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

[mathdad]

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?

 

[Prove It]

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.

 

[Greg]

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

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

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

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

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

$$72=2k$$

$$k=36$$

 

[mathdad]

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!