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# Inequalities watch

1. Can someone check to see if I am doing this right?

Factorise and Solve the Inequality:

4x^2 < 25

4x^2 - 25=0

(2x - 5)^2

(x - 5/2)=0

X= 5/2

-5/2 < x < 5/2

I am self teaching Maths, I have the questions and answers, I know the answer is correct, I would however feel at ease knowing my workings are correct.
2. (Original post by Shakaa)
Can someone check to see if I am doing this right?

Factorise and Solve the Inequality:

4x^2 < 25

4x^2 - 25=0

(2x - 5)^2

(x - 5/2)=0

X= 5/2

-5/2 < x < 5/2

I am self teaching Maths, I have the questions and answers, I know the answer is correct, I would however feel at ease knowing my workings are correct.
You sure this part is correct?

3. You could get to the answer in just two steps:

4x^2 < 25
x^2 < 25/4
|x| < 5/2 (|x| means the magnitude i.e. size of x)

Also for when you have the difference of two squares.
4. Doing it with your method, OP, it would be

4x²<25
<=> 4x²-25<0
<=> (2x-5)(2x+5)< 0

Now either you do the
x -inf - 5/2 5/2 +inf

2x-5 Negative | neg | pos
2x+5 Negative | pos | pos

4x²-25 Positive | neg | pos

Therefore 4x²-25<0 <=> x belongs to ]-5/2,5/2[

Either you know that for a polynomial P(x)=ax²+bx+c,
the polynomial is of the sign of a outside of the "roots", and of the contrary sign inside the roots

Here, a=4>0 so P(x)>0 for x belonging to ]-inf, -5/2[ u ]5/2,+inf[
and P(x)<0 for x E ]-5/2,5/2[

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