Question

To be solved 2x^2-5x+3≤ 0

Answer 1

first step: can you factor this quadratic ?

Answer 2

To solve it you can complete the square:
\[2x^2-5x+3\le0\]
First divide everything by 2:
\[x^2-\frac{ 5 }{ 2 }x+\frac{ 3 }{ 2 }\le0\]

Subtract 3/2 from both sides:
\[x^2-\frac{ 5 }{ 2 }x \le-\frac{ 3 }{ 2 }\]

Take half of the b-term "-5/2" and square it so half of it is -5/4 and if you square that you get 25/16. Add that to both sides:
\[x^2-\frac{ 5 }{ 2 }x+\frac{ 25 }{ 16 }\le -\frac{ 3 }{ 2 }+\frac{ 25 }{ 16 }\]

Simplify:
\[x^2-\frac{ 5 }{ 2 }x+\frac{ 25 }{ 16 }\le \frac{ 1 }{ 16 }\]
factor the right side to the half the b-term or -5/4 you found earlier:
\[(x-\frac{ 5 }{ 4 })(x-\frac{ 5 }{ 4 })\le \frac{ 1 }{ 16}\]
becomes:



\[(x-\frac{ 5 }{ 4 })^2\le \frac{ 1 }{ 16 }\]

Take the square root of both sides remember square roots are positive and negative:
\[x-\frac{ 5 }{ 4 }\le \pm \frac{ 1 }{ 4 }\]

Add 5/4 to both sides:
\[x \le \frac{ 3 }{ 2 }\]

or
\[x \le1\]

Answer 3

thanks

Answer 4

to solve
\[ 2x^2-5x+3≤ 0 \]
first factor it into
\[ (2x -3)(x-1) ≤ 0 \]
Now notice that the two terms on the left will be negative (i.e. less than zero) when one term is negative and the other is positive:
so you need either
2x-3≤ 0 and x-1 ≥ 0 --> x≤ 3/2 and x ≥ 1
or
2x-3 ≥ 0 and x-1 ≤ 0 --> x ≥ 3/2 and x ≤ 1 *** this case is not possible**

the answer is
1 ≤ x ≤ 3/2

Answer 5

@phi got the same answer i got but wrote it as a compound inequality!