maths how to solve problem like quadratic equation
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A quadratic equation is a polynomial whose highest power is the square of a variable (x2, y2etc.)
Definitions
A monomial is an algebraic expression with only one term in it.
Example: x3, 2x, y2, 3xyz etc.
A polynomial is an algebraic expression with more than one term in it.
Alternatively it can be stated as –
A polynomial is formed by adding/subtracting multiple monomials.
Example: x3+2y2+6x+10, 3x2+2x-1, 7y-2 etc.
A polynomial that contains two terms is called a binomial expression.
A polynomial that contains three terms is called a trinomial expression.
A standard quadratic equation looks like this:
ax2+bx+c = 0
Where a, b, c are numbers and a≥1.
a, b are called the coefficients of x2 and x respectively and c is called the constant.
The following are examples of some quadratic equations:
1) x2+5x+6 = 0 where a=1, b=5 and c=6.
2) x2+2x-3 = 0 where a=1, b=2 and c= -3
3) 3x2+2x = 1
→ 3x2+2x-1 = 0 where a=3, b=2 and c= -1
4) 9x2 = 4
→ 9x2-4 = 0 where a=9, b=0 and c= -4
For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.
For a quadratic equation ax2+bx+c = 0,
the sum of its roots = –b/a and the product of its roots = c/a.
A quadratic equation may be expressed as a product of two binomials.
For example, consider the following equation
x2-(a+b)x+ab = 0
x2-ax-bx+ab = 0
x(x-a)-b(x-a) = 0
(x-a)(x-b) = 0
x-a = 0 or x-b = 0
x = a or x=b
Here, a and b are called the roots of the given quadratic equation.
Now, let’s calculate the roots of an equation x2+5x+6 = 0.
We have to take two numbers adding which we get 5 and multiplying which we get 6. They are 2 and 3.
Let us express the middle term as an addition of 2x and 3x.
→ x2+2x+3x+6 = 0
→ x(x+2)+3(x+2) = 0
→ (x+2)(x+3) = 0
→ x+2 = 0 or x+3 = 0
→ x = -2 or x = -3
This method is called factoring.
We saw earlier that the sum of the roots is –b/a and the product of the roots is c/a. Let us verify that.
Sum of the roots for the equation x2+5x+6 = 0 is -5 and the product of the roots is 6.
The roots of this equation -2 and -3 when added give -5 and when multiplied give 6.
Solved examples of Quadratic equations
Let us solve some more examples using this method.
Problem 1: Solve for x: x2-3x-10 = 0
Solution:
Let us express -3x as a sum of -5x and +2x.
→ x2-5x+2x-10 = 0
→ x(x-5)+2(x-5) = 0
→ (x-5)(x+2) = 0
→ x-5 = 0 or x+2 = 0
→ x = 5 or x = -2
Problem 2: Solve for x: x2-18x+45 = 0
Solution:
The numbers which add up to -18 and give +45 when multiplied are -15 and -3.
Rewriting the equation,
→ x2-15x-3x+45 = 0
→ x(x-15)-3(x-15) = 0
→ (x-15) (x-3) = 0
→ x-15 = 0 or x-3 = 0
→ x = 15 or x = 3
Till now, the coefficient of x2 was 1. Let us see how to solve the equations where the coefficient of x2 is greater than 1.
Problem 3: Solve for x: 3x2+2x =1
Solution:
Rewriting our equation, we get 3x2+2x-1= 0
Here, the coefficient of x2 is 3. In these cases, we multiply the constant c with the coefficient of x2. Therefore, the product of the numbers we choose should be equal to -3 (-1*3).
Expressing 2x as a sum of +3x and –x
→ 3x2+3x-x-1 = 0
→ 3x(x+1)-1(x+1) = 0
→ (3x-1)(x+1) = 0
→ 3x-1 = 0 or x+1 = 0
→ x = 1/3 or x = -1
Definitions
A monomial is an algebraic expression with only one term in it.
Example: x3, 2x, y2, 3xyz etc.
A polynomial is an algebraic expression with more than one term in it.
Alternatively it can be stated as –
A polynomial is formed by adding/subtracting multiple monomials.
Example: x3+2y2+6x+10, 3x2+2x-1, 7y-2 etc.
A polynomial that contains two terms is called a binomial expression.
A polynomial that contains three terms is called a trinomial expression.
A standard quadratic equation looks like this:
ax2+bx+c = 0
Where a, b, c are numbers and a≥1.
a, b are called the coefficients of x2 and x respectively and c is called the constant.
The following are examples of some quadratic equations:
1) x2+5x+6 = 0 where a=1, b=5 and c=6.
2) x2+2x-3 = 0 where a=1, b=2 and c= -3
3) 3x2+2x = 1
→ 3x2+2x-1 = 0 where a=3, b=2 and c= -1
4) 9x2 = 4
→ 9x2-4 = 0 where a=9, b=0 and c= -4
For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.
For a quadratic equation ax2+bx+c = 0,
the sum of its roots = –b/a and the product of its roots = c/a.
A quadratic equation may be expressed as a product of two binomials.
For example, consider the following equation
x2-(a+b)x+ab = 0
x2-ax-bx+ab = 0
x(x-a)-b(x-a) = 0
(x-a)(x-b) = 0
x-a = 0 or x-b = 0
x = a or x=b
Here, a and b are called the roots of the given quadratic equation.
Now, let’s calculate the roots of an equation x2+5x+6 = 0.
We have to take two numbers adding which we get 5 and multiplying which we get 6. They are 2 and 3.
Let us express the middle term as an addition of 2x and 3x.
→ x2+2x+3x+6 = 0
→ x(x+2)+3(x+2) = 0
→ (x+2)(x+3) = 0
→ x+2 = 0 or x+3 = 0
→ x = -2 or x = -3
This method is called factoring.
We saw earlier that the sum of the roots is –b/a and the product of the roots is c/a. Let us verify that.
Sum of the roots for the equation x2+5x+6 = 0 is -5 and the product of the roots is 6.
The roots of this equation -2 and -3 when added give -5 and when multiplied give 6.
Solved examples of Quadratic equations
Let us solve some more examples using this method.
Problem 1: Solve for x: x2-3x-10 = 0
Solution:
Let us express -3x as a sum of -5x and +2x.
→ x2-5x+2x-10 = 0
→ x(x-5)+2(x-5) = 0
→ (x-5)(x+2) = 0
→ x-5 = 0 or x+2 = 0
→ x = 5 or x = -2
Problem 2: Solve for x: x2-18x+45 = 0
Solution:
The numbers which add up to -18 and give +45 when multiplied are -15 and -3.
Rewriting the equation,
→ x2-15x-3x+45 = 0
→ x(x-15)-3(x-15) = 0
→ (x-15) (x-3) = 0
→ x-15 = 0 or x-3 = 0
→ x = 15 or x = 3
Till now, the coefficient of x2 was 1. Let us see how to solve the equations where the coefficient of x2 is greater than 1.
Problem 3: Solve for x: 3x2+2x =1
Solution:
Rewriting our equation, we get 3x2+2x-1= 0
Here, the coefficient of x2 is 3. In these cases, we multiply the constant c with the coefficient of x2. Therefore, the product of the numbers we choose should be equal to -3 (-1*3).
Expressing 2x as a sum of +3x and –x
→ 3x2+3x-x-1 = 0
→ 3x(x+1)-1(x+1) = 0
→ (3x-1)(x+1) = 0
→ 3x-1 = 0 or x+1 = 0
→ x = 1/3 or x = -1
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