How to Find Values of k for Equations with One Real Root?

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SUMMARY

The discussion focuses on finding values of k for two quadratic equations that yield exactly one real root. The equations are 3x² + (√{2k})x + 6 = 0 and kx² + kx + 1 = 0. Participants confirm that the discriminant, calculated using the formula b² - 4ac, must equal zero to achieve this condition. This method is essential for determining the specific values of k that satisfy the requirement for a single real root.

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Find values of k such that the equation has exactly one real root.

1. 3x^2 + (sqrt{2k})x + 6 = 0

2. kx^2 + kx + 1 = 0

Question:

Do the questions above involve the discriminant?

If so, I must apply b^2 - 4ac = 0, right?
 
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RTCNTC said:
Find values of k such that the equation has exactly one real root.

1. 3x^2 + (sqrt{2k})x + 6 = 0

2. kx^2 + kx + 1 = 0

Question:

Do the questions above involve the discriminant?

If so, I must apply b^2 - 4ac = 0, right?

Yes . you can solve with discriminant and you must apply $b^2-4ac=0$
 
Not too bad.
 

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