JEE Main Eligibility Criteria 2025- Marks in Class 12th, Age Limit

# Quadratic Equation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• 50 Questions around this concept.

## Solve by difficulty

The equation  $\large \text { } x^2-4 x+[x]+3=x[x] \text {, where }[x] \text { : }$ denotes the greatest integer function, has

Let $S=\left\{x: x \in \mathbb{R}\right.$ and $\left.(\sqrt{3}+\sqrt{2})^{x^2-4}+(\sqrt{3}-\sqrt{2})^{x^2-4}=10\right\}$Then $n(S)$ is equal to

## Concepts Covered - 1

An expression of the form  f(x) = a0xn + a1xn-1 + a2xn-2 +... + an-1x + an , is called a polynomial expression.

Where x is variable and a0, a1, a2, …….. , an  are constant, known as coefficients and  a0 ≠ 0, n is non-negative integer,

Degree: The highest power of variable the polynomial expression is called the degree of the polynomial. In $a_0\cdot x^n + a_1\cdot x^{n-1} + ... +a_n$ , the highest power of x is n, so the degree of this polynomial is n.

If coefficients are real numbers then it is called real polynomial, and when they are complex numbers, then the polynomial is called complex polynomial.

Root of polynomial:

If f(x) is a polynomial, then f(x) = 0 is called polynomial equation.

The value of x for which the polynomial equation, f(x) = 0 is satisfied is called a root of the polynomial equation.

If x = α is a root of the equation f(x) = 0, then f(α) = 0.

Eg, x=2 is a root of x- 3x + 2 = 0, as x=2 satisfies this equation.

A polynomial equation of degree n has n roots (real or imaginary).

A polynomial equation in which the highest degree of a variable term is 2 is called quadratic equation.

Standard form of quadratic equation is ax2 + bx + c = 0

Where a, b and c are constants (they may be real or imaginary) and called the coefficients of the equation and $a\neq0$ (a is also called the leading coefficient).

Eg, -5x- 3x + 2 = 0, x2 = 0, (1 + i)x- 3x + 2i = 0

As degree of quadratic polynomial is 2, so it always has 2 roots (number of real roots + number of imaginary roots = 2)

The root of the quadratic equation is given by the formula:

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

Where D is called the discriminant of the quadratic equation, given by $D = b^2 - 4ac$ ,

Proof

$\\\mathrm{ax^2+bx+c=0}\\\mathrm{Take,\;'a'\;common}\\\mathrm{a\left ( x^2+\frac{b}{a}x+\frac{c}{a} \right )=0}\\\mathrm{a\left [ \left ( x+\frac{b}{2a} \right )^2-\frac{b^2}{4a^2}+\frac{c}{a} \right ]=0}\\\mathrm{\left ( x+\frac{b}{2a} \right )^2=\frac{b^2-4ac}{4a^2}}\\\mathrm{\left ( x+\frac{b}{2a} \right )=\pm\frac{\sqrt{b^2-4ac}}{2a}}\\\mathrm{x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}}\\\mathrm{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$