Solve x^2+7x>6+2x^3: Tips & Solutions

  • Thread starter Gughanath
  • Start date
In summary, the problem is to solve the inequality x^2 + 7x > 6 + 2x^3, which can be rewritten as -2x^3 + x^2 + 7x - 6 > 0. One possible way to solve this is by factoring the expression, which results in (x-1)(x+2)(2x-3) > 0. This implies that either x<0 or (2x^2 - x - 1) < 0, and since the expression must be negative, one of the terms must be negative.
  • #1
Gughanath
118
0
x^2+7x>6+2x^3...I tried many times..i can't solve this :grumpy: ...please help
 
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  • #2
x^2 + 7x > 6 + 2x^3 = -2x^3 + x^2 + 7x - 6 > 0

Try solving for:
-2x^3 + x^2 + 7x - 6 = 0
 
  • #3
Dr-NiKoN said:
x^2 + 7x > 6 + 2x^3 = -2x^3 + x^2 + 7x - 6 > 0

Try solving for:
-2x^3 + x^2 + 7x - 6 = 0

I can't solve that sorry...could you help any further?
 
  • #4
I noticed the sum of the coefficients is 1 so (x-1) is a factor:

[tex]2x^3-x^2-7x+6 = (x-1)(x+2)(2x-3)[/tex]

That should get you started!
 
  • #5
try it,
x^2 + 7x>6x + 2x^3
or, 2x^3 - x^2 -x<0
or, x(2x^2 - x - 1)<0
this implies, either x<0 or (2x^2 - x - 1) <0
as the expression is negative which is only possible if one of the term is negative.
 
  • #6
aekanshchumber said:
try it,
x^2 + 7x>6x + 2x^3
or, 2x^3 - x^2 -x<0
or, x(2x^2 - x - 1)<0
this implies, either x<0 or (2x^2 - x - 1) <0
as the expression is negative which is only possible if one of the term is negative.
You have read the problem incorrectly. It should read
[tex] x^2 + 7x > 6 + 2x^3 [/tex]
 
Last edited:

1. What is the first step in solving the inequality x^2+7x>6+2x^3?

The first step in solving this inequality is to rearrange the terms so that all the variables are on one side and all the constants are on the other side. In this case, we can subtract 6 from both sides to get x^2+7x-6>2x^3.

2. How do I determine the critical values for this inequality?

The critical values for this inequality are the values of x at which the inequality switches from being greater than to being less than. To find these values, set the expression x^2+7x-6 equal to 0 and solve for x using the quadratic formula. The resulting values will be the critical values.

3. Can I use the same approach for solving this inequality as I would for a regular equation?

No, inequalities require a slightly different approach than regular equations. In this case, we will need to consider the different possible scenarios based on the critical values we found in the previous step.

4. What are the different cases to consider when solving this inequality?

There are three cases to consider when solving this inequality:1. When x is less than the smallest critical value2. When x is between the two critical values3. When x is greater than the largest critical value

5. How do I express the final solution for this inequality?

The final solution for this inequality will be expressed in interval notation, using the critical values and the results of each case. For example, the solution may look like (-infinity, -3) U (1, infinity). It is important to check each interval to ensure that the inequality holds true for all values within it.

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