Problem 92
Question
Work in a small group to solve \(a x^{2}+b x+c>0\) for \(x\) in each case. a) \(b^{2}-4 a c=0\) and \(a>0\) b) \(b^{2}-4 a c=0\) and \(a<0\) c) \(b^{2}-4 a c<0\) and \(a>0\) d) \(b^{2}-4 a c<0\) and \(a<0\) e) \(b^{2}-4 a c>0\) and \(a>0\) f) \(b^{2}-4 a c>0\) and \(a<0\)
Step-by-Step Solution
Verified Answer
a) all x except \(-\frac{b}{2a}\), b) never, c) all x, d) never, e) outside roots, f) between roots.
1Step 1: Understanding the discriminant
The discriminant of a quadratic equation \(ax^2 + bx + c\) is given by \(b^2 - 4ac\). The sign of the discriminant determines the nature of the roots.
2Step 2: Analyze case a
For case a) \(b^2 - 4ac = 0\) and \(a > 0\): The quadratic equation has a repeated root at \(x = -\frac{b}{2a}\). Because \(a > 0\), the parabola opens upwards. \(ax^2 + bx + c > 0\) for all \(x\) except at the repeated root, where it equals zero.
3Step 3: Analyze case b
For case b) \(b^2 - 4ac = 0\) and \(a < 0\): The quadratic equation has a repeated root at \(x = -\frac{b}{2a}\). Because \(a < 0\), the parabola opens downwards. \(ax^2 + bx + c > 0\) never occurs since the graph is always non-positive.
4Step 4: Analyze case c
For case c) \(b^2 - 4ac < 0\) and \(a > 0\): The quadratic equation has no real roots and the parabola opens upwards. \(ax^2 + bx + c > 0\) for all \(x\).
5Step 5: Analyze case d
For case d) \(b^2 - 4ac < 0\) and \(a < 0\): The quadratic equation has no real roots and the parabola opens downwards. \(ax^2 + bx + c > 0\) never occurs since the graph is always negative.
6Step 6: Analyze case e
For case e) \(b^2 - 4ac > 0\) and \(a > 0\): The quadratic equation has two distinct real roots. Because \(a > 0\), the parabola opens upwards. \(ax^2 + bx + c > 0\) for all \(x\) values outside the interval between the roots.
7Step 7: Analyze case f
For case f) \(b^2 - 4ac > 0\) and \(a < 0\): The quadratic equation has two distinct real roots. Because \(a < 0\), the parabola opens downwards. \(ax^2 + bx + c > 0\) for all \(x\) values within the interval between the roots.
Key Concepts
DiscriminantParabola OrientationQuadratic Equation Roots
Discriminant
The discriminant is a key component in solving any quadratic equation. It is calculated using the formula \(b^2 - 4ac\). This value helps us understand the nature of the roots (solutions) to the quadratic equation \(ax^2 + bx + c\). The sign of the discriminant plays a crucial role:
- If the discriminant is greater than zero \( (b^2 - 4ac > 0)\), the quadratic equation has two distinct real roots.
- If the discriminant equals zero \( (b^2 - 4ac = 0)\), the quadratic equation has exactly one real repeated root.
- If the discriminant is less than zero \( (b^2 - 4ac < 0)\), the quadratic equation has no real roots.
Parabola Orientation
The orientation of a parabola is determined by the sign of the coefficient ‘a’ in the quadratic equation \(ax^2 + bx + c\). Here's how it works:
- If \(a > 0\), the parabola opens upwards, forming a 'U' shape. This is because the coefficient of the \(x^2\) term is positive.
- If \(a < 0\), the parabola opens downwards, forming an 'n' shape. This is because the coefficient of the \(x^2\) term is negative.
Quadratic Equation Roots
The roots of a quadratic equation, also known as the solutions or x-intercepts, are the values of x that make the equation equal to zero. Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we can find these roots. The discriminant inside the quadratic formula helps us determine how many roots we will have:
- If the discriminant is \0\, we have one repeated root \(x = -\frac{b}{2a} \)
- If the discriminant is greater than zero, we have two distinct real roots.
- If the discriminant is less than zero, there are no real roots.
Other exercises in this chapter
Problem 91
Find all real or imaginary solutions to each equation. Use the method of your choice. $$-x^{2}+x+6=0$$
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Solve each problem. Recovering an investment. The manager at Cream of the Crop bought a load of watermelons for \(\$ 200 .\) She priced the melons so that she w
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Find all real or imaginary solutions to each equation. Use the method of your choice. $$-x^{2}+x+12=0$$
View solution Problem 93
Find all real or imaginary solutions to each equation. Use the method of your choice. $$x^{2}-6 x+10=0$$
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