/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Determine a value for \(a\) such... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine a value for \(a\) such that one root of the equation \(a x^{2}+x-1=0\) is five times the other.

Short Answer

Expert verified
The value of \(a\) is \(-\frac{5}{36}\).

Step by step solution

01

Set up the equation based on root relation

Given that one root of the quadratic equation \(ax^2 + x - 1 = 0\) is five times the other, let \(r\) be the smaller root. Then the larger root is \(5r\). According to Vieta's formulas, the sum of the roots \(r + 5r\) equals \(-\frac{b}{a}\). Here, \(b = 1\), so \(6r = -\frac{1}{a}.\)
02

Express the product of the roots

According to Vieta’s formulas, the product of the roots of the equation is equal to \(\frac{c}{a}\). Thus, \(r \times 5r = \frac{-1}{a}.\) Simplifying this, we have \(5r^2 = \frac{-1}{a}.\)
03

Substitute the sum of the roots equation into the product equation

From Step 1, we have \(r = -\frac{1}{6a}.\) Substituting \(r = -\frac{1}{6a}\) into the product equation \(5r^2 = \frac{-1}{a},\) we get:\[ 5 \left(-\frac{1}{6a}\right)^2 = \frac{-1}{a} \]
04

Solve the equation for \(a\)

Now solve for \(a\) in the equation:\[ 5 \left(\frac{1}{36a^2}\right) = \frac{-1}{a} \]Simplify to:\[ \frac{5}{36a^2} = \frac{-1}{a} \]Cross multiply to find:\[ 5 = -36a \]So, \(a = -\frac{5}{36}.\)

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Key Concepts

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

Vieta's Formulas
Understanding Vieta's formulas makes solving quadratic equations much simpler. These formulas relate the coefficients of a polynomial to sums and products of its roots. In a quadratic equation of the form \(ax^2 + bx + c = 0\), Vieta's tell us:
  • The sum of the roots \(r_1 + r_2\) is equal to \(-\frac{b}{a}\).
  • The product of the roots \(r_1 \times r_2\) is \(\frac{c}{a}\).
These are especially helpful since they allow us to connect the polynomial's roots directly to the coefficients, reducing the need to solve the equation directly each time. For students, memorizing these relationships is vital for quickly navigating through problems involving roots and coefficients without full-blown calculations.
Roots of Polynomials
Roots of a polynomial are the values of \(x\) that make the polynomial expression equal to zero. In the quadratic equation \(ax^2 + bx + c = 0\), the roots can be found using the quadratic formula, by factoring, or using Vieta's formulas.
Roots indicate where the graph of a polynomial intersects the x-axis. Understanding the behavior of roots helps us to solve for unknowns effortlessly:
  • Knowing that one root is a multiple or specific relation to another helps to set up equations efficiently, as shown in our exercise.
  • Polynomial roots can show symmetry, such as being equal or opposite, impacting how we interpret the graph and solve the equation.
Equation Solving
Solving quadratic equations often requires a mix of algebraic skills and conceptual understanding. Let's break down that approach with the understanding from this exercise.
First, always attempt to simplify the given equation into a form where patterns or relationships, like those pointed out by Vieta's formulas, are easily spotted. Identifying these relationships, like one root being five times the other, allows you to substitute and form new equations.
Once these relationships are clear, substitute known values or expressions into your equation, and simplify further to find unknown coefficients or roots.
  • In this exercise, we found that one root being five times another leads to specific algebraic substitutions, simplifying solving for \(a\).
  • Always double-check your substitutions and simplifications—small algebra errors can lead to incorrect solutions early on.
By systematically working through these steps, equation solving becomes less daunting and more systematic.

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