Respuesta :
Step 1: Factor [tex]2 x^{2} - 3x -2[/tex]
1. Multiply 2 by -2, which is -4.
2. Ask: Which two numbers add up to -3 and multiply to -4?
3. Answer: 1 and -4
4. Rewrite [tex]-3x[/tex] as the sum of [tex]x[/tex] and [tex]-4x[/tex]
[tex]2 x^{2} +x-4x-2\ \textless \ 0[/tex]
Step 2: Factor out common terms in the first two terms, then in the last two terms.
[tex]x(2x+1)-2(2x+1)\ \textless \ 0[/tex]
Step 3: Factor out the common term [tex]2x+1[/tex]
[tex](2x+1)(x-2)\ \textless \ 0[/tex]
Step 4: Solve for [tex]x[/tex]
1. Ask: When will [tex](2x+1)(x-2)[/tex] equal zero?
2. Answer: When [tex]2x + 1 = 0[/tex] or [tex]x-2=0[/tex]
3. Solve each of the 2 equations above:
[tex]x=- \frac{1}{2} ,2[/tex]
Step 5: From the values of [tex]x[/tex] above, we have these 3 intervals to test.
x = < -1/2
-1/2 < x < 2
x > 2
Step 6: Pick a test point for each interval
For the interval [tex]x\ \textless \ - \frac{1}{2} [/tex]
Lets pick [tex]x=-1.[/tex] Then, [tex]2(-1) ^{2} -3 * -1 -2 \ \textless \ 0[/tex]
After simplifying, we get [tex]3\ \textless \ 0[/tex], Which is false.
Drop this interval.
For this interval [tex]- \frac{1}{2} \ \textless \ x\ \textless \ 2[/tex]
Lets pick [tex]x=0[/tex]. Then, [tex]2* 0^{2} - 3 * 0-2\ \textless \ 0[/tex]. After simplifying, we get [tex]-2\ \textless \ 0,[/tex] which is true. Keep this interval.
For the interval [tex]x\ \textgreater \ 2[/tex]
Lets pick [tex]x = 3.[/tex] Then, [tex]2 * 3 ^{2} -3*3-2\ \textless \ 0.[/tex] After simplifying, we get [tex]7\ \textless \ 0[/tex], Which is false. Drop this interval.
.Step 7: Therefore,
[tex]- \frac{1}{2} \ \textless \ x\ \textless \ 2[/tex]
Done! :)
1. Multiply 2 by -2, which is -4.
2. Ask: Which two numbers add up to -3 and multiply to -4?
3. Answer: 1 and -4
4. Rewrite [tex]-3x[/tex] as the sum of [tex]x[/tex] and [tex]-4x[/tex]
[tex]2 x^{2} +x-4x-2\ \textless \ 0[/tex]
Step 2: Factor out common terms in the first two terms, then in the last two terms.
[tex]x(2x+1)-2(2x+1)\ \textless \ 0[/tex]
Step 3: Factor out the common term [tex]2x+1[/tex]
[tex](2x+1)(x-2)\ \textless \ 0[/tex]
Step 4: Solve for [tex]x[/tex]
1. Ask: When will [tex](2x+1)(x-2)[/tex] equal zero?
2. Answer: When [tex]2x + 1 = 0[/tex] or [tex]x-2=0[/tex]
3. Solve each of the 2 equations above:
[tex]x=- \frac{1}{2} ,2[/tex]
Step 5: From the values of [tex]x[/tex] above, we have these 3 intervals to test.
x = < -1/2
-1/2 < x < 2
x > 2
Step 6: Pick a test point for each interval
For the interval [tex]x\ \textless \ - \frac{1}{2} [/tex]
Lets pick [tex]x=-1.[/tex] Then, [tex]2(-1) ^{2} -3 * -1 -2 \ \textless \ 0[/tex]
After simplifying, we get [tex]3\ \textless \ 0[/tex], Which is false.
Drop this interval.
For this interval [tex]- \frac{1}{2} \ \textless \ x\ \textless \ 2[/tex]
Lets pick [tex]x=0[/tex]. Then, [tex]2* 0^{2} - 3 * 0-2\ \textless \ 0[/tex]. After simplifying, we get [tex]-2\ \textless \ 0,[/tex] which is true. Keep this interval.
For the interval [tex]x\ \textgreater \ 2[/tex]
Lets pick [tex]x = 3.[/tex] Then, [tex]2 * 3 ^{2} -3*3-2\ \textless \ 0.[/tex] After simplifying, we get [tex]7\ \textless \ 0[/tex], Which is false. Drop this interval.
.Step 7: Therefore,
[tex]- \frac{1}{2} \ \textless \ x\ \textless \ 2[/tex]
Done! :)