2.3 Solve fine \( x ; \) \( 2.3 .1 \quad \frac{x(x-5)}{6}-1=0 \) \( 2.3 .2 \quad e=\sqrt{a+2 x} \) 2.4 Tabelo is currently four times as old as his dalghter. Linda, Six years from now Tabelo will be three times as old as Linda. Calculate Linda'a ago currently.
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To solve \( 2.3 .1 \), start by rearranging the equation: \[ \frac{x(x-5)}{6} - 1 = 0 \implies \frac{x(x-5)}{6} = 1 \implies x(x-5) = 6 \] This can be rewritten as a quadratic equation: \[ x^2 - 5x - 6 = 0 \] Factoring gives: \[ (x - 6)(x + 1) = 0 \] Thus, \( x = 6 \) or \( x = -1 \). Since age can't be negative, we have \( x = 6 \). Now for \( 2.3 .2 \): Given \( e = \sqrt{a + 2x} \), squaring both sides gives: \[ e^2 = a + 2x \implies a = e^2 - 2x \] Solving for \( a \) will depend on the value of \( x \) that you compute earlier. Now, addressing Tabelo and Linda, we denote Tabelo's current age as \( T \) and Linda's current age as \( L \). We have two equations: 1. \( T = 4L \) (Tabelo is currently four times Linda's age) 2. \( T + 6 = 3(L + 6) \) (In six years, Tabelo will be three times Linda's age) Substituting \( T \) from the first equation into the second: \[ 4L + 6 = 3(L + 6) \] Expanding gives: \[ 4L + 6 = 3L + 18 \implies 4L - 3L = 18 - 6 \implies L = 12 \] So, Linda's current age is \( 12 \) years old.
