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How do you solve $x\left( {2x + 7} \right) \geqslant 0$ ?

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint:In order to solve the question first we will use distributive property which is as follows:\[a(b + c) = ab + ac\],here a is distributed and multiplied with each of the numbers separated by operator and removes the bracket . We do not have to alter the inequation inside the brackets, just distribute the outer number with each of the inner numbers . Thereafter , we will find the critical points of the inequality . Now , we will factorize and set factors to zero , so that we can get the values of x and at
Lastly we will check the intervals between the critical points .

Step by Step Solution : We are given with the inequation ( that do not has only ‘ = ’ sign in it . ) . We will make this inequation quadratic by distributing the x with the terms in the parentheses .
We always have to ensure that the outside number is applied to all the terms inside the parentheses . By applying the distributive property we get ,
$x\left( {2x + 7} \right) \geqslant 0$
$(x)(2x) + (x)(7) \geqslant 0$
$\left( {2{x^2} + 7x} \right) \geqslant 0$
Next step is to determine the critical points of the inequality .
For that we need to make the inequality as equal to zero .
$2{x^2} + 7x = 0$
Now we are going to determine the critical points by factorizing this again ,
$x\left( {2x + 7} \right) = 0$
We can clearly see one of the value of x is $x = 0$and the other $2x + 7 = 0$need to solve completely
as ,
Subtract both side by 7 ,
$
2x + 7 - 7 = 0 - 7 \\
2x = - 7 \\
$
Dividing both sides by 2 , we get the value of x as $x = - \dfrac{7}{2}$ .
Therefore , we got two critical points from the equation $x = 0$ and $x = - \dfrac{7}{2}$.
Now we are going to check the intervals between the critical points .
$x \leqslant - \dfrac{7}{2}$( this works )
And $x \geqslant 0$( this works )
$ - \dfrac{7}{2} \leqslant x \leqslant 0$ ( this does not works )

Therefore, the answer is $x \leqslant - \dfrac{7}{2}$ And $x \geqslant 0$.

Note : We always have to ensure that the outside number is applied to all the terms inside the parentheses .
We use the distributive property generally when the two terms inside the parentheses cannot be added or operated because they are not the like terms .
The points on the graph are called to be critical points where the function 's rate of change is differentiable or changeable —either a change from increasing to decreasing or in some uncertain fashion .
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