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Chapter 4: Problem 34

If the roots of the equation \(a x^{2}+b x+c=0\) are of the form \(\frac{\alpha}{\alpha-1}\) and \(\frac{\alpha+1}{\alpha}\), then the value of \((a+b+c)^{2}\) is (A) \(b^{2}-2 a c\) (B) \(b^{2}-4 a c\) (C) \(4 b^{2}-a c\) (D) \(2 b^{2}-a c\)

Short Answer

Expert verified
The value of \((a + b + c)^2\) is \(b^2 - 4ac\).

Step by step solution

01

Identify the Roots

We are given that the roots of the quadratic equation are \(\frac{\alpha}{\alpha-1}\) and \(\frac{\alpha+1}{\alpha}\). We need to use these roots in relation to the quadratic equation \(ax^2 + bx + c = 0\).
02

Apply Vieta's Formulas

According to Vieta's formulas, the sum of the roots \(\left( \frac{\alpha}{\alpha-1} + \frac{\alpha+1}{\alpha} \right)\) must equal \(\frac{-b}{a}\) and the product of the roots \(\left( \frac{\alpha}{\alpha-1} \cdot \frac{\alpha+1}{\alpha} \right)\) must equal \(\frac{c}{a}\).
03

Calculate the Sum of the Roots

Calculate the sum of the given roots: \[ \frac{\alpha}{\alpha-1} + \frac{\alpha+1}{\alpha} = \frac{\alpha^2 + (\alpha-1)(\alpha+1)}{\alpha(\alpha-1)} = \frac{2\alpha^2-1}{\alpha^2-\alpha}. \] Let this equal \(\frac{-b}{a}\).
04

Calculate the Product of the Roots

Now, calculate the product of the roots: \[ \frac{\alpha}{\alpha-1} \cdot \frac{\alpha+1}{\alpha} = \frac{\alpha^2+\alpha}{\alpha(\alpha-1)} = \frac{\alpha+1}{\alpha-1}. \] Let this equal \(\frac{c}{a}\).
05

Find Relationships Between Coefficients

Given that \(a\times\) sum = \(-b\) and \(a\times\) product = \(c\), solve these equations to determine relationships among \(a, b,\) and \(c\). Simplify the relationships using common mathematical operations.
06

Calculate \((a+b+c)^2\)

Use the relationships derived in the previous step to calculate \((a + b + c)^2\). Find any simplifications by substituting \(a, b, c\) values as calculated.

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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 provide a powerful tool to relate the coefficients of a polynomial equation to its roots. For a quadratic equation of the form \(ax^2 + bx + c = 0\), Vieta's formulas tell us:
  • The sum of the roots \(x_1 + x_2 = -\frac{b}{a}\).
  • The product of the roots \(x_1 \cdot x_2 = \frac{c}{a}\).
By using these relationships, one can find connections between the roots and their corresponding coefficients, which can simplify problem-solving without actually finding the roots themselves.
Vieta's formulas are immensely helpful in exercises involving specific roots forms, like the given problem with unconventional fractions. They transform the task into simpler algebraic manipulations using the equation's coefficients.
Sum of Roots
The sum of the roots of a quadratic equation can be easily determined using Vieta鈥檚 formulas. In the equation \(ax^2 + bx + c = 0\), the sum of the roots \(x_1 + x_2\) is given by \[x_1 + x_2 = -\frac{b}{a}\]This is derived directly from rearranging the quadratic formula and shows the linear relationship between the sum of the roots and the coefficients.
In the original problem, after finding the expressions for the roots, you calculate the sum \[\frac{\alpha}{\alpha-1} + \frac{\alpha+1}{\alpha} = \frac{2\alpha^2-1}{\alpha^2-\alpha}\]and equate it to \(-\frac{b}{a}\). This step is crucial for unveiling relationships among the coefficients \(a\), \(b\), and \(c\).
Product of Roots
Like the sum of roots, Vieta's formulas also provide a straightforward way to find the product of roots for the quadratic equation \(ax^2 + bx + c = 0\). The formula states:\[x_1 \cdot x_2 = \frac{c}{a}\]This product is a direct expression of the relationship between the constant term \(c\) and the leading coefficient \(a\).
In our problem, the product of the roots is calculated as:\[\frac{\alpha}{\alpha-1} \cdot \frac{\alpha+1}{\alpha} = \frac{\alpha+1}{\alpha-1}\]This product is then set equal to \(\frac{c}{a}\), helping identify how the coefficients interact and solve for specific terms.
Relationship Between Coefficients
The relationship between the coefficients of a quadratic equation and its roots is an essential aspect of understanding quadratic equations. In the equation \(ax^2 + bx + c = 0\), each coefficient plays a unique role:
  • \(a\) affects the width and direction of the parabola.
  • \(b\) influences the vertex and axis of symmetry.
  • \(c\) is the y-intercept.
In problems like the one provided, where the roots have specific forms, understanding how the coefficients relate through Vieta's formulas leads to an equation system detailing how \(a, b, c\) interact.
Step 5 of the solution involved using the derived sum and product of roots to create equations that directly reflect these relationships, allowing further deduction of \((a + b + c)^2\) and simplification into one of the given choices.

