Help Finding Perfect Square Inequalities Practice Problems

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I need help finding specific problems for polynomial Inequalities. Every problem I’ve attempted with perfect squares has ended up wrong. I’ve realized my mistakes, and am ready to try my new skills on some practice problems. There is one problem: I can’t find any perfect Quadratic Equations that factor into perfect squares to practice on. This is frustrating, as I would like to master these types of Inequalities. These are the only problems I have trouble with, yet I cannot find any problems corasponding to my delemma. Could anyone please help me find some practice problems with Perfect Squares Inequalities?

An example of this would be: 9u^2-6u+1<0

Thank You.

 

[JeffM]

I need help finding specific problems for polynomial Inequalities. Every problem I’ve attempted with perfect squares has ended up wrong. I’ve realized my mistakes, and am ready to try my new skills on some practice problems. There is one problem: I can’t find any perfect Quadratic Equations that factor into perfect squares to practice on. This is frustrating, as I would like to master these types of Inequalities. These are the only problems I have trouble with, yet I cannot find any problems corasponding to my delemma. Could anyone please help me find some practice problems with Perfect Squares Inequalities?

An example of this would be: 9u^2-6u+1<0

Thank You.

You can easily make them up.

$\displaystyle (ax \pm b)^2 = ax^2 \pm 2abx + b^2.$

Substitute any integers for a and b, and you have the type of polynomial you want.

I suggest that you try, step by step,

$\displaystyle 4x^2 - 20x + 25 < 1$

show your work, and let us see where you are having trouble.

My advice on inequalties is to solve the related equality first.

Start by solving $\displaystyle 4x^2 - 20x + 25 = 1$

Now there may be no real solution, one real solution, or two real solutions to a quadratic equation. Think about what each of those cases means with respect to an inequality.

 

[spaceshowfeature1]

You can easily make them up.

$\displaystyle (ax \pm b)^2 = ax^2 \pm 2abx + b^2.$

Substitute any integers for a and b, and you have the type of polynomial you want.

I suggest that you try, step by step,

$\displaystyle 4x^2 - 20x + 25 < 1$

show your work, and let us see where you are having trouble.

My advice on inequalties is to solve the related equality first.

Start by solving $\displaystyle 4x^2 - 20x + 25 = 1$

Now there may be no real solution, one real solution, or two real solutions to a quadratic equation. Think about what each of those cases means with respect to an inequality.

Thank you! I will solve and use Symbolab. Have a great day!

 

[spaceshowfeature1]

You can easily make them up.

$\displaystyle (ax \pm b)^2 = ax^2 \pm 2abx + b^2.$

Substitute any integers for a and b, and you have the type of polynomial you want.

I suggest that you try, step by step,

$\displaystyle 4x^2 - 20x + 25 < 1$

show your work, and let us see where you are having trouble.

My advice on inequalties is to solve the related equality first.

Start by solving $\displaystyle 4x^2 - 20x + 25 = 1$

Now there may be no real solution, one real solution, or two real solutions to a quadratic equation. Think about what each of those cases means with respect to an inequality.

The answer I got for the problem you gave me is: 2<x<3 or in interval notation: (2,3)

 

[JeffM]

The answer I got for the problem you gave me is: 2<x<3 or in interval notation: (2,3)

You are correct, but you still are not showing your work. Just telling you whether your answer is right or wrong does not let us see what you are thinking. So let's walk through this example.

\(\displaystyle 4x^2 - 20x + 25 = 1 \implies 4x^2 - 20x + 24 = 0 \implies 4(x^2 - 5x + 6) = 0 \implies \\

4(x - 2)(x - 3) = 0 \implies x = 2 \text { or } x = 3.\)

Now why did I suggest starting with the equality? Because the function

$\displaystyle f(x) = 4x^2 - 20x + 25$

is continuous, it will be greater or less than a given number EVERYWHERE in domains to one side of a point of intersection. It will equal that number only at points of intersection. We have 2 points of intersection so we have 2 + 1 = 3 domains to test. So in the domain less than 2, pick a convenient value like x = 0.

$\displaystyle 4(0^2) - 20(0) + 25 = 25 > 1 \implies f(x) > 1 \text { if } x < 2.$

And we know that $\displaystyle f(x) = 1 \text { if } x = 2.$

Now pick a convenient value for x in the domain between 2 and 3 such as 2.5.

\(\displaystyle 4(2.5^2) - 20(2.5) + 25 = 4(6.25) - 50 + 25 = 25 - 25 = 0 < 1 \implies\\

f(x) < 1 \text { if } 2 < x < 3.\)

And we know that $\displaystyle f(x) = 1 \text { if } x = 3.$

Now pick a convenient value in the domain greater than 3 like 10.

$\displaystyle 4(10^2) - 20(10) + 25 = 400 - 200 + 25 = 225 > 1 \implies f(x) > 1 \text { if } x > 3.$

You now know the relationship between f(x) and 1 for every value of x and so can answer any question about inequalities between f(x) and 1.

Try this one. What is the domain where

$\displaystyle 4x^2 - 20x + 25 < -\ 3.$

 

[Denis]

My advice on inequalties is to solve the related equality first.

Start by solving $\displaystyle 4x^2 - 20x + 25 = 1$

AGREE...good advice!!

 

[Denis]

I will solve and use Symbolab.

You don't have a pencil?

 

[pka]

I need help finding specific problems for polynomial Inequalities. Every problem I’ve attempted with perfect squares has ended up wrong. I’ve realized my mistakes, and am ready to try my new skills on some practice problems. There is one problem: I can’t find any perfect Quadratic Equations that factor into perfect squares to practice on. This is frustrating, as I would like to master these types of Inequalities. These are the only problems I have trouble with, yet I cannot find any problems corasponding to my delemma. Could anyone please help me find some practice problems with Perfect Squares Inequalities?
An example of this would be: 9u^2-6u+1<0

I assume that we are working with real numbers.
Can you find a real number $\displaystyle x$ such that $\displaystyle x^2<0~?$
$\displaystyle 9u^2-6u+1<0$
$\displaystyle (3 u -1)^2 <0$

What does that mean?

 

[Steven G]

You don't have a pencil?

Come on Dennis, you're into HI finances so please buy the student a few pencils.

 

[Denis]

Come on Dennis, you're into HI finances so please buy the student a few pencils.

HOKAY...you buy him some paper...

 

[spaceshowfeature1]

HOKAY...you buy him some paper...

Thanks im good with Quadratic inequalities now, but still having problems with ones of higher power.