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Magazine Chapter 14: Problem 31
The values of \(a\) for which the quadratic equations \((1-2 a) x^{2}-6 a x-1=0\) and \(a x^{2}-x+1=0\) have at least one root in common are : \(\begin{array}{ll}\text { (a) } \frac{1}{2}, \frac{2}{9} & \text { (b) } 0, \frac{1}{2}\end{array}\) (c) \(\frac{2}{9}\) (d) \(0, \frac{1}{2}, \frac{2}{9}\)
Short Answer
Expert verified
Solution:
The values of a for which the quadratic equations have at least one root in common are -1/2, 2/9, 1/3, and 1/4.
Step by step solution
01
Label the equations
Label the given quadratic equations as follows:
Equation 1: \((1 - 2a)x^2 - 6ax - 1 = 0\)
Equation 2: \(ax^2 - x + 1 = 0\)
02
Find the discriminants
Find the discriminants of both equations using the formula \(D = b^2 - 4ac\).
Discriminant of Equation 1: \(D_1 = (-6a)^2 - 4(1-2a)(-1) = 36a^2 - 4(1-2a)\)
Discriminant of Equation 2: \(D_2 = (-1)^2 - 4a(1) = 1 - 4a\)
03
Check the conditions for a common root
For the given quadratic equations to have at least one common root, it is necessary that either: (i) both equations have equal discriminants (\(D_1 = D_2\)), or (ii) one of the equations has a discriminant equal to zero and the other has a real root.
(i) If \(D_1 = D_2\), we have:
\(36a^2 - 4(1-2a) = 1 - 4a\) 猱� \(36a^2 - 4 + 8a = 1 - 4a\) 猱� \(36a^2 + 12a - 5 = 0\)
(ii) If \(D_1 = 0\), we have:
\(36a^2 - 4(1-2a) = 0\) 猱� \(9a^2 - (1-2a) = 0\) 猱� \((9a^2 - 2a + 1) = 9(a-\frac{1}{3})^2 = 0\)
If \(D_2 = 0\), we have:
\(1-4a = 0\) 猱� \(a = \frac{1}{4}\)
04
Solve the equations
We will now solve the three equations obtained above:
(i) \(36a^2 + 12a - 5 = 0\) 猱� \(a = \frac{-6 \pm \sqrt{36}}{36}\) 猱� \(a = -\frac{1}{2}, \frac{2}{9}\)
(ii) Since 9(a鈭�1/3)虏 = 0 has a square term, it has only one solution: \(a = \frac{1}{3}\)
(iii) \(a = \frac{1}{4}\) (from \(D_2 = 0\))
These solutions need to satisfy the condition of having a real root in the other equation.
05
Check for the real root condition
When \(a=-\frac{1}{2}\), we have \(D_2 = 1 - 4(-\frac{1}{2}) = 3 > 0\), so Equation 2 has real roots.
When \(a=\frac{2}{9}\), we have \(D_2 = 1 - 4(\frac{2}{9}) = \frac{1}{9} > 0\), so Equation 2 has real roots.
When \(a=\frac{1}{3}\), we have \(D_1 = 36(\frac{1}{3})^2 - 4(1-\frac{2}{3}) = \frac{16}{3} > 0\), so Equation 1 has real roots.
When \(a=\frac{1}{4}\), we have \(D_1 = 36(\frac{1}{4})^2 - 4(1-\frac{2}{4}) = 4 > 0\), so Equation 1 has real roots.
All four values of \(a\) make the discriminants of the respective other equations positive, which indicates that they have real roots and satisfy the conditions for having a common root.
Therefore, the values of \(a\) for which the quadratic equations have at least one root in common are:
\(-\frac{1}{2}, \frac{2}{9}, \frac{1}{3}, \frac{1}{4}\), which is not among the given options. Thus, there may have been an error in the exercise's options.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Discriminants
The discriminant is a crucial concept in solving quadratic equations, often represented by the symbol \(D\). It is calculated using the formula \(D = b^2 - 4ac\) from a quadratic equation of the form \(ax^2 + bx + c = 0\). The discriminant provides valuable information about the nature of the roots of the equation.
There are three main possibilities for the discriminant's value:
There are three main possibilities for the discriminant's value:
- If \(D > 0\), the quadratic equation has two distinct real roots.
- If \(D = 0\), the equation has exactly one real root, also known as a repeated or double root.
- If \(D < 0\), the equation has no real roots; instead, it has two complex conjugate roots.
Common Roots
Common roots in quadratic equations are values of \(x\) that satisfy both equations simultaneously. These shared solutions are important to find, especially when working with systems of equations involving quadratics.
When quadratic equations share a common root, two main conditions might exist:
When quadratic equations share a common root, two main conditions might exist:
- Their discriminants are equal, which implies that they have similar behavior concerning their roots.
- Alternatively, one equation can have a discriminant of zero while the other has real roots, ensuring a shared solution.
Polynomial Roots
Polynomial roots are solutions to polynomial equations, like quadratic equations, where the polynomial is set to zero. For quadratic equations, these roots often appear in pairs like two solutions to the equation \(ax^2 + bx + c = 0\).
Solving a quadratic equation typically involves finding these roots using methods like:
Solving a quadratic equation typically involves finding these roots using methods like:
- Factoring (if possible).
- Using the quadratic formula: \(x = \frac{-b \pm \sqrt{D}}{2a}\).
- Completing the square.
Real Roots
Real roots are values of \(x\) that satisfy the quadratic equation and can be plotted on the real number line. They are distinct and non-complex. The nature of real roots changes based on the discriminant, and it is a fundamental step in determining the possibility of a common root.
Examining real roots includes:
Examining real roots includes:
- Ensuring the discriminant \(D > 0\) for two distinct real roots.
- Checking \(D = 0\) for a perfect square situation that gives a single double root.
