Problem 61
Question
If \(f(x)=a x^{2}+b x+c\) and \(r=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a},\) find \(f(r)\) without doing any algebra and explain how you arrived at your result.
Step-by-Step Solution
Verified Answer
The value of \(f(r)\) is 0.
1Step 1: Recognize root properties
The root of any quadratic function \(f(x) = ax^2 + bx + c\), according to the quadratic formula, is given by \(r = \frac{-b \pm \sqrt {b^2 - 4ac}}{2a}\). This means that substituting \(r\) in \(f(x)\)` will produce zero as a result. This is based on the property of roots which implies that if \(r\) is a root, then \(f(r)\) will be equal to zero.
2Step 2: Substitute root in function
Substitute \(r\) in \(f(x)\), the expression becomes \(f(r) = a r^{2} + b r + c\). Remember that we know \(r\) is a root so this will equal to zero.
3Step 3: Prove root is zero
Equating the equation to zero due to root property we can show that \(f(r)\) equals to zero: \(0 = ar^2 + br + c\).
Key Concepts
Quadratic FormulaRoots of EquationsProperties of Quadratic Equations
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of a quadratic equation. A quadratic equation is generally given in the standard form, \[ ax^2 + bx + c = 0 \]. The quadratic formula allows us to solve for the variable \( x \) in terms of its coefficients \( a \), \( b \), and \( c \):
- \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- If the discriminant is positive, the quadratic equation has two distinct real roots.
- If it is zero, the equation has exactly one real root (or a repeated root).
- If the discriminant is negative, the roots are complex conjugates.
Roots of Equations
The roots of an equation are the values of the variable that make the equation true. In the context of quadratic equations, these are the values of \( x \) for which \( ax^2 + bx + c = 0 \). These values are often called "solutions" or "zeros" of the equation.
When a quadratic equation is solved using the quadratic formula, the roots can be found as:
In our specific exercise, when you find \( r \) using the formula given, substituting it back into the function \( f(x) \) will always result in zero because it's a fundamental property of roots of a quadratic equation.
When a quadratic equation is solved using the quadratic formula, the roots can be found as:
- \[ x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \]
- \[ x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \]
In our specific exercise, when you find \( r \) using the formula given, substituting it back into the function \( f(x) \) will always result in zero because it's a fundamental property of roots of a quadratic equation.
Properties of Quadratic Equations
Quadratic equations have unique properties that help us understand more than just their roots. Here are some important features:
- The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of \( a \). If \( a > 0 \), it opens upwards; if \( a < 0 \), it opens downwards.
- The axis of symmetry of a quadratic function is found at \( x = \frac{-b}{2a} \), dividing the parabola into two mirror-image halves.
- The vertex of the parabola, which gives its maximum or minimum point, is located at the same \( x \) value as the axis of symmetry.
Other exercises in this chapter
Problem 60
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If 4 times a number is decreased by 3 and this difference is squared, the result is \(9 .\) Find the number \((s)\)
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