Problems on Quadratic Equations

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


Example 6: Solve: x/3 +3/x =  4 1/4

or,x2 + 9/3x = 17/4

or, 4x2 + 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


7. 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, a2 - 1/a + 48/7 = 0

or, a2 - 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



9 Grade Math

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