335 total views, 11 views today

## Introduction

Quadratic Equations have parabolic profiles. Parabolas are bilateral symmetrical along the base point.

Since quadratic equation curves look like parabolas, it means they have symmetric properties.

Know that quadratic equations have two solutions because the parabolic curves have two arms that transverse the x or y-axis at two points depending on the orientations of the curve a given quadratic equation gives.

Based on these, we say, quadratic equations have two roots: the roots are the two solutions.

**Picture Image used for illustrative purpose**

**Symmetric Properties of Quadratic Equations**

If we assume that a typical Quadratic Equation has two roots £ and $, then the quadratic equation could be defined as:

**y = x^2 – (£ + $ ) + (£ * $ );**

From the above equation,

**y is the dependent variable;**

**x is independent variable;**

**(£ + $ ) is sum of the roots;**

**(£ * $ ) being product of the roots.**

Note that the above equation applied to all forms of quadratic equations.

## Example 1:

Find the quadratic equation whose roots are 3 and -5.

**Solution:**

root 1, £ = 3; root 2, $ = -5

Sum of the roots: (£ + $) = 3 + -5 = 3 – 5 = -2;

Product of the roots: (£ * $) = 3 * -5 = -15;

Hence, the final quadratic equation when those values are substituted into the general quadratic equation becomes:

**y =x^2 + 2x -15; and this gives the quadratic equation whose roots are -2 and 5.**

## Example 2

The question to be solved here comes from NECO SSCE 2019.

If the roots of aquadratic equation are -2/3 and -3/2, find the equation.

**Solution**

Root 1 = -2/3; Root 2 = -3/2;

Sum of the roots:

-2/3 + -3/2 = (-4 + -9)/6

[Addition of fractions through LCM]

= -13/6

Product of the roots = -2/3 * -3/2 = 1;

Final equation is :

y = x^ 2 +(-13/6)x + 1;

i.e. y = x^2 – (13/6)x + 1;

Note that y is always 0 at solution points if the parabolic curve is vertically inclined. If it is horizontally inclined , then x is 0 while values of y are solutions.

## Example 3

The roots of a QE in x are -m and 2n. Find the equation.

**Solution**

Sum of roots = (2n – m)

Product of roots = -2mn

Equation is:

y = x^2 + (2n – m)x – 2mn;

Or x^2 + (2n – m)x – 2mn = 0

**Example 4**

If (x – 3) is a factor of the QE x^2 + kx – 21 = 0 where k is a constant, find the value of k.

**Solution**

If we assume the other root is (x + j),

then, (x -3)(x+j) = x^2 + kx -21 = 0 ;

If x – 3 is assumed as the zero valued factor,

then x – 3 = 0; x = 3.

Substitite x = 3 in the given QE, we have :

3^ 2 + 3k -21 = 0; hence, 9 + 3k – 21 = 0;

then, 3k = 21 – 9; which is 3k = 12; **therefore k = 4.**

If you have grey areas that requires more explanations, kindly drop it in the comment section of the post.

**Please share the post **

Advanced examples would come your way later. And if you have questions you find difficult to solve post them to my email profsmith24@gmail.com

Thanks for reading

## Leave a Reply