Question on: JAMB Mathematics - 2013
If S = \(\sqrt{t^2 - 4t + 4}\), find t in terms of S
A
S2 - 2
B
S + 2
C
S - 2
D
S2 + 2
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Correct Option: B
S = \(\sqrt{t^2 - 4t + 4}\)
S2 = t2 - 4t + 4
t2 - 4t + 4 - S2 = 0
Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Substituting, we have;
Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\)
\(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\)
\(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\)
\(t = \frac{4 \pm \sqrt{4S^2}}{2}\)
\(t = \frac{2(2 \pm S)}{2}\)
Hence t = 2 + S or t = 2 - S
S2 = t2 - 4t + 4
t2 - 4t + 4 - S2 = 0
Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Substituting, we have;
Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\)
\(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\)
\(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\)
\(t = \frac{4 \pm \sqrt{4S^2}}{2}\)
\(t = \frac{2(2 \pm S)}{2}\)
Hence t = 2 + S or t = 2 - S
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