New Arrivals

# 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:

(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:

(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:

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.