Pregunta
The roots of the equation \( f(x)=0 \) have the roots \( x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} \), for which volues of \( t \) wall the roots be equal?
Ask by Warner Turnbull. in South Africa
Feb 24,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
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The roots are equal when \( k = \frac{1}{2} \) or \( k = -\frac{1}{2} \).
Solución
The roots are given by
x = (–5 ± √(3 – 12k²)) / 4.
For the roots to be equal, the expression under the square root must be zero. That is, set
3 – 12k² = 0.
Solve for k as follows:
12k² = 3
k² = 3/12 = 1/4
k = 1/2 or k = –1/2.
So the roots are equal (i.e., they have a double root) when k = 1/2 or k = –1/2.
In that case, the double root becomes
x = (–5 ± 0) / 4 = –5/4.
Thus, the roots are equal when k is ±½.
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Simplifique esta solución Bonus Knowledge
To find the values of \( k \) for which the roots of the equation are equal, we need the discriminant of the roots to be zero. The discriminant from the equation is given by: \[ D = 3 - 12k^2. \] Setting the discriminant equal to zero gives us: \[ 3 - 12k^2 = 0. \] Solving for \( k^2 \) yields: \[ 12k^2 = 3 \\ k^2 = \frac{3}{12} = \frac{1}{4}. \] Taking the square root of both sides gives: \[ k = \pm \frac{1}{2}. \] Thus, the values of \( k \) for which the roots of the equation will be equal are \( k = \frac{1}{2} \) and \( k = -\frac{1}{2} \).
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