If $S = \sqrt{t^2 - 4t + 4}$, find t in terms of S

Assessment: JAMB UTME - Mathematics - 2013 Subject: General Mathematics

Question 1 Report

S2 - 2

S + 2

S - 2

S2 + 2

Answer Details

S = $\sqrt{t^{2} - 4 t + 4}$
S2 = t2 - 4t + 4
t2 - 4t + 4 - S2 = 0
Using $t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Substituting, we have;
Using $t = \frac{- (- 4) \pm \sqrt{(- 4)^{2} - 4 (1) (4 - S^{2})}}{2 (1)}$
$t = \frac{4 \pm \sqrt{16 - 4 (4 - S^{2})}}{2}$
$t = \frac{4 \pm \sqrt{16 - 16 + 4 S^{2}}}{2}$
$t = \frac{4 \pm \sqrt{4 S^{2}}}{2}$
$t = \frac{2 (2 \pm S)}{2}$
Hence t = 2 + S or t = 2 - S