Quadratic Inequalities

In summary, the conversation covers two quadratic equations and their respective solutions. For the first one, the possible range of k is determined by using the discriminant and considering the positive roots. For the second one, one positive and one negative root are given and the task is to find the values of t that satisfy the equation. The process involves rewriting the equations and using formulas to find the roots.
  • #1
lingling
22
0
1.
(a) If the roots of the equation 2(x)^2 + kx + 100 = 0 are positive,
find the possible range of k.
(b) If, in addition, one root is twice the other, find the roots and the value of k.

I have tried (a), but incorrect:
discriminate > 0
k^2 - (4)(2)(100) > 0
k^2 > 800
k > + or - 20(2)^1/2
What's wrong with my calculation?
The correct ans is:k is less than or equal to - 20 (2)^1/2
2. Find the values of t for which the quadratic equation
(x)^2 - tx + t + 3 = 0 has one positive root and one negative root.

>>> I have no idea to start doing it.
The correct answer is t < -3
Can anyone help? :blushing:
Wish you all have a Merry Christmas! :rofl:
 
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  • #2
For #1, since the problem does not ask for two distinct real roots, the discriminant can also be 0, ie: [itex]\Delta \geq 0[/itex].
Solving that, you have:
[tex]\left[ \begin{array}{l} k \geq 20 \sqrt{2} \\ k \leq -20 \sqrt{2} \end{array} \right. \ (1)[/tex] (Is this what you get?)
But since the root(s) must also be positive (let x1, and x2 be the roots of the equation), you must also have:
[tex]\left\{ \begin{array}{l} x_1 + x_2 > 0 \\ x_1 x_2 > 0 \end{array} \right.[/tex].
Using Viète's formulas, we have:
[tex]\left\{ \begin{array}{l} -\frac{k}{2} > 0 \\ 50 > 0 \end{array} \right. \ (2)[/tex].
So for what k, does that equation have positive root(s)?
-------------
If one root is twice the other, assume that x2 = 2 x1.
Then your quadratic equation must be in some form of:
[tex]\alpha(x - x_1) (x - x_2) = \alpha(x - x_1) (x - 2x_1) = \alpha x ^ 2 - 3 \alpha x_1 x + 2 \alpha x_1 ^ 2 = 0, \ \mbox{where } \alpha \mbox{ is some number.}[/tex]
That means: [tex]2x ^ 2 + kx + 100 = \alpha x ^ 2 - 3 \alpha x_1 x + 2 \alpha x_1 ^ 2[/tex], so what's [tex]\alpha , \ x_1, \ x_2, \ k[/tex]?
#2, If one root is positive, and one is negative then x1 x2 < 0, right? That means:
[tex]t + 3 < 0[/tex].
Remember that when we have: [tex]x_1 x_2 = \frac{c}{a} < 0 \Leftrightarrow ca < 0[/tex], we also have: [tex]\Delta = b ^ 2 - 4ac > 0[/tex], that means the equation must also have 2 roots. So here we don't need to find the k value for which the discriminant is greater than 0 (since it's already greater than 0). We just have to solve:
[tex]x_1 x_2 = \frac{c}{a} < 0[/tex] for k.
Can you go from here?
 
Last edited:
  • #3
(a) If the roots of the equation 2(x)^2 + kx + 100 = 0 are positive,
find the possible range of k.
(b) If, in addition, one root is twice the other, find the roots and the value of k.
I have tried (a), but incorrect:
I have tried (a), but incorrect:
discriminate > 0
k^2 - (4)(2)(100) > 0
k^2 > 800
k > + or - 20(2)^1/2
What's wrong with my calculation?
The correct ans is:k is less than or equal to - 20 (2)^1/2
This one's easy: the question asks you to find a range for k when you know the roots are positive. Your derivation doesn't involve the roots at all! (Let alone involve the fact the roots are positive) Thus, it should be no surprise that you got the wrong answer.


2. Find the values of t for which the quadratic equation
(x)^2 - tx + t + 3 = 0 has one positive root and one negative root.
>>> I have no idea to start doing it.
The correct answer is t < -3
Can anyone help?
Wish you all have a Merry Christmas!
For almost every mathematical problem (at least for every homework problem), there is an obvious way to start. It may seem like a trivial step, but it is very frequently useful.

That method is to simply rewrite the problem in terms of the definitions.

You're talking about two things: a positive root, and a negative root. So, you should give them names! (I'll use a and b)
Then, you should write down the formulas that say that a is a positive root of that equation. (I.E. the formula that says a is a root, and the formula that says a is positive)
Then, you do the same thing for b.

It might not always be the best place to start, but you should always be able to start the problem. (e.g. VietDao suggests using a particular theorem about the roots of a polynomial, rather than just starting with the formulas that say the roots really are roots)
 
Last edited:

1. What is a quadratic inequality?

A quadratic inequality is an inequality that contains a quadratic expression of the form ax^2 + bx + c, where a, b, and c are constants and x represents a variable. It can be solved by finding the values of x that satisfy the inequality.

2. How do you graph a quadratic inequality?

To graph a quadratic inequality, first graph the related quadratic function. Then, use a test point to determine which region of the graph should be shaded to represent the solution set. Finally, plot the boundary points and shade the appropriate region.

3. What is the difference between a quadratic equation and a quadratic inequality?

A quadratic equation is an equation that is set equal to zero and can be solved to find the values of x that make the equation true. A quadratic inequality, on the other hand, is an inequality that is not set equal to zero and can be solved to find the values of x that make the inequality true.

4. How do you solve a quadratic inequality?

To solve a quadratic inequality, first use algebraic methods to rearrange the inequality into the form (ax + b)(cx + d) < 0 or (ax + b)(cx + d) > 0. Then, use the Zero Product Property to solve for the values of x that make each factor positive or negative. Finally, combine the solution sets to find the values of x that satisfy the original inequality.

5. What are the important properties of quadratic inequalities?

The important properties of quadratic inequalities are:

  • They can have multiple solutions.
  • The solutions are represented by intervals on a number line.
  • The solutions are affected by the sign of the leading coefficient.
  • The solutions can be found by graphing the related quadratic function.
  • The solutions can be found algebraically by using the Zero Product Property.

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