We will discuss here about some of the problems on quadratic equations.
1. Solve: x^2 = 36
x^2 = 36
or, x^2  36=0
or, (x + 6)(x  6) = 0
So, one of x + 6 and x  6 must be zero
From x + 6 = 0, we get x = 6
From x  6 = 0, we get x = 6
Thus, the required solutions are x = ± 6
Keeping the expression involving the unknown quantity and the constant term on left and right side respectively and finding square root from both sides, we can solve the equation also.
As in the equation x^2 = 36, finding square root from both sides, we get x = ± 6.
2. Solve 2x^2  5x + 3 = 0
2x^2  5x + 3 = 0
or 2x^2  3x – 2x + 3=0
or, x (2x  3)  1 (2x  3)=0
or, (x  1)(2x  3) = 0
Therefore, one of (x  1) and (2x  3) must be zero.
when, x  1 = 0, x = 1
and when 2x  3 = 0, x = 3/2
Thus required solutions are x = 1, 3/2
3. Solve: 3x^2  x = 10
3x^2  x = 10
or, 3x^2  x  10 = 0
or, 3x^2  6x + 5x  10 = 0
or, 3x (x  2) + 5 (x  2) =0
or, (x  2)(3x + 5) = 0
Therefore, one of x  2 and 3x + 5 must be zero
When x  2 = 0, x = 2
and when 3x + 5 = 0; 3x = 5 or; x = 5/3
Therefore, required solutions are x= 5/3, 2
4. Solve: (x  7)(x  9) = 195
(x  7)(x  9) = 195
or, x^2  9x – 7x + 63 – 195 = O
or, x2  16x  132=0
or, x^2  22 x + 6x  132=0
or, x(x  22) + 6(x  22) = 0
or, (x  22)(x + 6) = 0
Therefore, one of x  22 and x + 6 must be zero.
When x  22, x = 22
when x + 6 = 0, x =  6
Required solutions are x= 6, 22
5. Solve: x/3 +3/x = 4 1/4
or, x^{2} + 9/3x = 17/4
or, 4x^{2} + 36 = 51x
or, 4x^2  51x + 36 = 0
or, 4x^2  48x  3x + 36 = 0
or, 4x(x 12) 3(x  12) = 0
or, (x  12)(4x 3) = 0
Therefore, one of (x  12) and (4x  3) must be zero.
When x  12 = 0, x = 12 when 4x 3 = 0,x = 3/4
6. Solve: x  3/x + 3  x + 3/x  3 + 6 6/7 = 0
Assuming x  3/x + 3 = a, the given equation can be written as:
a  1/a + 6 6/7 = 0
or, a^{2}  1/a + 48/7 = 0
or, a^{2}  1/a =  48/7
or, 7a^2  7 =  48a
or, 7a^2 + 48a  7 = 0
or,7a^2 + 49a  a  7 = 0
or, 7a(a + 7)  1 (a + 7) = 0
or,(a + 7)(7a  1) = 0
Therefore, 0ne of (a + 7) and (7a  1) must be zero.
a + 7 = 0 gives a = 7 and 7a  1 = 0 gives a = 1/7
From a = 7 we get x 3/x + 3 = 7
or, x – 3 = 7x  2 1
or, 8x = 18
Therefore, x = 18/8 =  9/4
Again, from a = 1/7, we get x  3/x + 3 = 1/ 7
or, 7x  21 = x + 3
or,6x = 24
Therefore, x = 4
Required solutions are x = 9/4, 4
`Quadratic Equation
Introduction to Quadratic Equation
Formation of Quadratic Equation in One Variable
General Properties of Quadratic Equation
Methods of Solving Quadratic Equations
Examine the Roots of a Quadratic Equation
Problems on Quadratic Equations
Quadratic Equations by Factoring
Word Problems Using Quadratic Formula
Examples on Quadratic Equations
Word Problems on Quadratic Equations by Factoring
Worksheet on Formation of Quadratic Equation in One Variable
Worksheet on Quadratic Formula
Worksheet on Nature of the Roots of a Quadratic Equation
Worksheet on Word Problems on Quadratic Equations by Factoring
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