Question
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Given \( f(x)=3(x-1)^{2}+5 \) and \( g(x)=3 \) Determine the values of \( k \) for which \( q(a)=g(x)+k \) has Two unequal real roots.

Ask by Robbins Clarke. in South Africa
Mar 15,2025

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Tutor-Verified Answer

Answer

The equation \( q(a) = g(x) + k \) has two unequal real roots when \( k > 2 \).

Solution

1. Start with the equation formed by setting \( f(x) \) equal to \( g(x)+k \): \[ 3(x-1)^2+5 = 3 + k \] 2. Rearrange the equation to isolate the squared term: \[ 3(x-1)^2 = k + 3 - 5 = k - 2 \] \[ (x-1)^2 = \frac{k-2}{3} \] 3. For the equation \( (x-1)^2=\frac{k-2}{3} \) to have two unequal real roots, the right-hand side must be strictly positive: \[ \frac{k-2}{3} > 0 \] 4. Since \(3\) is positive, the inequality reduces to: \[ k-2 > 0 \] \[ k > 2 \] Thus, the equation has two unequal real roots if and only if \( k > 2 \).

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The Deep Dive

To determine the values of \( k \) for which the equation \( q(x) = g(x) + k = 3 + k \) has two unequal real roots, we start by transforming the context a bit. Here \( q(x) \) is a constant function because \( g(x) = 3 \). Now, since \( q(x) \) is a constant, it cannot have any roots unless it is equal to a quadratic function that can potentially cross the x-axis. The function \( f(x) = 3(x-1)^2 + 5 \) is a parabola that opens upwards. The vertex of this parabola is at \( x = 1 \) with a minimum value of \( f(1) = 5 \). For \( q(x) \) to intersect \( f(x) \) and have two unequal real roots, we require that \( q(x) = 3+k \) must be less than the minimum value of \( f(x) \). Thus, we set up the inequality: \[ 3 + k < 5 \] Solving this gives: \[ k < 2 \] Therefore, for \( k < 2 \), the line \( y = q(x) = 3 + k \) intersects the parabola \( f(x) \) at two points, resulting in two unequal real roots. In conclusion, the values of \( k \) for which \( q(x) = g(x) + k \) has two unequal real roots are: \[ k < 2 \]

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