For what value of x is 6x^2 not less than (6x)^2 ?

[Sarah2391]

For what value of x is 6x squared not less than (6x)squared ?

(a) -6

(b) 0

(c) 1/6

(d) 1

(e) For no value of x

 

[stapel]

For what value of x is 6x squared not less than (6x)squared ?

You set up the stated inequality:

. . . . .6x[sup:y5rdbrf0]2[/sup:y5rdbrf0] > (6x)[sup:y5rdbrf0]2[/sup:y5rdbrf0]

...simplified:

. . . . .6x[sup:y5rdbrf0]2[/sup:y5rdbrf0] > 36x[sup:y5rdbrf0]2[/sup:y5rdbrf0]

...and put everythong over on one side:

. . . . .0 > 36x[sup:y5rdbrf0]2[/sup:y5rdbrf0] - 6x[sup:y5rdbrf0]2[/sup:y5rdbrf0]

. . . . .36x[sup:y5rdbrf0]2[/sup:y5rdbrf0] - 6x[sup:y5rdbrf0]2[/sup:y5rdbrf0] < 0

...and... then what?

Please be specific, starting with how you solved the quadratic. Thank you!

Eliz.

 

[Sarah2391]

I was actually able to get that far, I just came seem to finish out the problem, my brain is freezing
Can you show how to get to the final answer?

 

[stapel]

I was actually able to get that far...

In the future, kindly provide us with what you've done so far, so we don't flounder about, wondering what you need, pointlessly repeating what (you say now) you've already completed.

Since you now have a very simple quadratic in two quadratic terms, it is surprising that you were unable to see that these two terms can be combined. Your first step now would be to simplify 36x[sup:2k7jv966]2[/sup:2k7jv966] - 6x[sup:2k7jv966]2[/sup:2k7jv966]. Then, as mentioned earlier, solve the resulting quadratic for its zeroes.

This will divide the number line into intervals. Use whatever method you've learned (graphing, test points, tables of values, etc) to find where the quadratic is on the "right" side of the axis.

If you get stuck again, please SHOW what you have done, so we know where you're actually having the trouble. Thank you!

Eliz.

 

[Deleted member 4993]

For what value of x is 6x squared not less than (6x)squared ?,,Remember - this includes "equal to"

(a) -6

(b) 0

(c) 1/6

(d) 1

(e) For no value of x