Question

If $\sqrt{13 - a\sqrt{10} = \sqrt{8} + \sqrt{5}$ then a = .........}

• A. -5
• B. -6
• C. -4
• D. -2

Hint:

When dealing with equations involving surds (square roots), especially nested ones or sums/differences of surds, squaring both sides is a very common and effective first step. Remember the perfect square factors when simplifying radicals (e.g., $\sqrt{AB} = \sqrt{A}\sqrt{B}$). Comparing the rational and irrational parts on both sides after squaring is crucial for solving.

Answer 1

The Correct Option is C

Step 1: Simplify the right-hand side (RHS) or prepare to square both sides.
The given equation is \( \sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5} \).
To eliminate the outermost square root on the left side, we will square both sides of the equation. Step 2: Square both sides of the equation.
\[ (\sqrt{13 - a\sqrt{10}})^2 = (\sqrt{8} + \sqrt{5})^2 \] The left side simplifies to: \[ 13 - a\sqrt{10} \] The right side uses the identity \( (x + y)^2 = x^2 + y^2 + 2xy \), with \( x = \sqrt{8} \) and \( y = \sqrt{5} \): \[ (\sqrt{8})^2 + (\sqrt{5})^2 + 2\sqrt{8}\sqrt{5} = 8 + 5 + 2\sqrt{40} \] Step 3: Simplify the radical term on the RHS.
\[ 2\sqrt{40} = 2\sqrt{4 \cdot 10} = 2 \cdot 2\sqrt{10} = 4\sqrt{10} \] Now the RHS becomes: \[ 13 + 4\sqrt{10} \] Step 4: Equate the simplified left and right sides of the equation.
\[ 13 - a\sqrt{10} = 13 + 4\sqrt{10} \] Step 5: Solve for \( a \) by comparing coefficients.
Subtract 13 from both sides: \[ -a\sqrt{10} = 4\sqrt{10} \] Divide both sides by \( \sqrt{10} \): \[ -a = 4 \] Multiply by -1: \[ a = -4 \] Step 6: Final Answer.
\[ (3) \quad -4 \]