Question

Hey guys, try this one
if x satisfied 2^(3x^2)+2^[(x^2)+1]is greater or equal to 2^[(3x^2)+2]-5*2^(2x^2) find x

Answer 1

\[2^{3x ^{2}}+2^{x ^{2}+1} \ge 2^{3x ^{2}+2}-5*2^{2x ^{2}}\]

Answer 2

Let x^2 be a
\[2^{3a}+ 2^{a+1} \ge 2 ^{3a+2}-5 * 2^{2a}\]

Answer 3

\[2^{3a} + 2^{a+1} \ge 2^{3a+2} - 20 * 2^a\]\[2^{a+1}+20 * 2^a \ge 2^{3a+2}-2^{3a} \implies 2^a (2 + 20) \ge 2^{3a}(2^2-1)\]\[= 22 * 2^a \ge 2^{3a} * 3\]\[\frac{22}{3} \ge 2^{2a}\]Solve for a. then subsitute bak to a = x^2

Answer 4

Or you can let 2^(x^2)=a

\[a^3+2a \ge 4a^3-5a^2\]

Answer 5

\[-3a^3+5a^2+2a \ge 0\]\[a(-3a^2+5a+2) \ge 0 \]\[a(a-2)(3a+1) \ge 0 \]

So \[a \ge 2\]\[\frac{-1}{3} \le a \le 0\]

Answer 6

Sorry I didn't really write how I got the result. You solve a(a-2)(3a+1)=0 and then enter in value either side back into a(a-2)(3a+1)>=0 to see which way the inequalities go.

Answer 7

ok thx

Answer 8

i got \[x \in [-1,1]\]

Answer 9

Yes your answer is correct. There's a mistake in my answer. Which method did you use?

Answer 10

I see my error. I did not go far enough. I needed to substitute 2^(x^2)=a back in to solve and find the inequalities.

So \[2^{x^2} =2\]\[2^{x^2}=0\]\[2^{x^3}=\frac{-1}{3}\]

So the bottom two are invalid answers and the top one give +1 and -1 as answers

Answer 11

Good work MrBank

Answer 12

well from \[2^{3x ^{2}}+2^{x ^{2}+1}\ge2^{3x ^{2}+2}-5*2^{2x ^{2}}\]
\[2^{3x ^{2}}+2*2^{x ^{2}}\ge4*2^{3x ^{2}}-5*2^{2x ^{2}}\]
\[3*2^{3x ^{2}}-5*2^{2x ^{2}}-2*2^{x ^{2}}\le0\]
divide all by \[2x ^{2}\] which is surely positive

and so on haha
i see you've found out so i'll stop here.
whewww
thx again

Answer 13

\[2^{x ^{2}}\] my mistake in divide all

Answer 14

every one make mistakes some times but u fixed it good

Answer 15

thx sunset