Use algebra to find the set of values of x for which x(x-5)>36

Ok so the question was use algebra to find the set of values of x for which x(x-5)>36?. I multiplied out and put all like terms on one side, and used a graph to work out when it was >0.

The answer I got was x<-4, x>9 which is correct.

I am now stuck on part b. It says use your answer to part a to find the set of values of y for which y^2(y^2-5)>36. How do i do this? can you please show your working so that I can understand? Thank you

the answer to this question (part b) should be y <-3 , y > 3 - but how?

x(x - 5) > 36

so, x² - 5x - 36 > 0

i.e. (x - 9)(x + 4) > 0

=> x < -4 or x > 9

The second part means that y² < -4 or y² > 9

No real values exist for y² < -4 as y² ≥ 0

So, only y² > 9 will give solutions

=> y² - 9 > 0

i.e. (y - 3)(y + 3) > 0

so, y < -3 or y > 3

:)>

Y^2(y^2-5)>36,
or, y^2*y^2-y^2*5>36,
or, y^4-5y^2>36,
or,y^4-5y^2-36>0,
or,y^4-(9y^2-4y^2)-36>0,
or, y^4-9y^2+4y^2-36>0,
or, y^2(y^2-9)+4(y^2-9)>0,
or, (y^2+4)(y^2-9)>0,
that implies, either, y^2+4>0,
solving which we get ,y^2>-4, which gets irrational number.
Or, y^2-9>0,
solving which ,we get , y^2>9,
or, y^2>(+,-)3^2,
or, y>+3 or-3.

The square root of x should be the value for y. The square root of -4 is an imaginary number, and they are looking for solutions in the real numbers. So they are looking for numbers that, when squared, are more than 9.

Put y^2 = x
Since y^2 cannot be<-4, you can ignore this possibility.
The other answer is y^2>9, so +/-y>+/-3
Think about it, and get the answer either y <-3 or y > 3

Replace "x" in part (a) of the question with "y^2".

So (i) y^2 < -4 … which doesn't have any real solutions

Or (ii) y^2 > 9
y < -3, y > 3

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