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Solve the quadratic equation by using the quadratic formula. Find only real solutions. $$\frac{1}{2} t^{2}-4 t-3=0$$

Short Answer

Expert verified
The real solutions for the given quadratic equation are \( t = 8.69 \) and \( t = -0.69 \)

Step by step solution

01

Identify the Coefficients

For the quadratic equation \( \frac{1}{2} t^{2}-4 t-3=0 \), the coefficients are \( a= \frac{1}{2} \), \( b= -4 \), and \( c= -3 \).
02

Calculate the Discriminant

The discriminant is given by the formula \( D=b^{2}-4ac \). Substituting the values for \( a \), \( b \), and \( c \), we get: \( D=(-4)^{2}-4* \( \frac{1}{2} \)*(-3) = 16 + 6 = 22\).
03

Apply the Quadratic Formula

The quadratic formula is \( t= \frac{-b± \sqrt{D}}{2a} \). Given that D is 22, and \( a= \frac{1}{2}, b=-4 \), we can substitute these values.
04

Calculate the Real Solutions

Using substitutions, we have two possible solutions for \( t \): \( t= \frac{-(-4) + \sqrt{22}}{2* \( \frac{1}{2} \)} = 4+ \sqrt{22} = 4 \pm 4.694 \). Therefore, the real solutions for \( t \) are \( 8.69 \) and \( -0.69 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula provides a way to find the values of \( x \) that satisfy the equation. The quadratic formula is represented as:
\[\[\begin{equation}x = \frac{-b \pm \sqrt{D}}{2a}\end{equation}\]\]Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation. The symbol \( \pm \) means "plus or minus," which indicates that there will typically be two solutions. Using the quadratic formula helps solve the equation by calculating these two potential values of \( x \) where the equation equals zero.
This method is especially useful when factoring is difficult or when the coefficients are inconvenient. Always remember to correctly identify \( a \), \( b \), and \( c \) from the equation before applying the formula.
Real Solutions
Real solutions of quadratic equations are solutions that are real numbers, as opposed to imaginary numbers. They can be obtained using the quadratic formula. Whether these solutions are real or imaginary depends on the discriminant (\( D \)) that appears under the square root in the quadratic formula:
\[\sqrt{D}\]When the discriminant \( D \) is greater than zero, the quadratic equation has two distinct real solutions. If \( D \) is equal to zero, there is one real solution, which implies that the solutions are repeated, or "double solutions." When \( D \) is less than zero, the solutions are not real but rather imaginary, which means they involve the square root of a negative number.
Understanding real solutions helps in analyzing the nature of the quadratic equation and its solutions when applying the quadratic formula.
Discriminant Calculation
The discriminant is a key part of understanding quadratic equations and their solutions. It affects how many and what type of solutions an equation has. The discriminant \( D \) is calculated using the formula:
\[D = b^2 - 4ac\]In this formula, \( a \), \( b \), and \( c \) refer to the coefficients of the quadratic equation. Calculating \( D \) helps predict the nature of the solutions:
  • If \( D > 0 \), the quadratic equation has two distinct real solutions.
  • If \( D = 0 \), there is exactly one real solution, resulting in a double root where the parabola touches the x-axis.
  • If \( D < 0 \), the solutions are imaginary, indicating that the parabola does not intersect the x-axis.
Computing the discriminant is the first step to determining how the quadratic formula should be applied in order to find the solutions to a given quadratic equation.

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Most popular questions from this chapter

A ball is thrown directly upward from ground level at time \(t=0\) ( \(t\) is in seconds). At \(t=3,\) the ball reaches its maximum distance from the ground, which is 144 feet. Assume that the distance of the ball from the ground (in feet) at time \(t\) is given by a quadratic function \(d(t) .\) Find an expression for \(d(t)\) in the form \(d(t)=a(t-h)^{2}+k\) by performing the following steps. (a) From the given information, find the values of \(h\) and \(k\) and substitute them into the expression \(d(t)=a(t-h)^{2}+k\) (b) Now find \(a\). To do this, use the fact that at time \(t=0\) the ball is at ground level. This will give you an equation having just \(a\) as a variable. Solve for \(a\) (c) Now, substitute the value you found for \(a\) into the expression you found in part (a). (d) Check your answer. Is (3,144) the vertex of the associated parabola? Does the parabola pass through (0,0)\(?\)

The perimeter of a square is \(P(s)=4 s\) where \(s\) is the length of a side in inches. The function \(C(x)=2.54 x\) takes \(x\) inches as input and outputs the equivalent result in centimeters. Find \((C \circ P)(s)\) and explain what it represents.

In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=\frac{1}{x-3}, x \neq 3$$

This set of exercises will draw on the ideas presented in this section and your general math background. Explain what is wrong with the following steps for solving a radical equation. $$\begin{aligned}\sqrt{x+1}-2 &=0 \\\\(x+1)+4 &=0 \\\x &=-5\end{aligned}$$

Is it true that \((f g)(x)\) is the same as \((f \circ g)(x)\) for any functions \(f\) and \(g ?\) Explain.

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