# Determine the nature of the solutions of the equation. 2t^2-4t=0.

## Question:

Determine the nature of the solutions of the equation. {eq}2t^2-4t=0. {/eq}

Quadratic equations are the mathematical equations of degree 2 and composed of constants and variables.

General representation-

{eq}ax^{2}+bx+c = 0 \ a \neq 0 {/eq}

here a and b are the coefficients of the variable {eq}x^{2} {/eq} and {eq}x {/eq} respectively.

c is the numeral constant.

A mathematical equation has the number of solutions equal to the degree of the equation so the quadratic equation will have two solutions.

#### Nature of Solutions of a Quadratic Equation

For a quadratic equation {eq}\displaystyle ax^{2}+bx+c = 0 {/eq}

Discriminant {eq}(D) = b^{2}-4ac {/eq}

The nature of the solutions of a quadratic equation will depend on the value of the discriminant.

(1) If {eq}D > 0 {/eq} then the quadratic equation will have two real and distinct solutions

(2) if {eq}D < 0 {/eq} then the quadratic equation will have imaginary solutions hence it does not have any real solution.

(3) if {eq}D = 0 {/eq} then the quadratic equation will have two equal and real solutions.

The solution of a quadratic equation {eq}\displaystyle ax^{2}+bx+c =0 {/eq} is given by the quadratic formula as follows-

{eq}\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} {/eq}

{eq}\displaystyle 2t^{2}-4t = 0 --------(1) {/eq}

by comparing the equation(1) with the standard form {eq}\displaystyle at^{2}+bt+c = 0 {/eq} we have

{eq}a = 2 {/eq}

{eq}b = -4 {/eq}

{eq}c = 0 {/eq}

Now

Discriminant {eq}\displaystyle (D) = (-4)^{2}-4 \times 2 \times 0 {/eq}

{eq}\displaystyle D = 16-0 > 0 {/eq}

So the above quadratic equation will have two real and distinct solutions.