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Most popular questions from this chapter

If \(a \in R\) and the equation \(-3(x-[x])^{2}+2(x-[x])\) \(+a^{2}=0\) (where \([x]\) denotes the greatest integer \(\leq x\) ) has no integral solution, then all possible values of a lie in the interval (A) \((-1,0) \cup(0,1)\) (B) \((1,2)\) (C) \((-2,-1)\) (D) \((-\infty,-2) \cup(2, \infty)\)

In the following questions an Assertion (A) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason( \(\mathrm{R}\) ) is False (D) Assertion(A) is False, Reason(R) is True Assertion: If the equation \(x^{2}+2(k+1) x+9 k-5=0\) has only negative roots, then \(k \leq 6\) Reason: The equation \(f(x)=0\) will have both roots negative if and only if (i) Discriminant \(\geq 0\), (ii) Sum of roots \(<0\), (iii) Product of roots \(>0\)

If for real number \(a\), the equation \((a-2)(x-[x])^{2}+\) \(2(x-[x])+a^{2}=0\) (where \([x]\) denotes the greatest integer \(\leq x\) ) has no integral solution and has exactly one solution in \((2,3)\), then \(a\) lies in the interval (A) \((-1,2)\) (B) \((0,1)\) (C) \((-1,0)\) (D) \((2,3)\)

If \(x\) is real, then the minimum value of \(\frac{(a+x)(b+x)}{(c+x)}(x>-c)\), for \(a>c, b>c\) is (A) \((\sqrt{a-b}+\sqrt{c-b})^{2}\) (B) \((\sqrt{a-c}+\sqrt{b-c})^{2}\) (C) \((\sqrt{a-c}-\sqrt{b-c})^{2}\) (D) None of these

The maximum number of positive real roots of a polynomial equation \(f(x)=0\) is the number of changes of signs from positive to negative and negative to positive in \(f(x)\). For example, consider the equation \(f(x)=x^{3}+6 x^{2}+11 x-\) \(6=0\). The signs of the various terms are: $$ +++- $$ Clearly, there is only one change of sign in the given expression. So, the given equation has at most one positive Ireal root. The maximum number of negative real roots of a polynomial equation \(f(x)=0\) is the number of changes of signs from positive to negative and negative to positive in \(f(-x)\). For example, for the equation \(f(x)=x^{4}+x^{3}+x^{2}-x-1=\) 0 , there are three changes of signs in \(f(-x)\). So, the given equation has atmost three negative real roots. If \(f(x)\) and \(f(-x)\) do not have any changes of signs, the equation \(f(x)=0\) has no real roots. Now, consider the polynomial $$ P_{n}(x)=1+2 x+3 x^{2}+\ldots+(n+1) x^{n} . $$ If \(n\) is even, the number of real roots of \(P_{n}(x)\) is (A) 0 (B) \(n\) (C) 1 (D) None of these
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