1

If the system of equations $3x^2+2x-1<0$ and $(3a-2)x-a^2x+2<0$ possesses a solution, find the least natural number $a$

My attempt is as follows:-

$$3x^2+3x-x-1<0$$ $$3x(x+1)-1(x+1)<0$$ $$x\in\left(-1,\dfrac{1}{3}\right)$$

$$(-a^2+3a-2)x<-2$$ $$(a^2-3a+2)x>2$$

Case $1$: $a^2-3a+2<0$

$$a\in(1,2)$$

In this interval there is no natural number, hence no need to proceed further.

Case $2$: $a^2-3a+2\ge0$

$$a\in(-\infty,1]\cup[2,\infty)$$

Checking for $a=1$:

$0>2$, which is not possible

Checking for $a=2$

$(4-6+2)x>2$

$0>2$, which is not possible

$$x>\dfrac{2}{a^2-3a+2}$$ $$x\in\left(\dfrac{2}{a^2-3a+2},\infty\right)$$

If system of equations possess a solution, then $\dfrac{2}{a^2-3a+2}<\dfrac{1}{3}$

$$6<a^2-3a+2$$ $$a^2-3a-4>0$$ $$a^2-4a+a-4>0$$ $$a(a-4)+(a-4)>0$$ $$(a+1)(a-4)>0$$ $$a\in(-\infty,-1)\cup(4,\infty)$$

So $a=5$ should be the answer, but actual answer is $2$

Jean Marie
  • 81,803
user3290550
  • 3,452
  • 4
    There is clearly no solution to the second equation when $a = 2$, as you verified for yourself. The solution clearly does not match the question. I, however, think that the question statement might have a typo, and maybe a square was left off one of the $x$ terms in the second inequality. – user759562 Mar 21 '20 at 07:40
  • I would like to congratulate you : you are very rigorous, with a clear mathematical writing. – Jean Marie Mar 21 '20 at 08:04
  • If $$ \left(a^{2}-3 a+2\right) x>2 $$. So either $$ a^{2}-3 a+2>0 \text { and } x>0 $$ or $$ a^{2}-3 a+2<0 \text { and } x<0 $$. Because product of two number will be positive iff both are negative or positive. You did the mistake in second case by assuming the expression to be greater then or equal to 0. Because product of two number can't be greator then 2 if one is 0. Check it! –  Mar 21 '20 at 08:38
  • if $a=2$ then the system has no solution so $a=2$ cannot be the right answer – Lozenges Mar 21 '20 at 09:22

1 Answers1

1

Inequalities can be sumized in this system:

$x<-1$ and $x>\frac{1}{3}$

$a<=\frac{3\sqrt{x}-\sqrt{x+8}}{2\sqrt{x}}$

$a>=\frac{3\sqrt{x}+\sqrt{x+8}}{2\sqrt{x}}$

The least natural number is $a=4$ for $x=\frac{1}{3}$.

Other values are:

$a=5$ for $x=\frac{1}{6}$;

$a=6$ for $x=\frac{1}{10}$;

$a=7$ for $x=\frac{1}{15}$;

$a=8$ for $x=\frac{1}{21}$;

$a=9$ for $x=\frac{1}{28}$;

$a=10$ for $x=\frac{1}{36}$,

etc…

giuseppe mancò
  • 1,671