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:

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

$$ x^2 + 7x – 3x – 21 = -7 $$

$$ \Rightarrow x^2 + 4x – 21 = -7 $$

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:

The given quadratic equation is

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

Taking LCM (Least Common Multiple) of 3 and 7 that is 21, now multiply both sides of the equation (i) by 21, we have

$$ 21({x^2 + 4 \over 3}) – 21({x \over 7}) = 21(1) $$

$$ \Rightarrow 7(x^2 + 4) – 3x = 21 $$

$$ \Rightarrow 7x^2 + 28 – 3x = 21 $$

$$ \Rightarrow 7x^2 – 3x + 28 = 21 $$

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:

The given quadratic equation is

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

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.

$ \pmb{ (iv). \;\;\;\;\; {x + 4 \over x – 2} – {x – 2 \over x} + 4 = 0 } $

Solution:

The given quadratic equation is

$$ {x + 4 \over x – 2} – {x – 2 \over x} + 4 = 0 \;\;\;\;\; (i) $$

Multiplying both sides of equation (i) by $ {(x – 2) (x)} $ (i.e. LCM of  (x – 2) and x), we get

$$ (x – 2) (x){x + 4 \over x – 2} – (x – 2) (x){x – 2 \over x} + 4(x – 2) (x) = 0(x – 2) (x) $$

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

$$ \Rightarrow x^2 + 4x – (x^2 – 4x + 4) + 4x^2 – 8x = 0 $$

$$ \Rightarrow x^2 + 4x – x^2 + 4x – 4 + 4x^2 – 8x = 0 $$

$$ \Rightarrow 4x^2 – 4 = 0 $$

Taking 4 common in above equation, we have

$$ x^2 – 1 = 0 \;\;\;\;\; (ii) $$

Equation (ii) is in the standard form of the quadratic equation and it is also a pure quadratic equation because b = 0.

$ \pmb{ (v). \;\;\;\;\; {x + 3 \over x + 4} – {x – 5 \over x} = 1 } $

Solution:

The given quadratic equation is

$$ {x + 3 \over x + 4} – {x – 5 \over x} = 1 \;\;\;\;\; (i) $$

Multiplying both sides of equation (i) by $ {(x + 4) (x)} $ (i.e. LCM of  (x + 4) and x), we get

$$ (x + 4) (x){x + 3 \over x + 4} – (x + 4) (x){x – 5 \over x} = 1(x + 4) (x) $$

$$ \Rightarrow (x) (x + 3) – (x + 4) (x – 5) = x^2 + 4x $$

$$ \Rightarrow x^2 + 3x – (x^2 + 4x -5x – 20) = x^2 + 4x $$

$$ \Rightarrow x^2 + 3x – (x^2 – x – 20) = x^2 + 4x $$

$$ \Rightarrow x^2 + 3x – x^2 + x + 20 = x^2 + 4x $$

$$ \Rightarrow 4x + 20 = x^2 + 4x $$

$$ \Rightarrow x^2 + 4x – 4x – 20 = 0 $$

$$ \Rightarrow x^2 – 20 = 0 \;\;\;\;\; (ii) $$

Equation (ii) is in the standard form of the quadratic equation and it is also a pure quadratic equation because b = 0.

$ \pmb{ (vi). \;\;\;\;\; {x + 1 \over x + 2} + {x + 2 \over x+ 3} = {25 \over 12} } $

Solution:

The given quadratic equation is

$$ {x + 1 \over x + 2} + {x + 2 \over x+ 3} = {25 \over 12} \;\;\;\;\; (i) $$

Multiplying both sides of equation (i) by $ {(x + 2) (x + 3) (12)} $ (i.e. LCM of  (x + 2), (x + 3) and 12, we get

$$ (x + 2) (x + 3) (12){x + 1 \over x + 2} + (x + 2) (x + 3) (12){x + 2 \over x+ 3} = (x + 2) (x + 3) (12){25 \over 12} $$

$$ \Rightarrow (x + 3) (12) (x + 1) + (x + 2) (12) (x+ 2) = (x + 2) (x + 3) (25) $$

$$ \Rightarrow (12) (x + 3) (x + 1) + (12) (x + 2) (x+ 2) = (25) (x + 2) (x + 3) $$

$$ \Rightarrow (12) (x^2 + 3x + x + 3) + (12) (x^2 + 2x + 2x + 4) = (25) (x^2 + 2x + 3x + 6) $$

$$ \Rightarrow (12) (x^2 + 4x + 3) + (12) (x^2 + 4x + 4) = (25) (x^2 + 5x + 6) $$

$$ \Rightarrow 12x^2 + 48x + 36 + 12x^2 + 48x + 48 = 25x^2 + 125x + 150 $$

By simplifying the above equation, we have

$$ \Rightarrow 24x^2 + 96x + 84 = 25x^2 + 125x + 150 $$

$$ \Rightarrow 24x^2 + 96x + 84 – 25x^2 – 125x – 150 = 0 $$

$$ \Rightarrow – x^2 – 29x – 66 = 0 $$

By taking (-) common, we get

$$ \Rightarrow x^2 + 29x + 66 = 0 \;\;\;\;\; (ii) $$

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