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# What is the value of k?

What is the value of k, when parabola y=X^2 + (K+2) +2K +1 touches the x-axis?

ATTENTION!!!! correct equation

y = x² + (k + 2)x + 2k + 1

### 3 Answers

- Dragon.JadeLv 77 years agoFavorite Answer
Hello,

It bears repeating: you should have applied the well-known following formulas: "Hello", "Please" and "Thanks".

Anyway...

= = = = = = = = = = = = = = = = = =

y = x² + (k + 2)x + 2k + 1

If the curve touch the x-axis, then y must be nil:

y = 0

x² + (k + 2)x + 2k + 1 = 0

[x + (k + 2)/2]² – (k + 2)²/4 + 2k + 1 = 0 →→→ Completing the square.

[x + (k + 2)/2]² – (k² + 4k + 4)/4 + (8k + 4)/4 = 0

[x + (k + 2)/2]² – (k² + 4k + 4 – 8k – 4)/4 = 0

[x + (k + 2)/2]² – (k² – 4k)/4 = 0

Now three cases are possible:

♠ If 0<k<4

Then (k²–4k)<0

And

[x + (k + 2)/2]² – (k² – 4k)/4

Being the sum of a perfect square and a positive value will NEVER be nil.

♠ If k=0 or k=4,

Then (k²–4k)=0

And

[x + (k + 2)/2]²

Being a perfect square will be nil when x=-(k+2)/2.

So if k=0 (resp. k=4) the curve will touch the x-axis at only one point (-1;0) (resp. (-3;0)).

♠ If k<0 or k>4,

Then (k²–4k)>0

And

[x + (k + 2)/2]² – (k² – 4k)/4 = [x + (k + 2)/2]² – [√(k² – 4k)/2]²

Being the difference od two squares will be nil at two points of the x-axis.

So if k<0 or k>4, the curve will touch the x-axis at two points:

(-[k + 2 + √(k² – 4k)]/2;0) and (-[k + 2 – √(k² – 4k)]/2;0).

= = = = = = = = = = = = = = = = = =

Since you want the curve to "touch" the x-axis, the only values of k that result in that is:

k=0 and k=4.

Regards,

Dragon.Jade :-)

- DWReadLv 77 years ago
The question says "touches," not "crosses," so the vertex lies on the x-axis.

y = x² + (k+2)x + (2k+1)

= x² + (k+2)x + ((k+2)/2)² - ((k+2)/2)² + (2k+1)

= (x + (k+2)/2)² - (k² + 4k + 4)/4 + (2k+1)

= (x + (k+2)/2)² + (-k² - 4k - 4)/4 + (8k+4)/4

= (x + (k+2)/2)² + (-k² + 4k)/4

vertex ((k+2)/2, (-k² + 4k)/4)

(-k² + 4k)/4 = 0

k² - 4k = 0

k - 4 = 0

k = 4

- Anonymous7 years ago
You know what else it touches?! My heart!