Question

Statement 1 If the roots of the quadratic equation \(a x^{2}+b x+c=0\) lie between \(-2\) and 2, then both \((4 a+2 b+c)\) and \((4 a-2 b+c)\) must be positive. and Statement 2 If \(\alpha\) and \(\beta\) are the real roots of the equation \(a x^{2}+b x+c=0\), then \(\left(a x^{2}+b x+c\right)\) will have the same sign as that of a if \(\mathrm{x}\) is chosen as a number lying beyond \(\alpha\) and \(\beta\).

Short answer

#Summary# To summarize, we have proven both statements related to the quadratic equation, \(ax^2+bx+c=0\). For Statement 1, we ensured that both \((4a+2b+c)\) and \((4a-2b+c)\) must be positive for the roots to lie between -2 and 2 by examining the function at the endpoints and using the intermediate value theorem. For Statement 2, we proved that the sign of the quadratic equation would be the same as that of a if x is a number lying beyond the given real roots \(\alpha\) and \(\beta\). This was done by understanding the parabolic graph of the quadratic equation and its behavior depending on the sign of "a" and the position of x relative to the roots.

Step-by-step solution

Prove Statement 1

To prove Statement 1, we will consider the given quadratic equation, \(ax^2+bx+c=0\), and its discriminant. The discriminant is given by: \(D = b^2-4ac\) Since the roots lie between -2 and 2, let's examine the function at the endpoints, that is, at x = -2 and x = 2. First, let's find the function value at x = -2: \(f(-2) = a(-2)^2+b(-2)+c = 4a-2b+c\) Next, let's find the function value at x = 2: \(f(2) = a(2)^2+b(2)+c = 4a+2b+c\) Now, since the roots lie between -2 and 2 and the quadratic equation is continuous, by the intermediate value theorem, the function must have at least one real root between x = -2 and x = 2. This implies the function will change its sign from positive to negative or vice versa. Therefore, the function must have the same sign at both end points: \((4a-2b+c)(4a+2b+c) > 0\) This proves that both \((4a+2b+c)\) and \((4a-2b+c)\) must be positive.

Provide a brief explanation for Statement 1

For Statement 1, it is necessary to show that the given expressions will be positive to ensure that the roots of the quadratic equation lie between -2 and 2. By examining the function at x = -2 and x = 2, and using the intermediate value theorem, we have shown that both \((4a+2b+c)\) and \((4a-2b+c)\) must be positive.

Prove Statement 2

To prove Statement 2, let's assume that \(\alpha\) and \(\beta\) are the real roots of the given quadratic equation, \(ax^2+bx+c=0\). Now, let's consider x as a number lying beyond \(\alpha\) and \(\beta\), that is, either \(x < \alpha\) or \(x > \beta\). In this step, we will understand the sign of the quadratic equation depending on the sign of "a" and the position of x relative to the roots. If \(a > 0\), the quadratic equation will have a minimum value, and since \(\alpha\) and \(\beta\) are the real roots, the function will be positive for x < \(\alpha\) and x > \(\beta\). If \(a < 0\), the quadratic equation will have a maximum value, and since \(\alpha\) and \(\beta\) are the real roots, the function will be negative for x < \(\alpha\) and x > \(\beta\). This proves that \(f(x) = ax^2+bx+c\) will have the same sign as that of a if x is chosen as a number lying beyond \(\alpha\) and \(\beta\).

Provide a brief explanation for Statement 2

For Statement 2, we demonstrated that for any x lying beyond the real roots \(\alpha\) and \(\beta\), the quadratic equation will have the same sign as that of a. This is because, depending on the sign of "a," the parabolic graph representing the quadratic equation has a minimum or maximum value at the real roots, which determines its sign for the extreme values of x.

Key concepts

Quadratic Inequality

Understanding quadratic inequalities is important for solving various mathematical problems, including finding the range of values for which a quadratic function takes positive or negative values. A quadratic inequality looks like a quadratic equation but instead of the equals sign, it has an inequality symbol such as <, >, ≤, or ≥.

For example, the inequality \( ax^2 + bx + c > 0 \) asks for the set of all \( x \) values that make the quadratic expression positive. When analyzing quadratic inequalities, it is crucial to find the roots of the associated quadratic equation, \( ax^2 + bx + c = 0 \), because the roots divide the number line into intervals over which the quadratic expression does not change its sign.

This principle is seen in action when we analyze the problem statements from our exercise. If the roots of the quadratic equation are within certain bounds, such as between -2 and 2, we can use this information to deduce that the expression must be positive or negative at specific points to accommodate those roots. Therefore, having a grip on the concept of quadratic inequality is essential for proving statements such as Statement 1 in our exercise.

Discriminant of Quadratic Equations

The discriminant of a quadratic equation is a powerful tool that reveals information about the roots without actually solving the equation. The discriminant is given by \( D = b^2 - 4ac \) for the quadratic equation \( ax^2 + bx + c = 0 \).

The value of the discriminant can tell us if the roots are real or complex, and whether they are distinct or repeated. If \( D > 0 \), there are two distinct real roots; if \( D = 0 \), there is one repeated real root; and if \( D < 0 \), the roots are complex and not real.

In our exercise, we looked at the discriminant to ensure that the roots being between -2 and 2 implied certain properties of the quadratic expression at those boundaries. If a discriminant is positive as in our exercise, this gives us a starting point to further deduce the nature of the function at specific points, helping to prove statements like Statement 1 in the exercise.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus, which applies to continuous functions—functions without breaks or jumps. It states that if a function \( f \) is continuous on a closed interval \( [a, b] \), and \( N \) is any number between \( f(a) \) and \( f(b) \), then there is at least one number \( c \) in the interval \( [a, b] \) such that \( f(c) = N \).

For quadratic functions, which are always continuous, the IVT confirms that if the function takes on opposite signs at the endpoints of an interval, it must cross the x-axis at some point within that interval, which implies the existence of a root there.

This theorem is critical in our exercise to prove Statement 1. By showing that the function values at \( x = -2 \) and \( x = 2 \) are both positive and considering that the roots lie within this range, we can invoke the IVT to assert the continuity and sign change of the function, further supporting the conditions for the roots to fall between -2 and 2.