CAT Equations of the form of ax2 + by +c = 0 MCQ

Quick Facts

5 Questions around this concept.

Solve by difficulty

EASY

Which of the following is not a quadratic equation ?

$x^2-4x+2=0$

$x+\frac{1}{x}=1$

$x^2+\frac{1}{x}=4$

$x^2=4x$

View Solution

Which of the following are quadratic equation?

$(2x-1)(2x+1)=(x+4)$

$x(x+3)=(x+1)(x+2)$

$(x+2)^3=2x$

Both (a) and (b)

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Concepts Covered - 1

Equations of the form of ax2 + by +c = 0

A quadratic equation is an equation of degree 2. The general form is:

$\mathrm{\[ ax^2 + bx + c = 0 \]}$

Where:

$\mathrm{- $ a $, $ b $, and $ c $}$ are constants.

$\mathrm{- $ a \neq 0 $ (because \: \: if $ a = 0 $, \text{the equation becomes linear}).}$

The solutions to this equation, often referred to as the roots or zeros of the equation, can be found using various methods.

Methods to Solve:

1. Factorization: If the quadratic can be factored, this method involves expressing the equation in the form $\mathrm{\mathrm{$(x - p)(x - q) = 0$, where $ p $ and $ q $}}$ are the roots.

2. Completing the Square: Convert the equation into a perfect square trinomial.

3. Quadratic Formula:

$\mathrm{\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]}$

Application of Previous Concepts:

When we talk about the intersection of lines and curves:

- A quadratic equation can represent a parabola. If a line intersects this parabola, the point(s) of intersection can be found by solving the linear equation and the quadratic equation simultaneously.

Foundation Building Questions:

(Note: Again, I don't have access to specific previous year questions from management entrances post-2021, so I'll create a representative question based on the typical style of these exams.)

Question:

Solve for ( x) in the equation:

$\mathrm{\[ 2x^2 - 5x + 2 = 0 \]}$

Solution:

Using the Quadratic Formula:

$\mathrm{\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]}$

Where:

$\mathrm{$ a = 2, b = -5, $ and $ c = 2 $.}$

Plugging in the values:

$\mathrm{\[ x_1 = \frac{{5 + 1}}{{4}} = \frac{3}{2} \] \[ x_2 = \frac{{5 - 1}}{{4}} = 1 \]}$

So, the solutions are $\mathrm{$ x_1 = \frac{3}{2} $ and $ x_2 = 1 $.}$

Tips and Tricks:

1. Nature of Roots: Before solving, a quick check of the discriminant $\mathrm{($ b^2 - 4ac $)}$ can give insight into the nature of the roots. If it's positive, there are two distinct roots; if zero, the roots are equal; if negative, the roots are imaginary.

2. Common Factors: Always look for common factors to simplify the equation before diving into solutions.

3. Graphical Insight: A quick sketch or mental visualisation of the parabola can give insights into the number and nature of the roots.

4. Intersection with Lines: If you need to find where a quadratic equation (parabola) intersects a line, substitute the expression for (y) from the line equation into the quadratic equation and solve for ( x ).

Quadratic equations are fundamental in many areas of mathematics and physics. Understanding their properties and methods of solution is crucial for a variety of applications.