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How do you solve the inequality \[{x^3} - {x^2} - 6x > 0\]?

seo-qna
Last updated date: 16th Jun 2024
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Answer
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Hint: An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value. Here we need to solve for ‘x’ which is a variable. Solving the given inequality is very like solving equations and we do most of the same thing but we must pay attention to the direction of inequality\[( \leqslant , > )\]. We can take ‘x’ common and we will have a quadratic equation and we can solve it easily.

Complete step by step solution:
We find the factor of \[{x^3} - {x^2} - 6x\] and we evaluate for each term.
Taking ‘x’ common we have,
\[x({x^2} - x - 6) > 0\]
We can split the middle term in parentheses,
\[x({x^2} - 3x + 2x - 6) > 0\]
\[x\left( {x(x - 3) + 2(x - 3)} \right) > 0\]
\[x(x - 3)(x + 2) > 0\]
Since we have greater than sign it means that \[x \ne 0\].
If \[x > 0\] then other two factors must be greater than zero. That is
 \[(x - 3)(x + 2) > 0\]
That is \[x > 3\]. (That is if we have\[x < 3\] then will have \[(x - 3)\] is negative and since \[(x + 2)\] is positive then the product becomes negative which is contradiction to the fact \[x > 0\])
If \[x < 0\] then \[(x - 3)\] will be negative then \[(x + 2)\] must be positive. So that the product of \[x < 0\],\[(x - 3)\] and \[(x + 2)\] will be greater than zero.
Thus \[(x + 2) > 0\]
\[x > - 2\]

Hence the solution of \[{x^3} - {x^2} - 6x > 0\] are \[x > 3\] or \[ - 2 < x < 0\].

Note: We know that \[a \ne b\] says that ‘a’ is not equal to ‘b’. \[a > b\] means that ‘a’ is less than ‘b’. \[a < b\] means that ‘a’ is greater than ‘b’. These two are known as strict inequality. \[a \geqslant b\] means that ‘a’ is less than or equal to ‘b’. \[a \leqslant b\] means that ‘a’ is greater than or equal to ‘b’.
The direction of inequality do not change in these cases:
i) Add or subtract a number from both sides.
ii) Multiply or divide both sides by a positive number.
iii) Simplify a side.
iv) The direction of the inequality change in these cases:
v) Multiply or divide both sides by a negative number.
vi) Swapping left and right hand sides.