Solve the Quadratic Inequality: Understanding y = -2x² + 8x - 6

Quadratic Inequalities with Interval Testing

Look at the following function:

y=2x2+8x6 y=-2x^2+8x-6

Determine for which values of x x the following is true:

f(x)>0 f(x) > 0

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Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Look at the following function:

y=2x2+8x6 y=-2x^2+8x-6

Determine for which values of x x the following is true:

f(x)>0 f(x) > 0

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Determine the roots of the quadratic function.
  • Step 2: Split the number line based on these roots.
  • Step 3: Test intervals to check where the function is greater than zero.
  • Step 4: Identify the correct interval corresponding to the solution.

Now, let's work through each step:

Step 1: Identify the roots of the quadratic equation 2x2+8x6=0 -2x^2 + 8x - 6 = 0 . Using the quadratic formula, x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=2 a = -2 , b=8 b = 8 , and c=6 c = -6 .

Calculate the discriminant: b24ac=824(2)(6)=6448=16 b^2 - 4ac = 8^2 - 4(-2)(-6) = 64 - 48 = 16 .

The roots are: x=8±164=8±44 x = \frac{-8 \pm \sqrt{16}}{-4} = \frac{-8 \pm 4}{-4} .

Thus, x1=8+44=1 x_1 = \frac{-8 + 4}{-4} = 1 and x2=844=3 x_2 = \frac{-8 - 4}{-4} = 3 .

Step 2: The roots 1 and 3 split the number line into intervals: (,1) (-\infty, 1) , (1,3) (1, 3) , (3,) (3, \infty) .

Step 3: Test a sample point from each interval:

  • For x=0 x = 0 in (,1) (-\infty, 1) : f(0)=6 f(0) = -6 , which is not greater than 0.
  • For x=2 x = 2 in (1,3) (1, 3) : f(2)=2(2)2+8(2)6=86=2 f(2) = -2(2)^2 + 8(2) - 6 = 8 - 6 = 2 , which is greater than 0.
  • For x=4 x = 4 in (3,) (3, \infty) : f(4)=14 f(4) = -14 , which is not greater than 0.

Step 4: We conclude that the function 2x2+8x6 -2x^2 + 8x - 6 is greater than 0 only in the interval (1,3) (1, 3) .

Therefore, the solution to the problem is 1<x<3 1 < x < 3 .

3

Final Answer

1<x<3 1 < x < 3

Key Points to Remember

Essential concepts to master this topic
  • Roots: Set quadratic equal to zero and solve for x-values
  • Technique: Test x = 2 in interval (1,3): f(2) = 2 > 0
  • Check: Verify endpoints excluded: f(1) = f(3) = 0, not > 0 ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting to check the sign of the leading coefficient
    Don't assume the parabola opens upward when a = -2 is negative = wrong intervals! This makes you pick where the function is negative instead of positive. Always remember that negative leading coefficients create downward-opening parabolas, so the function is positive between the roots.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

AAABBBCCCX

FAQ

Everything you need to know about this question

Why do we need to find the roots first?

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The roots are where the function equals zero, which divides the number line into intervals. The function's sign (positive or negative) stays the same within each interval, so testing one point tells us about the whole interval.

How do I remember which interval has the positive values?

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Since the coefficient of x2 x^2 is negative (-2), this parabola opens downward. The function is positive between the roots and negative outside them.

What if I get the wrong interval when testing?

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Double-check your arithmetic! Make sure you substitute correctly: for x=2 x = 2 , calculate 2(2)2+8(2)6=8+166=2 -2(2)^2 + 8(2) - 6 = -8 + 16 - 6 = 2 . Since 2 > 0, the interval (1,3) is correct.

Why don't we include x = 1 and x = 3 in our answer?

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Because the inequality asks for f(x)>0 f(x) > 0 (strictly greater than). At the roots x=1 x = 1 and x=3 x = 3 , the function equals zero, not greater than zero.

Can I solve this by graphing instead?

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Yes! Graph the parabola and look where it's above the x-axis. You'll see it's positive between the roots x = 1 and x = 3, confirming our algebraic solution.

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