 2(1/3) = 9/3 = 3, which equals y^(1/2). Both equations are satisfied, so the value 117 for x² + y² is confirmed.

Why Other Options Are Wrong:
45, 109, 126, and 135 would result from incorrect solutions of the quadratic equation or from misinterpreting y^(−1/2). For example, treating y^(−1/2) as y^(1/2) or ignoring the positivity condition for x can lead to wrong values. Only 117 corresponds to the correct pair x = 6 and y = 9.

Common Pitfalls:
Students sometimes forget that y^(−1/2) means 1 divided by the square root of y, not negative square root of y. Another common error is keeping both roots of the quadratic in x even when the question states that x is positive. Introducing a simple substitution like t = y^(1/2) greatly reduces confusion and makes the equation linear in t.

Final Answer:
The required value of x² + y² is 117.