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.