Problem 70
Question
Find all solutions of the quadratic equation. Relate the solutions of the equation to the zeros of an appropriate quadratic function. $$2 t^{2}+8 t=-9$$
Step-by-Step Solution
Verified Answer
The solutions of the equation \(2 t^{2}+8 t+9 = 0\) are \( t = -2 \pm i\sqrt{2} \). These solutions correspond to the zeros of the quadratic function \( f(t) = 2t^{2} + 8t + 9 \). Due to the negative term under the square root, these solutions are complex or imaginary.
1Step 1: Rewrite the Equation
Firstly, rearrange the given equation in standard form, which is \( ax^2 + bx + c = 0 \), by moving the constant term to the right-hand side. Thus, the equation becomes \( 2t^{2} + 8t + 9 = 0 \).
2Step 2: Apply the Quadratic Formula
Substitute the coefficients into the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the coefficients of the equation \( a = 2, b = 8, c = 9 \) gives \( t = \frac{-8 \pm \sqrt{(8)^2 - 4*2*9}}{2*2} \).
3Step 3: Simplify the Expression
Evaluate the expression inside and under the square root then simplify the equation to get the results. Result will be \( t = \frac{-8 \pm \sqrt{64 - 72}}{4} = \frac{-8 \pm \sqrt{-8}}{4} \). Since the term under the square root is negative, the solutions are imaginary or complex.
4Step 4: Calculate the Solutions and Relate to Zeros of the Function
Since the solutions are complex, they can be expressed in the form of \( a + bi \). Here, \( t = \frac{-8 \pm i\sqrt{8}}{4} = -2 \pm i\sqrt{2} \). These are the solutions, and they also represent the zeros of the corresponding quadratic function \( f(t) = 2t^{2} + 8t + 9 \).
Key Concepts
Quadratic FormulaComplex SolutionsQuadratic Functions Zeros
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots, or solutions, of a quadratic equation of the form
\( ax^2 + bx + c = 0 \)
where \(a\), \(b\), and \(c\) are coefficients and \(a eq 0\). The formula is derived from completing the square and is given by:
\[ t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
This equation provides us with the solution for \(t\), which can yield two real solutions, one real solution, or two complex solutions depending on the value of the discriminant
\( \Delta = b^2 - 4ac \). A positive discriminant indicates two real and distinct solutions, a discriminant of zero indicates one real solution (a repeated root), and a negative discriminant signifies two complex solutions. This is an essential concept in algebra as it provides a systematic method for tackling all quadratic equations eventually leading up to understanding more complex functions in mathematics.
\( ax^2 + bx + c = 0 \)
where \(a\), \(b\), and \(c\) are coefficients and \(a eq 0\). The formula is derived from completing the square and is given by:
\[ t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
This equation provides us with the solution for \(t\), which can yield two real solutions, one real solution, or two complex solutions depending on the value of the discriminant
\( \Delta = b^2 - 4ac \). A positive discriminant indicates two real and distinct solutions, a discriminant of zero indicates one real solution (a repeated root), and a negative discriminant signifies two complex solutions. This is an essential concept in algebra as it provides a systematic method for tackling all quadratic equations eventually leading up to understanding more complex functions in mathematics.
Complex Solutions
Complex solutions occur when the discriminant of a quadratic equation is negative. This means that the square root of the discriminant involves taking the square root of a negative number, which is not possible with real numbers alone.
For the equation in question, the discriminant is \( 64 - 4*2*9 = -8 \), which is negative, leading to the square root of a negative number. In mathematics, we represent the square root of -1 with the imaginary unit \(i\). Therefore, the complex solutions can be written as \( a + bi \) or \( a - bi \) where \(a\) and \(b\) are real numbers and \(i\) denotes the imaginary unit.
For the given quadratic equation \( 2t^2 + 8t + 9 = 0 \), the complex solutions are \( -2 + i\sqrt{2} \) and \( -2 - i\sqrt{2} \). It's fundamental to understand complex solutions as they extend the concept of solving equations beyond the real number system, allowing for a complete set of solutions to quadratic equations
For the equation in question, the discriminant is \( 64 - 4*2*9 = -8 \), which is negative, leading to the square root of a negative number. In mathematics, we represent the square root of -1 with the imaginary unit \(i\). Therefore, the complex solutions can be written as \( a + bi \) or \( a - bi \) where \(a\) and \(b\) are real numbers and \(i\) denotes the imaginary unit.
For the given quadratic equation \( 2t^2 + 8t + 9 = 0 \), the complex solutions are \( -2 + i\sqrt{2} \) and \( -2 - i\sqrt{2} \). It's fundamental to understand complex solutions as they extend the concept of solving equations beyond the real number system, allowing for a complete set of solutions to quadratic equations
Quadratic Functions Zeros
The zeros of a quadratic function are the values of \(x\) (or another variable, like \(t\) in our exercise) for which the function's value is zero. In other words, they are the roots of the equation or the solutions where the graph of the quadratic function intersects or touches the x-axis.
For the quadratic function given by \( f(t) = 2t^2 + 8t + 9 \), our zeros correspond to the solutions to the quadratic equation \( 2t^2 + 8t + 9 = 0 \) that we have obtained using the quadratic formula. Since the discriminant was negative, our quadratic function has no real zeros, but two complex ones.
These complex zeros, \( -2 + i\sqrt{2} \) and \( -2 - i\sqrt{2} \), are critical in the study of quadratic functions as they expand the understanding of function behavior and symmetry. They demonstrate that although a quadratic function may not intersect the x-axis on the graph when visualized within the real plane, it does have points of intersection in the complex plane—illustrating the profound link between algebraic solutions and graphical interpretations.
For the quadratic function given by \( f(t) = 2t^2 + 8t + 9 \), our zeros correspond to the solutions to the quadratic equation \( 2t^2 + 8t + 9 = 0 \) that we have obtained using the quadratic formula. Since the discriminant was negative, our quadratic function has no real zeros, but two complex ones.
These complex zeros, \( -2 + i\sqrt{2} \) and \( -2 - i\sqrt{2} \), are critical in the study of quadratic functions as they expand the understanding of function behavior and symmetry. They demonstrate that although a quadratic function may not intersect the x-axis on the graph when visualized within the real plane, it does have points of intersection in the complex plane—illustrating the profound link between algebraic solutions and graphical interpretations.
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