How to solve for x? (2/9)x^2 + (4/3)x > 0.5^x

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In summary, the conversation is about finding where a parabola is greater than an exponential equation. The answer seems to be x>0, but the person is having trouble proving it algebraically. They are considering using numerical analysis and wondering if there are any unusual functions that could help solve the problem.
  • #1
Superposed_Cat
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Hi all, for what I'm currently doing I need to find where that parabola is greater than the exponential equation.This may seem simple but I can't seem to do anything other than rephrase it without getting a solution. Any help appreciated. it seems as though the answer is where x>0 but I can't prove it algebraicly.
 
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  • #2
Superposed_Cat said:
Hi all, for what I'm currently doing I need to find where that parabola is greater than the exponential equation.This may seem simple but I can't seem to do anything other than rephrase it without getting a solution. Any help appreciated. it seems as though the answer is where x>0 but I can't prove it algebraicly.
With over 300 posts, you should know by now not to put your equations in the title.

If you plot the functions on the two sides of the inequality you will see they cross near ##x=1/2##, not near ##x=0##. You can solve for it numerically.
 
  • #3
Numerical analysis gives about 0.4, you are right. But is it undoable algebraicaly?
 
  • #4
Superposed_Cat said:
Numerical analysis gives about 0.4, you are right. But is it undoable algebraicaly?

It is closer to .5 than .4. Yes, you can not solve it algebraically in terms of the usual functions.
 
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Thank you, like your signature by the way. "usual functions" is there an unusual one to do this job?
 

1. How do I solve for x in a quadratic inequality?

To solve for x in a quadratic inequality, you can follow these steps:

  • 1. Rewrite the inequality in standard form: ax^2 + bx + c > 0
  • 2. Factor the quadratic expression.
  • 3. Set each factor equal to 0 and solve for x.
  • 4. The solutions of the quadratic equation will determine the intervals on the number line where the inequality is true.
  • 5. Use these intervals to write the solution set for x.

2. Can I use the quadratic formula to solve a quadratic inequality?

Yes, you can use the quadratic formula to solve a quadratic inequality. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a. However, you will need to set the expression equal to 0 before using the formula.

3. What is the difference between solving a quadratic equation and a quadratic inequality?

The main difference between solving a quadratic equation and a quadratic inequality is that when solving an equation, you are finding the specific values of x that make the equation true. In contrast, when solving an inequality, you are finding the intervals of x values that make the inequality true.

4. How do I determine the intervals of x values for a quadratic inequality?

To determine the intervals of x values for a quadratic inequality, you can use the sign chart method. This involves factoring the quadratic expression and then using the sign of each factor to determine the intervals where the inequality is true. You can also graph the quadratic expression to visually determine the intervals.

5. Can I solve a quadratic inequality without factoring?

Yes, you can solve a quadratic inequality without factoring by using other methods such as the quadratic formula or completing the square. However, factoring is often the most efficient method for solving quadratic inequalities, especially when the coefficients are simple integers.

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