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Find the range of values of $x$ which satisfies the inequality $(2x+1)(3x-1)<14$.

I have done more similar sums and I know how to solve it. I tried this one too but my answer doesn't matches the book's answer.

I did in this way: Solving, i.e, After rearranging the equation and factorizing it, I get

$(3x+5)(2x-3)<0$

Is this right?

Anyways, then finding the range of values, I get: $x<-\frac{5}{3}$ or $ x>\frac{3}{2}$

But my book says the answer should be $-\frac{5}{3}<x<\frac{3}{2}$.

Did I do any mistake?

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  • $\begingroup$ It looks like you did everything right except realize that you are doing $(3x+5)(2x-3)<0$ instead of $(3x+5)(2x-3)\gt0$... Note that the LHS is a parabola facing "up" so it is increasing as $x$ moves away from the center... $\endgroup$ – abiessu Feb 28 '14 at 19:07
  • $\begingroup$ @abiessu I used the sign $<$....Do I have to change it to $>$? $\endgroup$ – Kiara Feb 28 '14 at 19:09
  • $\begingroup$ No, I'll try to explain better. $\endgroup$ – abiessu Feb 28 '14 at 19:10
  • $\begingroup$ @abiessu I still did not understand! $\endgroup$ – Kiara Feb 28 '14 at 19:19
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Consider the inequality

$$ab\lt 0\tag 1$$

This inequality holds exactly when $a\gt 0, b\lt 0$ or when $a\lt 0, b\gt 0$, but not when $a\gt 0, b\gt 0$ or $a\lt 0,b\lt 0$.

In this question, the inequality has been reduced to

$$(3x+5)(2x-3)\lt0$$

Next it is necessary to find out the comparisons of $3x+5\lt 0$ and $2x-3\lt 0$, which have been found as $x\lt -\frac 53$ and $x\lt \frac 32$.

Further, we have that $-\frac 53\lt \frac 32$. So what remains is to put this information together with the conditions satisfying $(1)$. In particular, we have $2x-3\lt 3x+5$ when $x\gt -\frac 53$. If $x\lt -\frac 53$ then both $2x-3$ and $3x+5$ are negative, which breaks the conditions on $(1)$, so we cannot have $x\lt -\frac 53$, and similarly if $x\gt \frac 32$ then both $2x-3$ and $3x+5$ are positive, which also breaks our conditions. But if $-\frac 53\lt x\lt \frac 32$ then $2x-3$ is negative and $3x+5$ is positive, and the necessary conditions are met to satisfy our inequality.

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Thinking purely geometrically, you might realize that your left hand side defines a parabola $y=(2x+1)(3x-1)$ opening up, while the right hand side is a constant. Again, from a purely geometric point of view, this must be an open interval (possibly empty).

enter image description here

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The inequality you obtained, $$(3x+5)(2x-3) < 0,$$ is correct. The next step is to find where the LHS equals zero: this gives you the $boundary$ of the intervals for which the inequality is satisfied. You will find $x = -5/3$ and $x = 3/2$. This means we should consider three intervals: $x < -5/3$, $-5/3 < x < 3/2$, and $x > 3/2$.

Pick test points from each interval. For example, pick $x = -2$. Plug it into the inequality. Is it true? $(3(-2) + 5)(2(-2) - 3) = (-1)(-7) = 7 > 0$, so $x = -2$ does not satisfy the inequality. Next, pick $x = 0$: we get $(3(0)+5)(2(0)-3) = (5)(-3) = -15 < 0$, which does satisfy the inequality, so we know that $-5/3 < x < 3/2$ is a solution to the given inequality. Finally, just to check, try picking $x = 2$. Does it work?

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You're basically looking for values when one of 3x+5,2x-3 is negative and the other one positive(or smaller then zero) Since when $$x>\frac{3}{2} \implies (3x+5)(2x−3)>0\\x<-\frac{5}{3} \implies (3x+5)(2x−3)>0$$ In first case both 3x+5 and 2x-3 are positive,and in second both are negative

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