If √a + √b = 5 and √a − √b = 1, what is the value of a and b?

Correct — B.
Treat √a and √b as the two unknowns.

Add the equations: 2√a = 5 + 1 = 6, so √a = 3
Subtract them: 2√b = 5 − 1 = 4, so √b = 2

Square each root: a = 3² = 9 and b = 2² = 4
Check: 3 + 2 = 5 and 3 − 2 = 1
So a = 9, b = 4 → option (b)

Why the others are wrong

• (a)
  a = 4, b = 1 — √4 + √1 = 3, not 5. The difference 2 − 1 = 1 does fit, so this pair passes the second equation and fails the first.
• (c)
  a = 16, b = 9 — √16 + √9 = 7, not 5. It meets √a − √b = 1 (4 − 3), which is why checking only the difference lets it through.
• (d)
  a = 25, b = 16 — √25 + √16 = 9, not 5. It comes from taking √a = 5, the value of the whole sum, and then √b = 4 from the difference.

Sum and difference of two unknowns: if u + v = S and u − v = D, then u = (S + D)⁄2 and v = (S − D)⁄2.

Here the unknowns are u = √a and v = √b. Solve for the roots first, then square them to get a and b.

Squaring the given equations straight away, a + b + 2√(ab) = 25, brings in √(ab) and only lengthens the working.

Key facts

• If u + v = S and u − v = D, then u = (S + D)⁄2 and v = (S − D)⁄2
• (√a + √b)(√a − √b) = a − b, so here a − b = 5 × 1 = 5
• Test a chosen pair in both equations, not just one

Common traps

• Reading 5 as the value of √a instead of √a + √b, which leads to a = 25, b = 16
• Checking the options against √a − √b = 1 alone, which every option here satisfies

No directly related past PYQ was found.