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Magazine Chapter 3: Problem 51
Solve each equation. For equations with real solutions, support your answers graphically. $$2 x^{2}-4 x=1$$
Short Answer
Expert verified
The solutions are \( x = 1 \pm \frac{\sqrt{6}}{2} \).
Step by step solution
01
Set the Equation to Zero
First, modify the equation to equal zero by subtracting 1 from both sides: \[ 2x^2 - 4x - 1 = 0 \]
02
Identify a, b, and c
Next, identify the coefficients in the quadratic equation \( ax^2 + bx + c = 0 \). Here, \( a = 2 \), \( b = -4 \), and \( c = -1 \).
03
Use the Quadratic Formula
Apply the quadratic formula to solve the equation: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute \( a = 2 \), \( b = -4 \), and \( c = -1 \).
04
Compute the Discriminant
Calculate the discriminant using the formula \( b^2 - 4ac \).\(-4^2 - 4(2)(-1) = 16 + 8 = 24\). The discriminant is positive, indicating two real solutions.
05
Solve for x
Substitute the discriminant back into the quadratic formula:\[ x = \frac{-(-4) \pm \sqrt{24}}{2(2)} \]Simplify under the square root \( \sqrt{24} = 2\sqrt{6} \), thus:\[ x = \frac{4 \pm 2\sqrt{6}}{4} \]Which simplifies to:\[ x = 1 \pm \frac{\sqrt{6}}{2} \]
06
Graphical Verification
Plot the function \( f(x) = 2x^2 - 4x - 1 \) on a graph. The points where the graph intersects the x-axis correspond to the roots found: \( x = 1 + \frac{\sqrt{6}}{2} \) and \( x = 1 - \frac{\sqrt{6}}{2} \). These confirm the two solutions found.
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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, which are equations of the form \( ax^2 + bx + c = 0 \). It's especially useful when factoring is complex or impossible. The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's how it works:
- The "\( b^2 - 4ac \)" part under the square root is called the discriminant. It tells us about the nature of the roots.
- The "\( -b \pm \)" denotes that there can be two potential solutions: one by adding the square root part and one by subtracting it.
- Divide the result by \( 2a \), which is twice the coefficient of \( x^2 \).
discriminant
The discriminant is a key element of the quadratic formula. It is given by \( D = b^2 - 4ac \) and plays a crucial role in determining the nature of the roots of a quadratic equation. Here's how it helps:
- If \( D > 0 \), there are two distinct real roots. This means the graph of the quadratic equation will intersect the x-axis at two points.
- If \( D = 0 \), there is exactly one real root, indicating that the graph touches the x-axis at one point (a "perfect square").
- If \( D < 0 \), there are no real roots. The graph does not intersect the x-axis at all, and instead, the solutions are complex numbers.
graphical solution
Finding a graphical solution means plotting the equation to visually identify its roots. For a quadratic equation, this involves drawing its parabola and observing where it crosses the x-axis. Each intersection point corresponds to a real solution for the equation.For the equation \( 2x^2 - 4x - 1 \), plot the function to see its shape:
- The highest-degree term \( 2x^2 \) indicates that the graph opens upwards since the coefficient is positive.
- The vertex form or axis of symmetry gives us a rough idea of where the graph is centered. You can find this by \( x = -\frac{b}{2a} \).
real solutions
Real solutions to a quadratic equation are the x-values where the parabola crosses the x-axis. For an equation like \( 2x^2 - 4x - 1 = 0 \), if the discriminant is positive, it assures us of real and distinct solutions.Here's what this tells us:
- Real solutions are straightforward to find using calculators or graphing tools, as these methods verify the exact points of intersection.
- In our problem, solutions \( x = 1 + \frac{\sqrt{6}}{2} \) and \( x = 1 - \frac{\sqrt{6}}{2} \) are those real solutions. You can also solve visually by observing the graph.
- Real solutions also imply tangible, valid answers in most practical real-world scenarios. This can relate to problems of areas, velocities, and more. When applying these in real situations, ensure measurements and computations reflect reality.
