QUADRATIC EQUATIONS –
A quadratic equation in x is an equation in which the highest power of x is 2. The standard form of a quadratic equation in x is ax² + bx +c = 0, where a, b & c are constants and a ≠ 0
Example.) 5x² + 7x = 12, 3x² - 7x = 10 and 3x² - 25 = 0 are quadratic equations in x.
Solution of a quadratic equation –
In general, a quadratic equation has two solutions, which are also called roots. The two solutions may be equal.
Let the roots of the equation ax² + bx + c be x = α, x = β, and x = γ then each of the values of the x must be satisfy the equation, that is - aα² + bα + c = 0, aβ² + bβ + c = 0, and aγ² + bγ + c = 0
Solving quadratic equations –
To solve the quadratic equations, we first factorize it and then use the principle of zero product, which states – if the product of two expressions (or numbers) is zero then at least one of the expressions must be zero. We can express this symbolically as follows.
If ab = 0 then a = 0 or b = 0 or both a = 0 & b = 0
Example.1) If (x – 3)(x – 5) = 0, then by the principal of the zero product, either (x – 3) = 0 or (x – 5) = 0
If, x – 3 = 0, then x = 3
If, x – 5 = 0, then x = 5
There are some steps for solving a quadratic equation are given below –
Step.1) Write the equation in the standard form ax² + bx + c = 0
Step.2) Factorize the quadratic expression on the LHS
Step.3) Set each factor equal to zero.
Step.4) Solve each of the resulting linear equations.
There are some example are given below for your better understanding –
Example.1) Solve the equation x² + 6x + 5 = 0
Ans.) To, begin with, let us factorize the LHS of the given equation with 0 on RHS
x² + 6x + 5 = 0
or x² + (5 + 1) x + 5 = 0
or x² + 5x + x + 5 = 0
or x (x + 5) + 1 (x + 5) = 0
(x + 1) (x + 5) = 0
By, the zero product principle, either x + 1 = 0, which gives x = - 1
Or, x + 5 = 0, which gives x = - 5
Therefore, the roots (or solution) of the equation are – 1 & - 5 (Ans.)
Example.2) Solve the quadratic equation 12a² + 16a + 5 = 0
Ans.) Given, 12a² + 16a + 5 = 0
To, begin with, let us factorize the LHS of the given equation with 0 on RHS
So, 12a² + 16a + 5 = 0
Or, 12a² + (10 + 6)a + 5 = 0
Or, 12a² + 10a + 6a + 5 = 0
Or, 2a (6a + 5) + 1 (6a + 5) = 0
Or, (6a + 5) (2a + 1) = 0
By, the zero product principle, either 2a + 1 = 0, which gives a = - 1/2
And, 6a + 5 = 0, which gives a = - 5/6
Therefore, the roots (or solution) of the equation are – 1/2 & - 5/6 (Ans.)
Example.3) Solve the equation 25m² - 49 = 0
Ans.) Given, 25m² - 49 = 0
To, begin with, let us factorize the LHS of the given equation with 0 on RHS
So, 25m² - 49 = 0
Or, (5m)² - (7)² = 0
Or, (5m + 7) (5m – 7) = 0
By, the zero product principle, either 5m + 7 = 0, which gives m = - 7/5
And, 5m - 7 = 0, which gives m = 7/5
Therefore, the roots (or solution) of the equation are – 7/5 & 7/5 (Ans.)
50
Example.4) Solve the equation x - --------- = 5, x ≠ 0
x
50
Ans.) Given, x - ---------- = 5
x
x² - 50
So, --------------- = 5
x
By, cross multiplication we find –
x² - 50 = 5x
or, x² - 5x – 50 = 0
or, x² - (10 – 5)x – 50 = 0
or, x² - 10x + 5x – 50 = 0
or, x (x – 10) + 5 (x – 10) = 0
or, (x – 10) (x + 5) = 0
By, the zero product principle, either x - 10 = 0, which gives x = 10
And, x + 5 = 0, which gives x = - 5
Therefore, the roots (or solution) of the equation are 10 & - 5 (Ans.)
x + 3 3x - 7
Example.5) Solve the equation ------------ = ------------
x + 2 2x – 3
x + 3 3x - 7
Ans.) Given, ------------- = -------------
x + 2 2x – 3
by, cross multiplication we find –
(x + 3) (2x – 3) = (3x – 7) (x + 2)
Or, 2x² + 6x – 3x – 9 = 3x² - 7x + 6x – 14
Or, 0 = (3x² - 2x²) – (7x + 6x) + (6x + 3x) – 14 + 9
Or, x² - 13x + 9x – 5 = 0
Or, x² - 4x – 5 = 0
Or, x² - (5 – 1)x – 5 = 0
Or, x² - 5x + x – 5 = 0
Or, x (x – 5) + 1 (x – 5) = 0
Or, (x – 5) (x + 1) = 0
By, the zero product principle, either x - 5 = 0, which gives x = 5
And, x + 1 = 0, which gives x = - 1
Therefore, the roots (or solution) of the equation are 5 & - 1 (Ans.)