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Chapter 5: Problem 62

If \(\alpha\) and \(\beta\) are the roots of the equation \(2 x(2 x+1)=1\), then \(\beta\) is equal to: [Sep. 06, 2020 (II)] (a) \(2 \alpha(\alpha+1)\) (b) \(-2 \alpha(\alpha+1)\) (c) \(2 \alpha(\alpha-1)\) (d) \(2 \alpha^{2}\)

Short Answer

Expert verified
\( \beta \) is \(-2 \alpha (\alpha + 1)\), so the answer is (b).

Step by step solution

01

Simplify the Given Equation

Start by rewriting the given equation \[ 2x(2x + 1) = 1 \]Expand the left side:\[ 4x^2 + 2x = 1 \]Rearrange it into a standard quadratic form:\[ 4x^2 + 2x - 1 = 0 \]
02

Use the Quadratic Formula to Find the Roots

For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are given by the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our equation, \( a = 4 \), \( b = 2 \), and \( c = -1 \). Plug these values into the formula:\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]Simplify inside the square root:\[ x = \frac{-2 \pm \sqrt{4 + 16}}{8} \]\[ x = \frac{-2 \pm \sqrt{20}}{8} \]Further simplify:\[ x = \frac{-2 \pm 2\sqrt{5}}{8} \]Divide by 2:\[ x = \frac{-1 \pm \sqrt{5}}{4} \]So the roots are \( \alpha = \frac{-1 + \sqrt{5}}{4} \) and \( \beta = \frac{-1 - \sqrt{5}}{4} \).
03

Check What Each Option Represents

Next, check the given options to see which can be equivalent to \( \beta \):(a) \( 2\alpha(\alpha + 1) \)(b) \( -2\alpha(\alpha + 1) \)(c) \( 2\alpha(\alpha - 1) \)(d) \( 2\alpha^2 \)
04

Substitute \( \alpha \) into Option B

Choose option (b) to match the form of \( \beta \). Substitute \( \alpha = \frac{-1 + \sqrt{5}}{4} \) into \(-2\alpha(\alpha + 1) \):Calculate \( \alpha + 1 \):\[ \alpha + 1 = \frac{-1 + \sqrt{5}}{4} + 1 = \frac{3 + \sqrt{5}}{4} \]Substitute in:\[ -2\alpha(\alpha + 1) = -2\left( \frac{-1 + \sqrt{5}}{4} \right)\left( \frac{3 + \sqrt{5}}{4} \right) \]\[ = -2\left(\frac{(-1+\sqrt{5})(3+\sqrt{5})}{16}\right) \]Calculate the product:\[ (-1+\sqrt{5})(3+\sqrt{5}) = -3 - \sqrt{5} + 3\sqrt{5} + 5 \]\[ = 2 + 2\sqrt{5} \]So substitute back:\[ = -2\left(\frac{2 + 2\sqrt{5}}{16}\right) \]\[ = -\frac{2\left(2 + 2\sqrt{5}\right)}{8} \]\[ = -\frac{-1 - \sqrt{5}}{4} \]Therefore, option (b) matches \( \beta \).

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Key Concepts

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

Quadratic Formula
Sometimes, finding the roots of a quadratic equation can be tricky. But, the quadratic formula makes this easier. This formula applies to any quadratic equation given in the standard form: \[ ax^2 + bx + c = 0 \]The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula gives you the two possible values for \( x \), which are the roots of the equation. Here's a simple breakdown:
  • \( a \), \( b \), and \( c \) are the coefficients from your quadratic equation.
  • The "+" and "-" signs before the square root mean you will actually do the calculation twice!
  • The part \( b^2 - 4ac \) is called the discriminant. It shows you if the solutions are real or imaginary.
Using the quadratic formula always leads you to the correct roots, no matter the quadratic equation.
Roots of Equations
The roots of an equation are the values of \( x \) that satisfy the equation. Simply put, they are the solutions. For polynomials, roots are where the function equals zero.
In a quadratic equation, there are always two roots because it is of degree 2. The roots could be:
  • Real and different - you have a neat pair of numbers, just like in our exercise example.
  • Real and identical - both roots are the same. This happens when the discriminant is zero.
  • Complex - when the discriminant is negative, the roots include imaginary numbers.
Knowing the roots is essential for various purposes, like graphing the polynomial or solving real-world problems. Once you find the roots using formulas or rearranging, you can explore how these solutions make the original equation zero.
Polynomial Equations
Polynomial equations are expressions involving variables raised to different powers. Each polynomial is a sum of terms, in standard form generally written as:
\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \]
  • The highest power of the variable is the degree of the polynomial. For example, a quadratic equation, with the form \( ax^2 + bx + c \), is a polynomial of degree 2.
  • Coefficients \(a_n, a_{n-1}, \ldots, a_0\) give each term weight, communicating how the variable terms contribute to forming the polynomial's graph or solution.
  • Understanding the structure helps predict the shape of the graph, such as parabolic (for quadratics).
Often, solving a polynomial equation means finding its roots. Techniques such as factoring, using the quadratic formula, or synthetic division come in handy, depending on the situation. By mastering these various methods, solving polynomial equations becomes more straightforward.

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Most popular questions from this chapter

If \(m\) is chosen in the quadratic equation \(\left(m^{2}+1\right) x^{2}-3 x+\left(m^{2}+1\right)^{2}=0\) such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is: [April 09, 2019 (II)] (a) \(10 \sqrt{5}\) (b) \(8 \sqrt{3}\) (c) \(8 \sqrt{5}\) (d) \(4 \sqrt{3}\)

If \(|z-3+2 i| \leq 4\) then the difference between the greatest value and the least value of \(|z|\) is \([\) Online April \(\mathbf{1 5}, \mathbf{2 0 1 8}]\) (a) \(\sqrt{13}\) (b) \(2 \sqrt{13}\) (c) 8 (d) \(4+\sqrt{13}\)

If the equation, \(x^{2}+b x+45=0(b \in R)\) has conjugate complex roots and they satisfy \(|z+1|=2 \sqrt{10}\), then: [Jan. 8, 2020 (I)] (a) \(b^{2}-b=30\) (b) \(b^{2}+b=72\) (c) \(b^{2}-b=42\) (d) \(b^{2}+b=12\)

Difference between the corresponding roots of \(x^{2}+a x+b=0\) and \(x^{2}+b x+a=0\) is same and \(a \neq b\), then (a) \(a+b+4=0\) (b) \(a+b-4=0\) (c) \(a-b-4=0\) (d) \(a-b+4=0\)

If \((1-p)\) is a root of quadratic equation \(x^{2}+p x+(1-p)=0\) then its root are (a) \(-1,2\) (b) \(-1,1\) (c) \(0,-1\) (d) 0,1
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