Problem 34
Question
Use the quadratic formula to solve the following. $$-25 x_{2}+10 x-1=0$$
Step-by-Step Solution
Verified Answer
The solution is \( x = \frac{1}{5} \).
1Step 1: Identify Coefficients
The quadratic equation is given in the standard form \( ax^2 + bx + c = 0 \).For the given equation \(-25x^2 + 10x - 1 = 0\),- \(a = -25\)- \(b = 10\)- \(c = -1\)
2Step 2: Write the Quadratic Formula
The quadratic formula is used to find the roots of a quadratic equation and is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3Step 3: Calculate the Discriminant
The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\).Substitute the values:\(D = (10)^2 - 4(-25)(-1) = 100 - 100 = 0\).
4Step 4: Substitute into Quadratic Formula
Since the discriminant \(D = 0\), there is one real solution.Substitute \(a = -25\), \(b = 10\), and \(c = -1\) into the quadratic formula:\[ x = \frac{-10 \pm \sqrt{0}}{2(-25)} \]
5Step 5: Simplify the Expression
Simplify the expression:\[ x = \frac{-10}{-50} = \frac{1}{5} \].The equation has a single repeated root.
Key Concepts
Quadratic EquationDiscriminantRoots of a Quadratic
Quadratic Equation
A quadratic equation is a type of polynomial equation of degree 2. It takes the standard form: \( ax^2 + bx + c = 0 \).
Here, *a*, *b*, and *c* are constants, with \( a eq 0 \), because if *a* were zero, the equation would not be quadratic but linear.
Quadratic equations are unique because they graph as parabolas, symmetrical curves that open up or down depending on the sign of *a*.
This last method is especially useful when the equation does not factor neatly, as it works for any quadratic equation.
Here, *a*, *b*, and *c* are constants, with \( a eq 0 \), because if *a* were zero, the equation would not be quadratic but linear.
Quadratic equations are unique because they graph as parabolas, symmetrical curves that open up or down depending on the sign of *a*.
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.
This last method is especially useful when the equation does not factor neatly, as it works for any quadratic equation.
Discriminant
The discriminant is a special value calculated from the coefficients of a quadratic equation and determines the nature and number of the roots. Calculated using the formula \( D = b^2 - 4ac \), the discriminant gives insight into the solutions without solving the entire equation.
It also provides insight into the graph of the corresponding quadratic function, as it indicates whether the parabola crosses the x-axis (real roots) or only touches it (a repeated root).
- If \( D > 0 \), there are two distinct real roots.
- If \( D = 0 \), there is exactly one real root, known as a repeated or double root.
- If \( D < 0 \), there are no real roots, but two complex conjugate roots.
It also provides insight into the graph of the corresponding quadratic function, as it indicates whether the parabola crosses the x-axis (real roots) or only touches it (a repeated root).
Roots of a Quadratic
The roots of a quadratic equation are the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). These roots can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula takes into account the discriminant \( b^2 - 4ac \) to determine the number and type of roots:
- With two different roots if the discriminant is positive.
- With one root (repeated root) if the discriminant is zero, as seen in the given problem because \( D = 0 \). This means the parabola just touches the x-axis at one point.
- With complex roots if the discriminant is negative, indicating the parabola does not intersect the x-axis.
Other exercises in this chapter
Problem 33
Graph. Find the vertex and the y-intercept. In addition, find the \(x\) - intercepts if they exist. $$ y=x 2-2 x+1 $$
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Solve by completing the square. $$x_{2}-1.6 x-0.8=0$$
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Perform the operations. $$ (6+5 i)(6-5 i) $$
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Solve by completing the square. $$x 2-23 x-13=0$$
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