- #1
Gughanath
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x^2+7x>6+2x^3...I tried many times..i can't solve this :grumpy: ...please help
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
You have read the problem incorrectly. It should readaekanshchumber 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.
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.
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.
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.
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
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.