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Chapter 9: Problem 28

Write a quadratic equation in \(x\) with the given solutions. \(\sqrt{5}\) and \(-\sqrt{5}\)

Short Answer

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Question: Write a quadratic equation with the solutions \(\sqrt{5}\) and \(-\sqrt{5}\). Answer: \(x^2 - 5 = 0\)

Step by step solution

01

Recall the Vieta's formulas for roots of a quadratic equation

Vieta's formulas say that if a quadratic equation \(ax^2 + bx + c = 0\) has roots \(r\) and \(s\), then: \(r + s = -\frac{b}{a}\) and \(rs = \frac{c}{a}\)
02

Plug in the given solutions

Since \(\sqrt{5}\) and \(-\sqrt{5}\) are the solutions, we can plug them into the formulas as \(r\) and \(s\). Thus, we get: \(\sqrt{5} - \sqrt{5}= -\frac{b}{a}\) and \((\sqrt{5})(-\sqrt{5}) = \frac{c}{a}\)
03

Simplify the equations

The first equation simplifies to \(0=-\frac{b}{a}\). Multiplying both sides by \(a\) gives us \(b=0\). The second equation simplifies to \(-5=\frac{c}{a}\). Now, let's adopt \(a = 1\) as it will not affect the roots. Then the equation becomes \(-5=c\).
04

Write down the quadratic equation

Now that we have our coefficients, we can write the quadratic equation: \(1x^2 + 0x - 5 = 0\) So the quadratic equation with the given solutions is: $$x^2 - 5 = 0$$

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Key Concepts

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

Vieta's Formulas
Vieta's formulas are a beautiful aspect of algebra connecting the coefficients of a polynomial to the sums and products of its roots. When dealing with a quadratic equation of the form \(ax^2 + bx + c = 0\), these formulas become particularly handy in understanding the relationship between coefficients and roots.
  • Sum of roots: \( r + s = -\frac{b}{a} \)
  • Product of roots: \( rs = \frac{c}{a} \)
Recall that roots are simply the values of \(x\) for which the equation becomes zero. Vieta鈥檚 formulas allow us to derive or construct the equation based on its solutions.In exercises, such as constructing a quadratic from given roots, Vieta鈥檚 formulas let us quickly establish the necessary symmetry and balance between the algebraic components.
Roots of Quadratic Equations
Finding the roots of quadratic equations can seem intimidating initially, but it transforms into a straightforward process with practice. Roots of a quadratic equation are values of \(x\) that make the equation true, i.e., they make \(ax^2 + bx + c = 0\).
  • The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is a general method to find roots when the equation is not easily factorable.
  • Roots can also be found by rewriting the equation in factored form \((x - r)(x - s) = 0\), providing a direct path to the solutions: \(r\) and \(s\).
In our problem, the given roots were \(\sqrt{5}\) and \(-\sqrt{5}\). With the help of Vieta's formulas, the equation becomes real and printable: \(x^2 - 5 = 0\). This equation clearly reflects the roots, maintaining a strong visual connection between a quadratic equation and its solutions.
Simplifying Quadratic Expressions
Simplifying quadratic expressions involves reducing equations to a form that's easier to handle or evaluate. This often means reducing complex expressions down to their simplest, most intuitive descriptions.
  • One common simplification step is preferring to set \(a = 1\) in the standard form \(ax^2 + bx + c = 0\), unless a specific problem demands otherwise.
  • After establishing \(b\) and \(c\) using Vieta's formulas, rewriting the equation using these simplified values enhances both readability and usability.
In this particular exercise, simplifying took us from complex roots expressions \((\sqrt{5})(-\sqrt{5}) = -5\) to simply adopting \(a = 1\), resulting in \(c = -5\) and consequently, \(x^2 - 5 = 0\). This concise representation makes it easier to visualize and solves problems, demonstrating the sheer utility of simplification.

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Most popular questions from this chapter

Put the quadratic function in factored form, and use the factored form to sketch a graph of the function without a calculator. $$ y=x^{2}+8 x+12 $$

A ball is thrown straight upward from the ground. Its height above the ground in meters after \(t\) seconds is given by \(-4.9 t^{2}+30 t+c\) (a) Find the constant \(c\). (b) Find the values of \(t\) that make the height zero and give a practical interpretation of each value.

In Exercises 48-53, use the discriminant to say whether the equation has two, one, or no solutions. $$ 7 x^{2}-x-8=0 $$

A flawed approach to solving the equation \(z-2 \sqrt{z}=8\) is shown below: $$ \begin{aligned} z-2 \sqrt{z} &=8 \\ z-8 &=2 \sqrt{z} \\ (z-8)^{2} &=(2 \sqrt{z})^{2} z^{2}-16 z+64 &=4 z \\ z^{2}-20 z+64 &=0 \\ (z-4)(z-16) &=0 \\ z &=4,16 \end{aligned}$$ Identify and account for the flaw, specifying the step (1)-(7) where it is introduced.

Solve the quadratic equations in Exercises 23-28 or state that there are no solutions. $$ 5 x-2 x^{2}-5=0 $$
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