If $P = \sqrt{QR\left(1+\frac{3t}{R}\right)}$, make R the subject of the formula.

Assessment: WAEC SSCE - General Mathematics - 2003 (Objective) Subject: General Mathematics

Question 1 Report

$R = \frac{3Qt}{P^2 - Q}$

$R = \frac{P^2 – 3t}{Q+1}$

$R = \frac{P^2 + 3t}{Q - 1}$

$R = \frac{P^2-3Qt}{Q}$

Answer Details

To make R the subject of the formula, we need to isolate R on one side of the equation. Starting with the given equation: $$P = \sqrt{QR\left(1+\frac{3t}{R}\right)}$$ Squaring both sides: $$P^2 = QR\left(1+\frac{3t}{R}\right)$$ Expanding the right side: $$P^2 = QR + 3tQ/R$$ Subtracting 3tQ/R from both sides: $$P^2 - 3tQ/R = QR$$ Dividing both sides by Q: $$\frac{P^2 - 3tQ}{Q} = R$$ Therefore, the answer is: $$R = \frac{P^2 - 3tQ}{Q}$$ So the correct option is: - $R = \frac{P^2-3Qt}{Q}$

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

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If \(P = \sqrt{QR\left(1+\frac{3t}{R}\right)}\), make R the subject of the formula.

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If \(P = \sqrt{QR\left(1+\frac{3t}{R}\right)}\), make R the subject of the formula.

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