Friends, today we all are going to learn the basic concept behind one of the most interesting and important topic of Grade VIII mathematics that is quadratic equations. Before proceeding further let's talk about Binomial formulas first. The Binomial formulas are:
(a + b)2 = a(a + b) + b(a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2
Now I am going to discuss about quadratic equations. A polynomial equation of the second order is known as Quadratic equation. The general form of quadratic equation is:
ax² + bx + c = 0
where x is a variable and a, b and c are constants. Here a is quadratic coefficient, b is a linear coefficient and c is a constant term or we can say that it is a free term. To solve any quadratic problem or equation refers to find the value of variable. Let's take y that makes the equation true. To understand it better and in more deeply let's take an example:
(y – 1)2 = 25
As we can say that it is not a quadratic equation, we can convert it into an equivalent equation that is in that form, by suitable operations on both the sides of the equation.
(y + 1)2 = 25 expand
y2 – 2y + 1 = 25
y2 – 2y – 24 = 0 a = 1, b = -2, c = -24
The last solution or equation is in the standard form, where a, b and c having the given values.
But the second equation can be solved much more easily than the first one (ax² + bx + c = 0 ).
(y + 1)2 = 25 root
y – 1 = ±5 +1
y = 1 ±5 consider both cases the answer
y = 6 or y = -4 (these are the two solutions of the equation.) Students needs to verify this by substituting these values in the exact equation. If y = 6 we get 52 = 25 and if y = 4
then we got (-5)2 = 25.
Note the symbol ± in the above sequence of the equation. The square root value of the number 25 is positive by conventions and equals +5. Meanwhile, our task at that stage is not just evaluating a square root value as such, but our main focus should be there to answer the question for what values of y does (y – 1)2 equal 25?
There are two such values y – 1 = -5 and y – 1 = + 5, and student needs to consider both the possibilities.
Now to understand it more wisely, consider the more general equation
(y – r)2 = s (***)
where r and s are considered known and as before y need to be determined. The above equation can be solved just like we solve one before this.
Here r and s are considered known and as earlier y needs to be determined.
We can solve this in the same way as we can solve the above solved problem.
(y – r)2 = s root
y – r = ±root s +r
y = r ±root s is the required answer
The most important way to solve a quadratic equation is to convert them to the above mentioned form. This process is known as completing the square. It is mainly based on the first and second binomial formulas.
Let's take an example to understand the basic concept behind and how this works with our equation in standard form:
y2 – 2y – 24 = 0
If in the equation, the constant term was 1 instead of -24 than it would be a perfect square. To make it perfect square what we just need to do is add 25 on both the sides and get the desired value
y2 – 2y + 1 = 25
It can be rewritten as
(y – 1)2 = 25
Now the one thing is
(y – r)2 = y2 – 2yr + r2
What students need to do is to simply look at the factor of y, than halve it and in next step square it and in final step add the appropriate constant that makes the constant equal to that desired value. Another simplest of the way to find out that constant value is to subtract whatever constant is there and in final stage add the desired value. To solve it in this manner, work the leading coefficient (multiplying y2) must equal to 1. If it doesn't happen than we need to divide the first by the leading coefficient on both sides.
Now I am going to discuss about quadratic formula. Now what we need to know is what actually a quadratic formula is:
Take the standard form of the quadratic equation:
(a + b)2 = a2 + b2 + 2ab First Binomial formula
(a – b)2 = a2 + b2 -2ab Second Binomial formula
(a2 – b2) = (a + b)(a – b ) Third Binomial formula
Here a and b are the variables, or they can be even more general expressions. In the first and second binomial formulas, expression on the left are perfect squares while the expression on the left hand side of the third formula is the difference of two squares. One of the important thing to notice is that the first and second binomial formulas are equivalent. What student needs to do is to replace b with -b to get from one to the other.
We can use distributive law to verify the binomial formulas straight from left to right. For example:(a + b)2 = a(a + b) + b(a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2
Now I am going to discuss about quadratic equations. A polynomial equation of the second order is known as Quadratic equation. The general form of quadratic equation is:
ax² + bx + c = 0
where x is a variable and a, b and c are constants. Here a is quadratic coefficient, b is a linear coefficient and c is a constant term or we can say that it is a free term. To solve any quadratic problem or equation refers to find the value of variable. Let's take y that makes the equation true. To understand it better and in more deeply let's take an example:
(y – 1)2 = 25
As we can say that it is not a quadratic equation, we can convert it into an equivalent equation that is in that form, by suitable operations on both the sides of the equation.
y2 – 2y + 1 = 25
y2 – 2y – 24 = 0 a = 1, b = -2, c = -24
The last solution or equation is in the standard form, where a, b and c having the given values.
