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Chapter 5: Problem 134

Let \(a, b \in \mathrm{R}, a \neq 0\) be such that the equation, \(a x^{2}-2 b x+5=0\) has a repeated root \(\alpha\), which is also a root of the equation, \(x^{2}-2 b x-10=0\). If \(\beta\) is the other root of this equation, then \(\alpha^{2}+\beta^{2}\) is equal to : [Jan. 9, \(\mathbf{2 0 2 0}\) (II)] (a) 25 (b) 26 (c) 28 (d) 24

Short Answer

Expert verified
The value of \(\alpha^2 + \beta^2\) is 25.

Step by step solution

01

Understanding Repeated Roots

For the equation \(ax^2 - 2bx + 5 = 0\) to have a repeated root \(\alpha\), the discriminant must be zero. This means that \((2b)^2 - 4\cdot a\cdot 5 = 0\).
02

Calculate the Discriminant for Repeated Root

Set the discriminant \((2b)^2 - 20a = 0\) equal to zero to find \(a\) in terms of \(b\). Simplifying gives \(4b^2 = 20a\), or \(a = \frac{4b^2}{20} = \frac{b^2}{5}\).
03

Identify \(\alpha\) as a Root of the Second Equation

We know \(\alpha\) is also a root of \(x^2 - 2bx - 10 = 0\). To find \(\alpha\), apply the quadratic formula: \(x = \frac{2b \pm \sqrt{(2b)^2 + 4\cdot 10}}{2}\). This simplifies to \(x = b \pm \sqrt{b^2 + 10}\). Since \(\alpha\) is repeated in the first equation, we choose \(\alpha = b + \sqrt{b^2 + 10}\).
04

Find \(\beta\), the Other Root

From the quadratic equation \(x^2 - 2bx - 10 = 0\), the sum of roots \(\alpha + \beta = 2b\). Since \(\alpha = b + \sqrt{b^2 + 10}\), substitute to get \(\beta = 2b - \alpha = 2b - (b + \sqrt{b^2 + 10}) = b - \sqrt{b^2 + 10}\).
05

Calculate \(\alpha^2 + \beta^2\)

Use the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\). We have \(\alpha + \beta = 2b\) and \(\alpha \beta = -10\). So, \(\alpha^2 + \beta^2 = (2b)^2 - 2(-10) = 4b^2 + 20\).
06

Use Constraints to Find Exact Value

Since from Step 2, \(a = \frac{b^2}{5} \implies 0 = (2b)^2 - 20a = 0\), thus \(4b^2 = 20a = 20\cdot \frac{b^2}{5}\). So \(\alpha^2 + \beta^2 = 4b^2 + 20 = 25\). Therefore, \((\alpha^2 + \beta^2) = 25\).

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Key Concepts

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

Repeated Roots
Repeated roots occur in a quadratic equation when it has only one unique solution rather than two distinct ones. This happens because the two solutions of the equation are the same, essentially creating a double solution.
For example, for a quadratic equation in the standard form of \( ax^2 + bx + c = 0 \), the equation will have repeated roots if the discriminant (expressed as \( b^2 - 4ac \)) equals zero.
  • If the discriminant is zero, it means that there is exactly one repeated solution or root, rather than two distinct solutions.
  • The repeated root \( \alpha \) can be directly found using the formula: \( x = -\frac{b}{2a} \).
In the context of our exercise, the equation \( ax^2 - 2bx + 5 = 0 \) has repeated roots when the discriminant \((2b)^2 - 20a = 0\). This leads to the equations being solved for specific values where one root (repeatedly occurring one) is \( \alpha \).
Roots of Quadratic Equation
The roots of a quadratic equation are the values of \( x \) that make the equation true, essentially where the graph of the quadratic intersects the x-axis.
To solve for the roots of a quadratic equation in the form \( ax^2 + bx + c = 0 \), you can use the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]
  • "\( \pm \)" here signifies that there are usually two solutions (roots) from this formula, one using the plus sign, and one using the minus sign.
  • When using this formula, if the discriminant \( b^2 - 4ac \) is positive, it indicates two distinct roots; if zero, the roots are repeated (as mentioned earlier); and if negative, the roots are complex numbers (imaginary).
In the specific problem we looked at, after determining the value of \( b \) that renders the discriminant zero, \( \alpha \), the repeated root, is found to also satisfy a second equation providing evidence of its repeated nature and helping find \( \beta \) as the other root for additional calculations.
Discriminant of a Quadratic Equation
The discriminant of a quadratic equation鈥攅xpressed as \( b^2 - 4ac \)鈥攑lays a critical role in determining the nature of the roots. The value of the discriminant can tell us not just about the character of the roots but also their quantity.
  • A positive discriminant indicates two distinct real roots.
  • A zero discriminant indicates a repeated root, implying a "bounce" on the x-axis rather than an intersection, making the graph touch the axis but not cross it.
  • A negative discriminant indicates complex roots, meaning the solutions are not real numbers.
In the exercise at hand, we set the discriminant \((2b)^2 - 20a = 0\) to find the conditions under which the quadratic has repeated roots and derived the relations between \( a \) and \( b \). This allowed us to further analyze the other polynomial and calculate \( \alpha^2 + \beta^2 \). These steps are crucial for simplifying polynomial relationships and understanding root properties and equations.

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Most popular questions from this chapter

Let \(p, q\) and \(r\) be real numbers \((\mathrm{p} \neq \mathrm{q}, \mathrm{r} \neq 0)\), such that the roots of the equation \(\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\) are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to. [Online April 16, 2018] (a) \(p^{2}+q^{2}+r^{2}\) (b) \(p^{2}+q^{2}\) (c) \(2\left(p^{2}+q^{2}\right)\) (d) \(\frac{p^{2}+q^{2}}{2}\)

If \(f(x)\) is a quadratic expression such that \(f(a)+f(b)=0\) and \(-1\) is a root of \(f(x)=0\), then the other root of \(f(x)=0\) is \([\) Online April 15, 2018] (a) \(-\frac{5}{8}\) (b) \(-\frac{8}{5}\) (c) \(\frac{5}{8}\) (d) \(\frac{8}{5}\)

Product of real roots of the equation \(t^{2} x^{2}+|x|+9=0\) (a) is always positive (b) is always negative (c) does not exist (d) none of these

Let \(\alpha, \beta\) be real and \(z\) be a complex number. If \(z^{2}+\alpha z+\beta=0\) has two distinct roots on the line \(\operatorname{Re} z=1\), then it is necessary that : (a) \(\beta \in(-1,0)\) (b) \(|\beta|=1\) (c) \(\beta \in(1, \infty)\) (d) \(\beta \in(0,1)\)

If \(\alpha\) and \(\beta\) be the roots of the equation \(x^{2}-2 x+2=0\), then the least value of \(n\) for which \(\left(\frac{\alpha}{\beta}\right)^{n}=1\) is : [April 8, 2019(I)] (a) 2 (b) 5 (c) 4 (d) 3
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