Help needed please

[ButterflyEma]

Hello everyone. I have problem solving two inequalities. I have an idea and I solved it but the answer is not correct. Could you please help me how to get to the answer and where I’m making mistake. In pictures below you can see how I did it and correct answers.

I really need help with this one. Thank you

 

[HallsofIvy]

Unfortunately, by posting pictures of hand written equations rather than typing in the problem, you have made it difficult for others to help you. I THINK the inequality you are concerned with is \(\displaystyle x^2> x\sqrt{x^2+ 5x+ 4}\). For continuous functions, the region where "A< B" is always separated from the region where "A> B" by a point where "A= B".

In other words, I would start by solving the equation \(\displaystyle x^2= x\sqrt{x^2+ 5x+ 4}\). Obviously x= 0 is a solution since that makes both sides 0. If x is not 0 we can divide by 0 to get \(\displaystyle x= \sqrt{x^2+ 5x+ 4}\). Get rid of the square root by squaring both sides: \(\displaystyle x^2= x^2+ 5x+ 4\). Subtracting \(\displaystyle x^2\) from both sides, 5x+ 4= 0 so x= -4/5.

But squaring both sides of the equation can introduce solutions that do not work for the original equation so I need to check that answer in the original equation. The original equation is \(\displaystyle x^2= x\sqrt{x^2+ 5x+ 4}\) and it is clear that x cannot be negative: the left side is a square so is positive while the right side is a negative number, x, times a square root which is positive.

So x= 0 is the only solution to the equation, to find solutions to the inequality we only need to check one number on either side of x= 0. If x< 0, that is, if x is negative, the left side, \(\displaystyle x^2\) is positive while the right side, \(\displaystyle x\sqrt{x^2+ 5x+ 4}\) is a negative number times a positive number whichj is negative. YES, a positive number is always larger than a negative number so the every negative number satisfies this inequality.

For the positive numbers, try x= 1. The left side is \(\displaystyle 1^2= 1\) while the right side, \(\displaystyle 1\sqrt{1^2+ 5(1)+ 4}=\sqrt{10}> 1\). That is enough to tell us that NO positive number is a solution.

The solution set is "the set of all negative numbers".

 

[ButterflyEma]

Thank you so much for this answer of yours. But unfortunately the right answer is X e (-inf, -3) U (-2,0). Which yes, is a set of negative numbers but I still dont know how to get to those exact numbers.

And that is first equation and the second picture is another one. Basicly two different problems but I believe they have simmilar solution.

But thank you either way. I understand you point of view

 

[lex]

[MATH]x^2>x\sqrt{x^2+5x+4}[/MATH]A few observations first:
(A) For this to exist (as a real number) we need [MATH](x+1)(x+4)≥0[/MATH]i.e. [MATH]\boxed{ \hspace2ex x≤-4 \text{ or } x≥-1\}}\hspace2ex[/MATH] Restriction (1)

(B) As [MATH]x^2[/MATH] and [MATH]\sqrt{x^2+5x+4}[/MATH] are both ≥0
when [MATH]\boldsymbol{x}[/MATH] is negative, then [MATH]x^2[/MATH] is positive and then [MATH]x^2>x\sqrt{x^2+5x+4} [/MATH]
[MATH] \text{Now find when} \hspace2ex x^2 =x \sqrt{x^2+5x+4}\\ x(x-\sqrt{x^2+5x+4})=0\\ \begin{align*} x=0 \text{ or } x &= \sqrt{x^2+5x+4}\\ & \text{but this } \rightarrow x≥0 \hspace1ex \text{, but then clearly } \sqrt{x^2+5x+4}>x\\ &\text{so no solutions}\\ \end{align*}[/MATH]i.e. only solution is [MATH]x=0[/MATH]
Now we need to know which region (x<0 or x>0) satisfies the inequality.
From (B) above, we know that [MATH]x<0 \rightarrow x^2>x\sqrt{x^2+5x+4}[/MATH]So [MATH]\boxed{ \hspace2ex x<0\hspace2ex}[/MATH] Restriction (2)
Putting these restrictions (1) and (2) together:
[MATH](-\infty,-4] \cup [-1,0)[/MATH]
The "correct solution" you are quoting, is for a different question: [MATH]x^2>x\sqrt{x^2+5x+\boldsymbol{6}}[/MATH]and should be [MATH](-\infty,-3] \cup [-2,0)[/MATH]

 

[lex]

[MATH]\frac{1}{x}(1-\sqrt{1-9x^2})<1 \hspace2ex[/MATH] (*)

A few observations first:
(A) For this to exist (as a real number) we need [MATH]1-9x^2≥0 \text{ and }x≠0[/MATH]Clearly [MATH]\rightarrow[/MATH][MATH]\boxed{ \hspace2ex x≠0 \text{ and } -\frac{1}{3}≤x≤\frac{1}{3}}\hspace2ex[/MATH] Restriction (1)

(B) Clearly[MATH](1-\sqrt{1-9x^2})\in (0,1][/MATH], so when [MATH]\boldsymbol{x}[/MATH] is negative, then [MATH]\frac{1}{x}(1-\sqrt{1-9x^2})<1[/MATH]since [MATH]\frac{1}{x}<0[/MATH] and [MATH](1-\sqrt{1-9x^2})>0[/MATH]
[MATH] \text{Now find when} \hspace2ex \frac{1}{x}(1-\sqrt{1-9x^2})=1\\ 1-x=\sqrt{1-9x^2}\\ \rightarrow x^2-2x+1=1-9x^2\\ \rightarrow 10x^2-2x=0\\ \rightarrow x(5x-1)=0\\ x=0 \text { or }x=\frac{1}{5}[/MATH](even though x≠0 in our question, we will consider the regions around 0, since the function [MATH]\frac{1}{x}(1-\sqrt{1-9x^2})[/MATH] is discontinuous).
Examining the regions, against the inequality (*); from B we know x<0 is included, and we find that [MATH]0<x<\frac{1}{5}[/MATH] is included.

[MATH]\therefore \boxed{ \hspace2ex x<\frac{1}{5}\hspace2ex}[/MATH] Restriction (2)
Putting these restrictions (1) and (2) together:
[MATH]-\frac{1}{3}≤x<\frac{1}{5}[/MATH], and [MATH]x≠0[/MATH]i.e. [MATH][-\frac{1}{3},0) \cup (0,\frac{1}{5})[/MATH]

 

[Steven G]

Don't you see that what you call the answer MUST be wrong. You correctly concluded that x<=-4 and x>=-1 to make what is under the radical positive.

The solution says that can be any number in (-infty,-3) or (-2,0)
Well, -3.5 IS in (-infty,_3) but is -3.5 is NOT less -4 nor bigger than -1.
Also, -1.5 is in (-2,0) while -1.5 is not less than -4 or is it greater than -1.

 

[Otis]

Unfortunately, by [not] typing in the problem, you have made helping difficult [for others] …

Thank goodness, you were willing to help.