Find k: 3x^2 + sqrt{2k}x + 6 = 0

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Discussion Overview

The discussion revolves around finding the value of k in the quadratic equation 3x^2 + sqrt{2k}x + 6 = 0 such that the equation has exactly one root. The focus is on the use of the discriminant and alternative methods of analysis.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the discriminant must equal zero for the equation to have exactly one root, leading to the equation 2k - 72 = 0.
  • Participants calculate k as 36 based on the discriminant approach, confirming the steps taken.
  • Others present an alternative method involving the vertex of the quadratic, arriving at the same conclusion that k = 36.
  • There is a reiteration of the calculations and methods used, with some participants affirming the correctness of the results.

Areas of Agreement / Disagreement

Participants generally agree on the value of k being 36, with multiple methods leading to the same conclusion. However, the discussion does not explore any alternative values or methods that might yield different results.

Contextual Notes

The discussion relies on the assumption that the discriminant is the appropriate method for determining the number of roots, and it does not address potential limitations or alternative interpretations of the problem.

mathdad
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Find the value of k such that the equation has exactly one root.

3x^2 + sqrt{2k}x + 6 = 0

This question involves the discriminant, right?

b^2 - 4ac = 0

(sqrt{2k}^2 - 4(3)(6) = 0

2k - 72 = 0

2k = 72

k = 72/2

k = 36

Correct?
 
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RTCNTC said:
Find the value of k such that the equation has exactly one root.

3x^2 + sqrt{2k}x + 6 = 0

This question involves the discriminant, right?

b^2 - 4ac = 0

(sqrt{2k}^2 - 4(3)(6) = 0

2k - 72 = 0

2k = 72

k = 72/2

k = 36

Correct?

Yes correct.
 
Alternatively,

$$3x^2+\sqrt{2k}x+6=0$$

$$3\left(x^2+\frac{\sqrt{2k}}{3}x\right)+6=0$$

$$3\left(x+\frac{\sqrt{2k}}{6}\right)^2+6-3\left(\frac{\sqrt{2k}}{6}\right)^2=0$$

If the equation has exactly one real root then the vertex "touches" (is tangent to) the $x$-axis, so

$$6-3\left(\frac{\sqrt{2k}}{6}\right)^2=0$$

$$6=3\left(\frac{\sqrt{2k}}{6}\right)^2$$

$$6=3\frac{2k}{36}$$

$$2=\frac{2k}{36}$$

$$72=2k$$

$$k=36$$
 
greg1313 said:
Alternatively,

$$3x^2+\sqrt{2k}x+6=0$$

$$3\left(x^2+\frac{\sqrt{2k}}{3}x\right)+6=0$$

$$3\left(x+\frac{\sqrt{2k}}{6}\right)^2+6-3\left(\frac{\sqrt{2k}}{6}\right)^2=0$$

If the equation has exactly one real root then the vertex "touches" (is tangent to) the $x$-axis, so

$$6-3\left(\frac{\sqrt{2k}}{6}\right)^2=0$$

$$6=3\left(\frac{\sqrt{2k}}{6}\right)^2$$

$$6=3\frac{2k}{36}$$

$$2=\frac{2k}{36}$$

$$72=2k$$

$$k=36$$

Nicely done!
 

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