Today our topic of discussion is Quadratic Equations in One Variable.
Quadratic Equations in One Variable
Quadratic Equations in One Variable
Equations of the form ax + bx + c = 0 (where, a, b, c are constants and a + 0) are called the quadratic equation in one variable. Left hand side of a quadratic equation is a polynomial of second degree, right hand side is generally taken to be zero.
Length and breadth of a rectangular region of area 12 square em. are respectively z cm. and (x-1) cm.
∴ area of the rectangular region i = x(x – 1) square cm.
By the question, x(x – 1) = 12 or, x² – x – 12 = 0
x is the variable in the equation and highest power of z is 2. Such equation is a quadratic equation. The equation, which has the highest degree 2 of the variable, is called the quadratic equation.
In class VIII, we have factorized the quadratic expressions in one variable of the forms x²+ px + q and a * x²+ bx + c Here, we shall solve the equations of the forms x² + px + q = 0 and a * x² + bx + c = 0 by factorizing the left hand side and by finding the value of the variable.
An important law of real numbers is applied to the method of factorization. The law is as follows:
If the product of two quantities is equal to zero, either only one of the quantities or both quantities will be zero. That is, if the product of two quantities a and b
Le, ab = 0, a = 0,b = 0, both a =0: and b = 0
Example 8. Solve: (x + 2)(x – 3) = 0
Solution: (x + 2)(x – 3) = 0
x + 2 = 0 or, x – 3 = 0
If x + 2 = 0, x = – 2
Again, if x – 3 = 0, x = 3
* solution is x = – 2 or, x = 3
Example 9. Find the solution set: y² = sqrt(3) * y
Solution: y² = (√3) * y
or, y² – (√3) * y = 0
or, y(y – (√3) = 0 therefore y = 0 or y – (√3) = 0
Again, if y – (√3) = 0 y = (√3)
Solution set is ( 0, (√3))
Example 10.
Solve and write the solution set: x – 4 = (x – 4)/x
Solution: x – 4 = (x – 4)/x
or, x(x – 4) = x – 4 [by cross multiplication]
or, x(x – 4) – (x – 4) = 0 [by transposition]
or, (x – 4)(x – 1) = 0
.. x – 4 = 0 or x – 1 = 0
If x – 4 = 0 x = 4
Again, if x – 1 =0.x=1
∴Solution set is {1,4}
Example 11. Solve: ((x + a)/(x – a)²- 5((x + a)/(x – a)) + 6 = 0
Solution: ((x + a)/(x – a)² – 5((x + a)/(x – a)) + 6 =0…(1)
Let, (x + a)/(x – a) = y
.. from(1) we get, y ² – 5y + 6 = 0
or, y ² – 2y – 3y + 6 = 0
or, y(y – 2) – 3(y – 2) = 0
or, (y – 2)(y – 3) = 0
∴ when, y – 2 = 0 y = 2 or, if y – 3 = 0 y = 3
Now, if y = 2
(x + a)/(x – a) = 2/1 [putting the value of y]
or, x + a = 2(x – a) [by cross multiplication]
or, x + a = 2x – 2a
or, 2x – x = a + 2a
or, x = 3a
Again, when y = 3
(x + a)/(x – a) = 3/1
or, x + a = 3(x – a) [by cross multiplication]
or, x + a = 3x – 3a
or, 3x – x = a + 3a
or, x = 2a
Solution is x = 2a or, x = 3a
Read more: