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# Pythagoras Theorem

Quadratic equation is an equation of the form  + bx + c = 0 where a, b and c are constants and x is a variable, and a, is not equal to zero, otherwise it will be a linear equation. (Seward, K. (2009). A quadratic equation has two solutions and there are four steps involved.

If a, is the height, b the base length and c the diagonal length (hypotenuse) for a right angled triangle, Pythagorean Theorem states that  + =, that is the length of hypotenuse squared has the same area as the area of the other two sides combined, (Lindauer, H.). The theory is very applicable in daily lives. This paper explores the solutions to quadratic equation and application of the Pythagoras theorem.

High school assignment – quadratic equation and Pythagoras theorem

1. Two.
2. From the Quadratic formula, (Dawkins, P), D or Δ (discriminant) refers to the expression underneath the square root. .

(a)    If D is greater than zero, the equation has two solutions:   D =     or   .

(b)   If D is zero, the solution is   D = -b/2a.

(c)    If D is less than zero, the solution is not defined; the answer will have imaginary numbers.

Pythagorean Theorem

The theorem states that  + =  where a, is the height, b the base length and c the diagonal length (hypotenuse) for a right angled triangle, (Lindauer, H.). The Theorem is much applicable in daily lives. For example; painter intents to work on a building 4 meters tall where at the base of the wall there are flowers planted 3 meters from the wall. To get to the roof, he needs a ladder at least 5 meters tall, placed 3 meters from the base of the wall.

1. Steps for solving quadratic equation

Step 1 - make the coefficient of  to be one, by dividing through the equation by a.

Step 2 – separate variable   and x from the constants by subtracting   both sides,   + x = -

Step 3 – Divide the coefficient of x by 2, and square it. Then add this square to both sides of the equation, to make it a perfect square.

Step 4 - put in brackets, the results and combine the like terms.

Step 5– the equation in step 4 is solved by taking square roots of both sides of the equation, to get the value of the x.  (Seward, K. (2009).

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