QUADRATIC EQUATIONS
Quadratic Equations:
It is defined as the two degree equation Equations having two values for variables which may be real and imaginary.
These values known as roots of equation.
Consider the general quadratic equation
ax2 + bx + c = 0
Example: x2 + 3x – 4 = 0
Quadratic formula:
ax2 + bx + c = 0
Example: x2 + 3x – 4 = 0
Quadratic formula:
Methods of Solving Quadratic Equations
There are three main methods for solving quadratic equations:
- Factorization method
- Completing the square method
- Quadratic Equation Formula
Factorization
Example 1: Solve the equation: x2 + 3x – 4 = 0
Solution: This method is also known as splitting the middle term method. Here, a = 1, b = 3, c = -4. Let us multiply a and c = 1 * (-4) = -4. Next, the middle term is split into two terms. We do it such that the product of the new coefficients equals the product of a and c.
Solution: This method is also known as splitting the middle term method. Here, a = 1, b = 3, c = -4. Let us multiply a and c = 1 * (-4) = -4. Next, the middle term is split into two terms. We do it such that the product of the new coefficients equals the product of a and c.
We have to get 3 here. Consider (+4) and (-1) as the factors, whose multiplication is -4 and sum is 3. Hence, we write x2 + 3x – 4 = 0 as x2 + 4x – x – 4 = 0. Thus, we can factorise the terms as: (x+4)(x-1) = 0. For any two quantities a and b, if a×b = 0, we must have either a = 0, b = 0 or a = b = 0.
Thus we have either (x+4) = 0 or (x-1) = 0 or both are = 0. This gives x+4 = 0 or x-1 = 0. Solving these equations for x gives: x=-4 or x=1. This method is convenient but is not applicable to every equation. In those cases, we can use the other methods as discussed below.
Completing the Square Method
Each quadratic equation has a square term. If we could get two square terms on two sides of the quality sign, we will again get a linear equation. Let us see an example first.
Example 2: Let us consider the equation, 2x2=12x+54, the following table illustrates how to solve a quadratic equation, step by step by completing the square.
Solution: Let us write the equation 2x2=12x+54. In the standard form, we can write it as: 2x2 – 12x – 54 = 0. Next let us get all the terms with x2 or x in them to one side of the equation: 2x2 – 12 = 54
Solution: Let us write the equation 2x2=12x+54. In the standard form, we can write it as: 2x2 – 12x – 54 = 0. Next let us get all the terms with x2 or x in them to one side of the equation: 2x2 – 12 = 54
In the next step, we have to make sure that the coefficient of x2 is 1. So dividing throughout by the coefficient of x2, we have: 2x2/2 – 12x/2 = 54/2 or x2 – 6x = 27. Next, we make the left hand side a complete square by adding (6/2)2 = 9 i.e. (b/2)2 where ‘b’ is the new coefficient of ‘x’, to both sides as: x2 – 6x + 9 = 27 + 9 or x2 – 2×3×x + 32 = 36. Now we can write it as a binomial square:
- (x-3)2 = 36; Take square root of both sides
- x – 3 = ±6; Which gives us these equations:
- x = (3+6) or x = (3-6) or x = 9 or x = -3
This is known as the method of completing the squares.
Quadratic Equation Formula
There are equations that can’t be reduced using the above two methods. For such equations, a more powerful method is required. A method that will work for every quadratic equation. This is the general quadratic equation formula. We define it as follows: If ax2 + bx + c = 0 is a quadratic equation, then the value of x is given by the following formula:
Just plug in the values of a, b and c, and do the calculations. The quantity in the square root is called the discriminant or D. The below image illustrates the best use of a quadratic equation.
Sridharacharya:
Example 3: Solve: x2 + 2x + 1 = 0
Solution: Given that a=1, b=2, c=1, and
Discriminant = b2 − 4ac = 22 − 4×1×1 = 0
Using the quadratic formula, x = (−2 ± √0)/2 = −2/2
Therefore, x = − 1
Discriminant = b2 − 4ac = 22 − 4×1×1 = 0
Using the quadratic formula, x = (−2 ± √0)/2 = −2/2
Therefore, x = − 1
Roots of a Quadratic Equation
The number of roots of a polynomial equation is equal to its degree. Hence, a quadratic equation has 2 roots. Let α and β be the roots of the general form of the quadratic equation :ax2 + bx + c = 0. We can write:
α = (-b-√b2-4ac)/2a and β = (-b+√b2-4ac)/2a
Here a, b, and c are real and rational. Hence, the nature of the roots α and β of equation ax2 + bx + c = 0 depends on the quantity or expression (b2 – 4ac) under the square root sign. We say this because the root of a negative number can’t be any real number. Say x2 = -1 is a quadratic equation. There is no real number whose square is negative. Therefore for this equation, there are no real number solutions.
Hence, the expression (b2 – 4ac) is called the discriminant of the quadratic equation ax2 + bx + c = 0. Its value determines the nature of roots as we shall see. Depending on the values of the discriminant, we shall see some cases about the nature of roots of different quadratic equations.
Let us recall the general solution, α = (-b-√b2-4ac)/2a and β = (-b+√b2-4ac)/2a
- Case I: b2 – 4ac > 0
When a, b, and c are real numbers, a ≠ 0 and the discriminant is positive, then the roots α and β of the quadratic equation ax2 +bx+ c = 0 are real and unequal.
- Case II: b2– 4ac = 0
When a, b, and c are real numbers, a ≠ 0 and the discriminant is zero, then the roots α and β of the quadratic equation ax2+ bx + c = 0 are real and equal.
- Case III: b2– 4ac < 0
When a, b, and c are real numbers, a ≠ 0 and the discriminant is negative, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are unequal and not real. In this case, we say that the roots are imaginary.
- Case IV: b2 – 4ac > 0 and perfect square
When a, b, and c are real numbers, a ≠ 0 and the discriminant is positive and perfect square, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real, rational and unequal.
- Case V: b2 – 4ac > 0 and not perfect square
When a, b, and c are real numbers, a ≠ 0 and the discriminant is positive but not a perfect square then the roots of the quadratic equation ax2 + bx + c = 0 are real, irrational and unequal.
Here the roots α and β form a pair of irrational conjugates.
Here the roots α and β form a pair of irrational conjugates.
- Case VI: b2 – 4ac > 0 is perfect square and a or b is irrational
When a, b, and c are real numbers, a ≠ 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation ax2 + bx + c = 0 are irrational.
b2 – 4ac > 0 | Real and unequal |
b2 – 4ac = 0 | Real and equal |
b2 – 4ac < 0 | Unequal and Imaginary |
b2 – 4ac > 0 (is a perfect square) | Real, rational and unequal |
b2 – 4ac > 0 (is not a perfect square) | Real, irrational and unequal |
b2 – 4ac > 0 (is aperfect square and a or b is irrational) | Irrational |
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