Sum and Sum-of-Squares of Two Numbers

Given the sum and the sum of squares of two numbers, the elegant route uses the identity $(x+y)^2 = x^2 + 2xy + y^2$ to get the product $xy$. Knowing the sum and product, the numbers are the roots of a quadratic.

Two numbers have a sum of $15$, and the sum of their squares is $125$. Find the two numbers. [5]

The working

Let the numbers be $x$ and $y$, so $x + y = 15$ and $x^2 + y^2 = 125$.

Step 1, square the sum and use the identity: $(x + y)^2 = x^2 + 2xy + y^2 \implies 15^2 = 125 + 2xy \quad \text{(M1)}$

Step 2, solve for the product $xy$: $225 = 125 + 2xy \implies 2xy = 100 \implies xy = 50 \quad \text{(A1)}$

Step 3, form the quadratic with sum $15$ and product $50$ (roots $x$, $y$): $t^2 - 15t + 50 = 0 \quad \text{(M1)}$

Step 4, factorise and solve: $(t - 5)(t - 10) = 0 \implies t = 5 \ \text{or} \ t = 10$

So the two numbers are $\boxed{5 \text{ and } 10}$. (A1, A1)

Check: $5 + 10 = 15$ and $25 + 100 = 125$. ✓

Where the marks are won and lost

• The identity $(x+y)^2 = x^2 + 2xy + y^2$ is the key: it links the two given facts to the product.
• Once you have sum $= 15$ and product $= 50$, use $t^2 - (\text{sum})t + (\text{product}) = 0$.
• A quick check in both original conditions confirms the pair.

Common mistakes

• Writing $(x+y)^2 = x^2 + y^2$ (forgetting the $2xy$).
• Direct substitution $y = 15 - x$ into $x^2 + y^2 = 125$ also works but is heavier and error-prone.
• Sign slip in the quadratic: it is $t^2 - 15t + 50$, with product positive.

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