Exercise 1.1 (Math 10)

Question 1.   Write the following quadratic equations in the standard form and point out pure quadratic equations.

(i).          (x + 7)(x − 3) = −7

Solution:

The given quadratic equation is

        (x + 7)(x − 3) = −7          (i)

Multiplying the expressions (x + 7) and (x − 3) in equation (i) with each other, we have

        x² + 7x − 3x − 21 = − 7

⇒     x² + 4x − 21 = − 7

Now adding 7 on both sides of the above equation and get

        x² + 4x − 21 + 7 = − 7 + 7

⇒     x² + 4x − 14 = 0          (ii)

Equation (ii) is in the standard form of the quadratic equation but it is not a pure quadratic equation because b ≠ 0.

 

(ii).          (x² + 4)/3  −  x/7 = 1

Solution:

The given quadratic equation is

        (x² + 4)/3  −  x/7 = 1          (i)

By solving the expression (x² + 4)/3  −  x/7  in equation (i), we have

        (7x² + 28 − 3x)/21 = 1

Now multiplying both sides of the above equation by 21, we get

        7x² + 28 − 3x = 21

⇒     7x² − 3x + 28 = 21

Now subtracting 21 from both sides of the above equation and get

        7x² − 3x + 28 − 21 = 21 − 21

⇒     7x² − 3x + 7 = 0          (ii)

Equation (ii) is in the standard form of the quadratic equation but it is not a pure quadratic equation because b ≠ 0.

 

(iii).         x/(x + 1)  +  (x + 1)/x = 6

Solution:

The given quadratic equation is

        x/(x + 1)  +  (x + 1)/x = 6          (i)

By solving the expression  x/(x + 1)  +  (x + 1)/x  in equation (i), we have

         x² + x² + 2x + 1    /    x(x + 1)  = 6

⇒     2x² + 2x + 1    /    x² + x  = 6

Now multiplying both sides of the above equation by x² + x, we get

       2x² + 2x + 1 = 6(x² + x)

⇒     2x² + 2x + 1 = 6x² + 6x

⇒     6x² − 2x² + 6x − 2x − 1 = 0

⇒     4x² + 4x − 1 = 0          (ii)

Equation (ii) is in the standard form of the quadratic equation but it is not a pure quadratic equation because b ≠ 0.