Problem 88 - A Complete Resource Book in Mathematics for JEE Main

Question

If the two equations \(a x^{2}+b x+c=0\) and \(2 x^{2}-3 x+\) \(4=0\) have a common root, then (A) \(6 a=4 b=-3 c\) (B) \(3 a=-4 b=3 c\) (C) \(6 a=-4 b=3 c\) (D) None of these

Step-by-Step Solution

Answer

(C) 6a = -4b = 3c.

Step 1: Identifying the common root condition

If two quadratic equations \(a x^2 + bx + c = 0\) and \(2 x^2 - 3 x + 4 = 0\) have a common root, then this common root, say \(\alpha\), should satisfy both equations. Therefore, \(a \alpha^2 + b \alpha + c = 0\) and \(2 \alpha^2 - 3 \alpha + 4 = 0\).

Step 2: Expressing one variable in terms of the other

To find a relationship between \(a\), \(b\), and \(c\), we solve one equation for the common root. Substitute \(\alpha\) from the second equation into the first equation by expressing \(\alpha^2\) in terms of \(\alpha\):From \(2 \alpha^2 - 3 \alpha + 4 = 0\), solve for \(\alpha^2\):\[ \alpha^2 = \frac{3\alpha - 4}{2} \]

Step 3: Substituting to find the relationships

Replace \(\alpha^2\) in the first equation \(a \alpha^2 + b \alpha + c = 0\) with the expression found:\[ a \left( \frac{3\alpha - 4}{2} \right) + b \alpha + c = 0 \]Which simplifies to:\[ \frac{3a\alpha}{2} - 2a + b \alpha + c = 0 \]

Step 4: Simplifying the equation

Multiply the entire equation by 2 to clear the fraction:\[ 3a\alpha - 4a + 2b\alpha + 2c = 0 \]Collecting terms involving \(\alpha\):\[ (3a + 2b)\alpha - 4a + 2c = 0 \]

Step 5: Considering coefficients

For the above equation to be satisfied for any \(\alpha\), the coefficients of \(\alpha\) and the constant term must both be zero. Thus, we have:1. \(3a + 2b = 0\) 2. \(-4a + 2c = 0\) Solving these equations will give us the relationships between \(a\), \(b\), and \(c\).

Step 6: Solving the system of equations

From the equations: \(3a + 2b = 0\) implies \(b = -\frac{3}{2}a\). \(-4a + 2c = 0\) implies \(c = 2a\). Find the proportionalities: - From \(b = -\frac{3}{2}a\), multiply by 4: \(4b = -6a\). - From \(c = 2a\), multiply by 3: \(3c = 6a\).

Step 7: Assigning correct proportionality to options

We found:\- \(6a = 4b\)- \(3c = 6a\)This matches option \(C\): \(6a = -4b = 3c\). Therefore, the relationship is satisfied.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree 2, typically written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This is because if \(a = 0\), the equation would not be quadratic, but rather a linear equation. Quadratic equations often have two solutions or roots, which can be real or complex numbers.
To find these roots, we can use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The term inside the square root, \(b^2 - 4ac\), is called the discriminant, and it determines the nature of the roots:

• If the discriminant is greater than zero, the equation has two distinct real roots.
• If the discriminant is zero, both roots are equal, meaning the equation has a repeated real root.
• If the discriminant is less than zero, the equation has two complex roots.

Understanding quadratic equations is crucial for solving many mathematical problems, such as those involving motion, geometry, and optimization.

Roots and Coefficients

The roots of a quadratic equation are the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\). The relationship between the roots \(\alpha\) and \(\beta\), and the coefficients \(a\), \(b\), and \(c\) is important in understanding these equations fully. This relationship is derived from Vieta's formulas:

• The sum of the roots \((\alpha + \beta)\) is equal to \(-\frac{b}{a}\).
• The product of the roots \((\alpha \beta)\) is equal to \(\frac{c}{a}\).

These formulas provide a deep insight into the nature of the roots based on the equation's coefficients. Knowing the coefficients lets us determine the fundamental properties of the roots without actually solving the equation. This concept is particularly useful when comparing different quadratic equations or when manipulating equations to find a common root, as shown in our original exercise where consistency between two equations' roots was sought.

Simultaneous Equations

Simultaneous equations are a set of equations containing multiple variables. The solutions to these equations are the values that satisfy all of the equations at once. In the context of quadratic equations, finding a common root often involves solving simultaneous equations.
Consider two equations:1. \(ax^2 + bx + c = 0\)2. \(2x^2 - 3x + 4 = 0\)
If these two have a common root, we need to find the values of the coefficients in such a way that there is a mutual solution for both equations. Solving them simultaneously involves substituting the common root found in one equation into another, thus forming a relationship between their coefficients.
In our example, the goal was to express the variables in terms of one another, comparing their respective forms to achieve the required conditions for a common root. This process often involves simplification and elimination, leading to a clearer understanding of how the equations relate to each other.