Question

Solve

Answer 1

\(\large \color{black}{\begin{align}
&x^2+5|x|+6=0\\~\\
&x\in \mathbb{R}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

Eyeball problem. No Solution.

\(x^{2}\) is at least zero (0).
\(5|x|\) is at least zero (0).
The entire expression CANNOT be less than 6.

Done!

Sometimes, they just jump out at you. No need to fret over them.

Answer 3

is there any other way to do this

Answer 4

If you graphed it, you would get:
https://www.desmos.com/calculator/3bps8xaezl

and you can tell that there will never be an x value that will make y=0 because the smallest value of y is 6.

Answer 5

And to explain what tkhunny said for others who don't understand what he said right away like myself.

We have `+ 6` in the equation.
So that means `x^2 + x|5|` has to equal `-6` in order for the whole equation to be negative

However, there will never be any x value that will make it -6.
And since we have `x^2`, any value of x that goes into that will be positive
And then there is `5|x|` any x value plugged in will also come out as positive.

So thus `x^2+5|x|` will never be negative. So it can not have any solutions.

Answer 6

Here's another way.

$$
x^2+5|x|+6=0
$$
means
for \(x\ge0\)
$$
x^2+5|x|+6=x^2+5x+6=0
$$
Which has solutions \(x=-3,x=-2\). But \(x\ge0\) so there is no solution.
for \(x\le0\)
$$
x^2+5|x|+6=x^2-5x+6=0
$$
Which has solutions \(x=3,x=2\). But \(x\le0\) so there is no solution.

Answer 7

Hi
Use following tool to solve this problem
http://www.acalculator.com/log.html

Answer 8

Another way? Why?!

Shall we use the brute force of 7th grade algebra or whatever app someone wrote or shall we think and use good judgment and just be done?

Note: On a GRE, LSAT, MCAT, or other significant examination, you WILL want the fast way.

Answer 9

teacher wont give me marks

Answer 10

Shame on that teacher. If you were my child, or a student under my tutelage, I would have words with that teacher.

Logic is as valid as any other construct. Shameful. Simply shameful.