I have a test tomorrow and can't solve this problem

[Anne_Boleyn]

Can anyone please help?

2x² - x + 1 < 9Edit: This is what I've done so far

2x² - x - 8 < 0
2 [ x² - 1/2x - 4] < 0
2(x - 1/2)² - 17/4 < 0

 

[quasar987]

This line is incorrect:

2 [ x² - 1/2x - 4] = 2(x - 1/2)² - 17/4

Start by getting the right one, then we'll go foward.

 

[Anne_Boleyn]

Then should it be: 2 [ x² - 1/2x - 4] = 2(x - 1/2)² + 15/4

I'm really lost

 

[quasar987]

The formula you should memorize if you have difficulty with this is

$$ax^2+bx+c= a\left[ \left( x+\frac{b}{2a} \right) ^2 - \left( \frac{b}{2a} \right) ^2 + \frac{c}{a} \right]$$

Apply this formula to your equation. (i.e. plug a=2, b=-1 and c=-8)

 

[sdekivit]

Can anyone please help?

2x² - x + 1 < 9


Edit: This is what I've done so far

2x² - x - 8 < 0
2 [ x² - 1/2x - 4] < 0
2(x - 1/2)² - 17/4 < 0

another method is to solve $2x^{2} - x - 8 = 0$ by dividing both sides by 2 and solve the quadratic equation using

$$x = \frac {-b ^{+}_{-} {\sqrt {b^{2} - 4ac}}} {2a}$$

and then investigate for what x this inequality is valid.

 

[VietDao29]

Can anyone please help?

2x² - x + 1 < 9Edit: This is what I've done so far

2x² - x - 8 < 0
2 [ x² - 1/2x - 4] < 0
2(x - 1/2)² - 17/4 < 0

Uhmm, let f(x) = ax² + bx + c, and let the equation f(x) = 0 has 2 real roots x₁, and x₂, such that x₁ < x₂. We choose a arbitrary x⁰ value. We have:
$$af(x_0) < 0 \Leftrightarrow x_0 \in ]x_1 , \ x_2[$$
$$af(x_0) > 0 \Leftrightarrow x_0 \notin ]x_1 , \ x_2[$$
Or in other words:
x₀ is in the interval ]x₁, x₂[ if and only if the leading coefficient a, and f(x₀) takes different signs, i.e if a is positive, then f(x₀) is negative, and vice versa.
x₀ is not in the interval ]x₁, x₂[ if and only if the leading coefficient a, and f(x₀) takes the same sign, i.e if a is positive, then f(x₀) is also positive, and vice versa.
---------------
I'll give you an example:
Example:
Solve the inequality:
x² + 5x + 4 < 0
<=> (x + 4) (x + 1) < 0
-4, and -1 are the 2 roots for this equation.
Let f(x) = x² + 5x + 4
The leading coefficient of f(x) is 1 > 0
And you have to solve for f(x) < 0, i.e, you will have to find all the value x such that f(x) and the coefficient take different signs, so the solution is
<=> -4 < x < -1
Can you get this? :)