Quadratic inequalities

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Fred Johnston

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Apr 28, 2007
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4
Hello,
The following method was used in my Bob Jones Algebra II math book to solve quadratic inequalities:
1. Complete the square & factor the perfect square trinomial.
2. Take the square root of each side, placing abs. value signs around the root.
3. Use the compound inequality theorem to solve the resulting compound inequality.

I've been checking math sites on the internet, and they use a completely different method which involves factoring, finding the zeroes, and then choosing the right section on the number line.

I can't find any site that teaches it the way my math book does, and I'm wondering if it's an accepted method or does everyone use the method I've been finding on the sites.
Thanks so much for your help!
 
May I suggest that you simply post a question that is bothering you? We will try to help out with a standard way of solving. If it is substantially different from the Jones text we will try to explain the difference.
 
Here's a sample problem: x^2 + 2x - 5 > 0.

I completed the square & factored the resulting perfect Sq. Tri. to get (x+1)^2 > 6.
Then, I took the square root of both sides to get |x+1| > sqrt(6) .

I applied the Compound Ineq. Th. to get x+1 < -sqrt(6) or x+1 > sqrt(6)

final answer x < -1-sqrt(6) or x>-1 + sqrt(6)

Is this method correct or do I need to use the factor method then find the zeroes & determine which section on the number line applies. I'm curious to know why I haven't seen the above method on any of the math sites. Is there a catch to it?

Thanks for your help!
 
Here's a sample problem: x^2 + 2x - 5 > 0.

I completed the square & factored the resulting perfect Sq. Tri. to get (x+1)^2 > 6.
Then, I took the square root of both sides to get |x+1| > sqrt(6) .

I applied the Compound Ineq. Th. to get x+1 < -sqrt(6) or x+1 > sqrt(6)

final answer x < -1-sqrt(6) or x>-1 + sqrt(6)

Is this method correct or do I need to use the factor method then find the zeroes & determine which section on the number line applies. I'm curious to know why I haven't seen the above method on any of the math sites. Is there a catch to it?

Thanks for your help!
 
That does produce the correct answer.
So what are you saying. If it is easy to factor then do it that way.
But the way you did it does work just fine.
 
Fred Johnston said:
The following method was used in my Bob Jones Algebra II math book to solve quadratic inequalities...
Click to expand...
Also posted here.
 
Thanks to pka and stapel for answering my question! I appreciate your time very much.

Have a good weekend.
 
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