But the second equation can be solved much more easily than the first one (ax² + bx + c = 0 ).
(y + 1)2 = 25 root
y – 1 = ±5 +1
y = 1 ±5 consider both cases the answer
y = 6 or y = -4 (these are the two solutions of the equation.) Students needs to verify this by substituting these values in the exact equation. If y = 6 we get 52 = 25 and if y = 4
then we got (-5)2 = 25.
There are two such values y – 1 = -5 and y – 1 = + 5, and student needs to consider both the possibilities.
Now to understand it more wisely, consider the more general equation
(y – r)2 = s (***)
where r and s are considered known and as before y need to be determined. The above equation can be solved just like we solve one before this.
Here r and s are considered known and as earlier y needs to be determined.
We can solve this in the same way as we can solve the above solved problem.
(y – r)2 = s root
y – r = ±root s +r
y = r ±root s is the required answer
The most important way to solve a quadratic equation is to convert them to the above mentioned form. This process is known as completing the square. It is mainly based on the first and second binomial formulas.
Let's take an example to understand the basic concept behind and how this works with our equation in standard form:
y2 – 2y – 24 = 0
If in the equation, the constant term was 1 instead of -24 than it would be a perfect square. To make it perfect square what we just need to do is add 25 on both the sides and get the desired value
y2 – 2y + 1 = 25
It can be rewritten as
(y – 1)2 = 25
Now the one thing is
(y – r)2 = y2 – 2yr + r2
What students need to do is to simply look at the factor of y, than halve it and in next step square it and in final step add the appropriate constant that makes the constant equal to that desired value. Another simplest of the way to find out that constant value is to subtract whatever constant is there and in final stage add the desired value. To solve it in this manner, work the leading coefficient (multiplying y2) must equal to 1. If it doesn't happen than we need to divide the first by the leading coefficient on both sides.
Now I am going to discuss about quadratic formula. Now what we need to know is what actually a quadratic formula is:
Take the standard form of the quadratic equation:
The general form of quadratic equation is:
ax² + bx + c = 0
where x is a variable and a, b and c are constants.
We already study that we can solve this general equation by completing the square exactly like we would solve it if the coefficients assumed specific values.
Quadratic equations can be solved by using following methods : factoring , completing the squares, graphing, Newton's method, and with the help of Quadratic formula.
The Quadratic formula. Quadratic equation is ax² + bx + c = 0 and it has the solutions
here the expression under the square root sign is known as discriminant of the quadratic equation. Discriminant is denoted by the upper case Greek delta.- Delta = squared b – 4ac .
- If the discriminant is zero then there is only one exact real root, also known as double root.
- X = -b/2a.
- The ‘±’ symbol indicate as ‘plus or minus’, which means that we need to work out the formula twice, once with a plus sign in that position, then again with a minus sign.
- Clearly there are three cases while finding a discriminant:
- D > 0 there are two real solutions
- D = 0 there is one real solution
- D < 0 there is a associated complex pair of solutions.
- If you wants to apply the quadratic formula to a particular quadratic equation, what we need to do is just convert the equation to standard form and substitute the appropriate values of a, b and c in the formula. In between this is the more efficient and less error flat to make sure that student understand the binomial formulas and the basic concept behind completing the square and the most important is to solve a quadratic equation from scratch. In addition to these there are some of the formulas like the binomial formulas, the distributive law and different rules for manipulating powers, which needs to be remembered by the students.
- Let's take an example to understand it better, we need to solve the following equation
- x² - 4x - 5 = 0 , here no coefficient is written before x so we can use 1 as a coefficient of x. Now here a = 1, b = -4, and c = -5 now substitute this values in the above equation we get two values for the same that is x = 5 or x = -1.
- The technique used for graphing quadratic equations is the same as for graphing linear equations. The most basic quadratic equation is y = x2.. A quadratic graph is a parabola it can be generated by using quadratic math help calculator.
- For solving quadratic equations we can also use math helper or solvers available over internet. For graphing quadratic equations we need to graph a proper parabola which is quite difficult , so we can use math helper which can provide us the useful x and y intercepts through which it becomes easy to graph a proper parabola of a particular quadratic equation.
- Now in next class we all are moving forward and going to see the problems involving quadratic equations and try to solve them in a better and faster manner. In addition to these we are also going to understand the quadratic equation graphs.
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