Quadratic Equation in VIII Grade

Hello friends, in today's class we are going to learn what a quadratic equation is and how can we solve it.

 

A Quadratic equation is a second order univariate polynomial equation. A general quadratic equation can be written as a x² + b x + c, here x is the variable and a,b and c are the constants with a ≠ 0. There are two solutions of the quadratic equations, either both are real or both are complex.

There are various methods of solving quadratic equations but the ones that we have to learn are as follows-

1. factoring
2. completing the square
3. by using the quadratic formula.

Let's start solving the quadratic equations by using the above methods one by one.

 

We start by using the factoring method, let's understand by using the general formula.

The general equation is ax² + bx + c. We follow some general steps to solve the quadratc equation.

1.multiply the constants a and c without forgetting their signs.

2.Write down b, with its sign.

3.Write all the possible factors of the products of ac and find which pairs add up to give b.

4.Now rewrite the x terms as the sum of two terms with these selected numbers as the coefficients.

So by this we can find the solution of any quadratic equation.

For Example –

1. Solve x² + 5 x + 6.

Answer – the possible combination of the middle term is 1x+4x and 2x+3x.

The product of a and c is 6 x 1= 6. So the second combination is the suitable option for this quadratic expression. So now it can be written as x² + 3 x + 2 x + 6.

Now we take x common from the first two terms and 2 from the last two terms.

Now the quadratic equation becomes x ( x + 3 ) + 2 ( x + 3 ).

This time we take (x + 3) common, so the equation now becomes (x + 2 ) ( x+ 3 ).

the value of x = -3, -2.

The solution for the quadratic equation x² + 5 x + 6 is x = -2, -3.

2. Solve x² – 3 = 2 x.

We bring 2x on the left hand side. So now the equation is x² – 2 x – 3.

now breaking the middle term we see that the possible combination is -1x -1x and -3x + x.

The product of a and c is -3. So the second combination is the suitable one for this quadratic expression.

Now the equation becomes x² – 3 x + 1 x – 3.

now by taking x common from the first two terms and 1 from the last two terms we get,

x ( x – 3 ) + 1 ( x – 3 ). Now taking x - 3 common from both the terms. We get

(x + 1 ) ( x – 3 ).

x = -1, 3.

so the solution to the quadratic equation x² - 2 x – 3 is x = -1, 3.

3. Solve x² - 9.

Answer- we can see that it is in a² - b² form, so we can break it as ( a + b )( a – b ).

So our equation becomes ( x + 3 ) ( x – 3 ).

now x = -3 , 3.

So the solution to the quadratic equation x² - 9 is x = -3, 3.

 

2. By completing the squares.

For the general equation a x² + b x + c.

In this method we follow these steps.

1. take the constant to one side and make the coefficient of x² = 1.
   

2. add the square of the half of coefficient of x on both the sides in the expression as shown above.
Now we can easily find out the value of x.
Let's clear the doubts with the help of an example.
1. Solve x² - 5 x + 6.
we follow the same general steps.
= x² - 5 x = -6.
add the square of the half of coefficient of x on both sides.
= x² - 5 x + 25/4 = 25/4 – 6.
so the left hand side becomes
= (x - 5/2)² = 25/4 – 6.
= (x – 5/2) = ± ¼.
Now value of x = 5/2 + ¼ or 5/2 – ¼.
x = 3, 2.
2. Solve x² – 3 = 2 x.
= x² - 2 x = 3.
= x² - 2 x + 1 = 3+1.
= (x -1)² = 4.
= (x -1) = ± 2.
= x = 1+ 2 or 1-2.
= x = 3, -1.
3. solve x² - 9.
In this we can see that the coefficient of x = 0, so we bring the constant to the right hand side.
= x² = 9.
= x = ± 3
= x = -3, +3.
Let's now use the third method to find the solution to the quadratic expressions that is by using the quadratic formula.


3. For the general equation a x² + b x + c.

The value of


The two values are using the positive and the negative signs.
But how did we got this formula?
We got this by using completing the squares method, which we have just learned.
Let's derive it for our general equation,
1.


2.


3.



4.






So this is how we got our general formula. Let's now solve some questions using this quadratic formula.
1. Solve – .




Here a = 2, b = -3 and c = 1/2. now we put these values in the formula and find the values of x.




So we get two values of x.
2. Solve-

for x by using the quadratic formula. Here a = 1, b = 2 and c = 10 .

so we put these values in the formula.

So the value of x becomes,

 

Now as we have solved the quadratic equations by using the different methods, so let's now solve some word problems on Quadratic equations.

By using the same general formulas we can solve the word problems on quadratic equations, What we have to do is just form the quadratic expression from the given problem and the rest could be found using the above three formulas.

Let's solve some of them for more understanding.

 

1. Question - The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.

Answer - The formula for the area of a rectangle is length x breadth.

So if width is assumed as x than the other side i.e. the breadth will be x+6 as per the given condition.

So the area is x ( x + 6 ), and the area is given as 91.

So x ( x + 6 ) = 91.

now the quadratic equation is x² + 6 x – 91 = 0.

so now by solving this equation we will get x = 7, -13.

since the length of the triangle cannot be negative so the breadth of the triangle is 7 and its length is 7 + 6 = 13.

2. Question - The product of two consecutive odd integers is 1 less than four times their sum. Find the two integers?

Answer – let us assume the first number to be n. Since the second number is the consecutive odd one so the value of second number will be n + 2.

now we implement the given problem.

The product of two numbers will be n(n+2), this is equal to 1 less than 4 times its sum,

the sum is 2 n + 2, 4 time of it 4 x (2 n + 2 ), and it is 1 less = 4 ( 2 n + 2 ) -1.

so the final quadratic equation is n² + 2 n = 8 n + 8 – 1.

= n² - 6 n – 7.

By solving the above quadratic equation we get the values of x as 7, 9 and -1, 1.

So these are the two possible combinations possible for the numbers.

3. Question-The hypotenuse of a right triangle is 6 more than the shorter leg. The longer leg is three more than the shorter leg. Find the length of the shorter leg.

Answer – for solving these equations we should know about pythagoras theorem.

According to pythagoras theorem (side)² + (side)² = (hypotenuse)²

Take one side as x, than the other side is x + 3 and the hypotenuse is x + 6.

now by using the pythagoras theorem.

x² + ( x + 3 )² = ( x + 6 )²

so we get, x² - 6 x – 27.

now by solving this we will get x = -3, 9. since the side cannot be negative so the value of will be 9.

Eighth grade quadratic equations

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 = 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.

(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:

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 = x^2.. 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.