Real Roots of the Quadratic Equation for all Alpha Values

[crays]

Show that the equation x² + (3$$\alpha$$ - 2)x + $$\alpha$$($$\alpha$$ - 1) = 0 has real roots for all values of α $$\in$$ IR.

How do i do this? Do i just use the
b² + 4ac $$\geq$$ 0 ?

 

[dirk_mec1]

Yes that should be correct.

 

[crays]

But i can't factorize it out. i get 5[tex]\alpha[tex]² - 2[tex]\alpha[tex] + 4 [tex] \geq[tex] 0

Do i use the formula? and write [tex]\alpha[tex] ranges that will satisfy it?


here's another question, show that x² - x + 1 has the same sign for all values of x.
How do i show?

Just x² - x [tex]\geq[tex] 0 will do ?

 

[malawi_glenn]

you can show that by drawing the graph for the function f(x) = x^2 - x + 1 and check that the function never crosses the x-axis.

 

[crays]

How do i draw a graph without knowing its roots?

 

[malawi_glenn]

How do i draw a graph without knowing its roots?

so you don't know how to draw graphs using differential calculus?

otherwise, just find the roots to x^2 - x + 1 , (it is very simple) - if you want.

 

[crays]

Sorry for the late reply. Yes i don't know how to do so.
But X^2 + x - 1, i use the formula to find the roots? then just draw the graph?

 

[rock.freak667]

Sorry for the late reply. Yes i don't know how to do so.
But X^2 + x - 1, i use the formula to find the roots? then just draw the graph?

$$x^2-x+1$$
you could just use the disciminant or find the minimum point or putting it in the completing the square form.

 

[Defennder]

Which question are you doing now? For the first one, just apply the discriminant condition and you'll get a quadratic equation in terms of $\alpha$. From there, you should use the discriminant condition again for the alpha quadratic to see if the quadratic equation for alpha can ever be negative.

The second question is answered much the same. You have x^2 - x +1. What condition must the discriminant satisfy such that f(x) does not cross the x-axis? Check also that the graph is above the x-axis to begin with.

 

[crays]

so do i just write x^2 - x >= 0 ? In that condition, it will never touches. I still don't get how should i draw a graph. If i use completing the square, i found (x - 1/2)^2 + 1 + 1/4

ps. I got it already, thanks guys.

 

[Defennder]

Where did x^2 - x come from? The discriminant is b^2 - 4ac. There should be no x in the discriminant, only numbers. As for graph sketching, note that the graph is either a u-shape or n-shape depending on the sign of the coefficient of the x^2 term. That should help you visualise if the f(x) is always positive or negative.

 

[crays]

Erm with the discriminant i found -3. So i write -3 < 0 ?

 

[Defennder]

What does that tell you about f(x)? More specifically what does that tell you about how the graph of f(x) should look like?

 

[crays]

it means the value is below -3? the graph should looks like a U right ?

 

[Defennder]

Yes and you also need to add in what the discriminant tells you.

 

[crays]

thanks. Mind if i ask another question?
Determine the condition to be satisfied by k such that the expression 2x^2 + 6x + 1 + k(x^2 + 2) is positive for all x .
Do i use discriminant for this?

 

[dirk_mec1]

Note that

$$2x^2>0$$
$$kx^2>0,\ k \geq 0$$

and so on. You could also use the derative.

 

[Defennder]

thanks. Mind if i ask another question?
Determine the condition to be satisfied by k such that the expression 2x^2 + 6x + 1 + k(x^2 + 2) is positive for all x .
Do i use discriminant for this?

Yes you can discriminant for this. I assume the reason why you didn't use calculus is because you haven't learned that yet.

 

[crays]

erm, is differentiation and integration called calculus?
I tried using the discriminant for it. But the value is off too.

 

[Defennder]

erm, is differentiation and integration called calculus?
I tried using the discriminant for it. But the value is off too.

I don't know what you mean by 'off'. Can you post your working here?

 

[crays]

Working :
2x² + 6x + 1 + k(x² + 2) > 0
(2 + k)x² + 6x + 2k + 1 > 0

b² - 4ac > 0 [is positive for all x]

36 - 4(2 + k)(2k + 1) > 0
2k² - 5k + 7 > 0

(2k - 7)(k + 1) > 0
k > 7/2 or k > -1.

Answer is
K > 1

 

[Defennder]

What is the condition the discriminant has to satisfy in order for the quadratic to not cross the axis?

 

[crays]

not cross the axis would be ... less than 0. <0.

 

[Defennder]

Well, so what did you do in your working with respect to this?