Given, √(x + 3) + √(x − 2) = 5, then find the value of x.

Correct — B. The numbers under the roots differ by (x + 3) − (x − 2) = 5. Pair the sum with its conjugate.

(√(x + 3) + √(x − 2)) × (√(x + 3) − √(x − 2)) = 5
Divide by the given sum 5: √(x + 3) − √(x − 2) = 1
Add the two equations: 2√(x + 3) = 6, so √(x + 3) = 3
Square: x + 3 = 9, so x = 6 → option (b)

Check: √9 + √4 = 3 + 2 = 5, as required.

Why the others are wrong

• (a)
  5 — x = 5 gives √8 + √3 ≈ 2.83 + 1.73 = 4.56, short of 5. The left side grows as x grows, so the root must be larger than 5.
• (c)
  7 — x = 7 gives √10 + √5 ≈ 3.16 + 2.24 = 5.40, already past 5.
• (d)
  8 — x = 8 gives √11 + √6 ≈ 3.32 + 2.45 = 5.77, too large. The sum keeps rising with x, so nothing above 6 can work.

When two square roots are added and the numbers under them differ by a constant k, use sum × difference = k: (√p + √q)(√p − √q) = p − q.

So the difference of the roots is k ÷ sum. Adding the sum and the difference isolates one root, without squaring twice.

Here k = 5 and the sum is 5, so the roots are 3 and 2. Both x + 3 = 9 and x − 2 = 4 give x = 6.

Key facts

• (√p + √q)(√p − √q) = p − q.
• If √p + √q = S and p − q = k, then √p − √q = k ⁄ S.
• Substitute a radical equation's answer back: squaring can create roots the original equation does not have.

Common traps

• Squaring the left side as (x + 3) + (x − 2) and dropping the cross term 2√((x + 3)(x − 2)). That gives x = 12, not even an option.
• Stopping at the difference √(x + 3) − √(x − 2) = 1 instead of adding it to the given sum.

No directly related past PYQ was found.