Solving Quadratic Equations and Simultaneous Equations

The quadratic formula is as defined in the diagram above. By applying the quadratic formula, you can find the values of x that make the quadratic equation equal to zero.

Some alternative methods to solve a quadratic are:

Using quadratic formula (In Diagram above)

Method of Completing the Square

Graphing Method to Find the Roots

The quadratic formula helps in understanding and graphing parabolic curves. By solving a quadratic equation, you can find the x-intercepts, also known as the roots or zeros of the equation. These x-intercepts provide information about the points where the parabola intersects the x-axis. Additionally, the quadratic formula can be used to determine the vertex of the parabola, which provides valuable information about its shape and orientation.

Did you know: The golden ratio is found as the positive solution of the quadratic equation

x^{2}-x-1=0. The equations of the circle and the other conic sections—ellipses, parabolas, and hyperbolas—are quadratic equations in two variables.

The primary purpose of the quadratic formula is to determine the roots or solutions of a quadratic equation. The roots represent the values of x that satisfy the equation and make it true. By applying the quadratic formula, you can find the values of x that make the quadratic equation equal to zero.

Quadratic equations arise in various real-life situations, such as physics, engineering, finance, and optimization problems. The quadratic formula provides a systematic way to find the solutions to these problems.

Overall, the quadratic formula is a fundamental tool in algebra and provides insights into the behavior of quadratic equations. By manipulating the equation and applying the quadratic formula, you can understand the relationship between the coefficients (a, b, and c) and the solutions (roots) of the equation. This understanding extends to more advanced mathematical concepts, such as complex numbers and the nature of quadratic functions.

An ancient calculation: Rules for quadratic equations were given in The Nine Chapters on the Mathematical Art, a Chinese treatise on mathematics created in the 10th-2nd century BCE. These were later refined by a Greek mathematician around 300 BCE.

Uses of Quadratics in Simultaneous Equation

In the JC math curriculum quadratics can be used in a simultaneous equation. For example, in a pair of linear and non linear equation the value of y can be found using a substitution method:

Without using a calculator, solve the simultaneous equations,

Equation 1: x + y = 1

Equation 2: 2x2-y2=2

Steps to solve the simultaneous equation by Substitution method

• (A) Make X or Y the subject in the linear equation
• (B) Substitute equation 1 into equation 2
• (C) Solve the value for X
• (D) Using the value of X solve for the value of Y.

The answer following the step above should be as follows:

(A)
y = 1-x

(B)
2x2-(1-x)2= 2

(C)
2x2– (1-2x+x2) = 2

2x2 – 1 + 2x -x2 =2

x2+x-3= 0 (example of a quadratic equation which follows the format ax2+bx+c=0)

(x+3)(x-1)= 0

x=3 or x=1

(D)

When x= -3, y= 1-(-3) = 4

When x= -1, y=1-1= 0

Solving Simultaneous Equations with a Graphing Calculator

For Casio Calculators, simultaneous equation can be used to solve the equation:

Press applications, and select the option Plysmlt2 (5).

Press any key to enter the main menu and select simultaneous equation solver (2)

Equations on the graph are presented in a matrix form, the left most number is the value for x, the middle number is the value for y the last value is the total

Lastly, the values of x and y will be shown on the right as x1 and x2 respectively.

Other Simultaneous Equations for practise

A rectangle has a perimeter of 36m, and the square of the length of its diagonal is 170m2 . Find the dimensions of this rectangle.

Solution:

Firstly, let X and Y be the length and width of the rectangle. (equation 2)

x + x + y + y = 36 (Length + Length + Width + Width)

2x + 2y = 36

x + y = 18

Hypotenuse (diagonal of the triangle) can be found using the pythagoras theorem. In this case let the hypotenuse be Z, Z2 = X2 +Y2 (Equation 1)

z2 = 170

Substitute equation 2 into equation 1

X2 + (18-x)2 = 170

X2 + 324 – 36x + x2 = 170

2x2 – 36x +154 = 0

X2– 18x + 77 = 0

x=7 or x=11

When x=7, y=11

When x=11, y=7

The dimensions of the rectangle is 11m x 7m.