New Arrivals

# Exercise 1.1 (Math 10)

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

$\pmb{ (i). \;\;\;\;\; (x + 7) (x – 3) = -7}$

### Solution:

$$(x + 7) (x – 3) = -7 \;\;\;\;\; (i)$$

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

$$x^2 + 7x – 3x – 21 = -7$$

$$\Rightarrow x^2 + 4x – 21 = -7$$

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

$$x^2 + 4x – 21 + 7 = -7 + 7$$

$$\Rightarrow x^2 + 4x – 14 = 0 \;\;\;\;\; (ii)$$

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

$\pmb{ (ii). \;\;\;\;\; {x^2 + 4 \over 3} – {x \over 7} = 1}$

### Solution:

$${x^2 + 4 \over 3} – {x \over 7} = 1 \;\;\;\;\; (i)$$

By solving the expression ${x^2 + 4 \over 3} – {x \over 7}$ in equation (i), we have

$${7x^2 + 28 – 3x \over 21} = 1$$

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

$$7x^2 + 28 – 3x = 21$$

$$\Rightarrow 7x^2 – 3x + 28 = 21$$

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

$$7x^2 – 3x + 28 – 21 = 21 – 21$$

$$\Rightarrow 7x^2 – 3x + 7 = 0 \;\;\;\;\; (ii)$$

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

$\pmb{ (iii). \;\;\;\;\; {x \over x + 1} + {x + 1 \over x} = 6 }$

### Solution:

$${x \over x + 1} + {x + 1 \over x} = 6 \;\;\;\;\; (i)$$

By solving the expression ${x \over x + 1} + {x + 1 \over x}$ in equation (i), we have

$${x^2 + x^2 + 2x + 1 \over x(x + 1)} = 6$$

$$\Rightarrow {2x^2 + 2x + 1 \over x^2 + x} = 6$$

Now multiplying both sides of the above equation by $x^2 + x$, we get

$$2x^2 + 2x + 1 = 6(x^2 + x)$$

$$\Rightarrow 2x^2 + 2x + 1 = 6x^2 + 6x$$

$$\Rightarrow 6x^2 – 2x^2 + 6x – 2x – 1 = 0$$

$$\Rightarrow 4x^2 + 4x – 1 = 0 \;\;\;\;\; (ii)$$

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