Question

(1200/x + 2) (x-10) -1200=60 solve the quadratic equation for x using factorisation

Answer 1

Answer: x = 100 or x = -60

Step-by-step explanation:

Here, the given expression is,

$(\frac{1200}{x}+2)(x-10)-1200=60$

⇒ $\frac{1200(x-10)}{x}+2(x-10)=60+1200$

⇒ $\frac{1200x-12000}{x}+2x-20=1260$

⇒ $\frac{1200x-12000+2x^2-20x}{x}=1260$

⇒ $1200x-12000+2x^2-20x=1260x$

⇒ $1180x-12000+2x^2-1260x=0$

⇒ $-80x+2x^2-12000=0$

⇒ $2x^2-80x-12000=0$

⇒ $x^2-40x-6000=0$

⇒ $x^2-100x+60x-6000=0$   ( By splitting middle term )

⇒ $x(x-100)+60(x-100)=0$

⇒ $(x+60)(x-100)=0$

Hence, x = -60 or x = 100

Answer 2

Step-by-step explanation:

Here, the given expression is,

(\frac{1200}{x}+2)(x-10)-1200=60(

x

1200

+2)(x−10)−1200=60

⇒ \frac{1200(x-10)}{x}+2(x-10)=60+1200

x

1200(x−10)

+2(x−10)=60+1200

⇒ \frac{1200x-12000}{x}+2x-20=1260

x

1200x−12000

+2x−20=1260

⇒ \frac{1200x-12000+2x^2-20x}{x}=1260

x

1200x−12000+2x

2

−20x

=1260

⇒ 1200x-12000+2x^2-20x=1260x1200x−12000+2x

2

−20x=1260x

⇒ 1180x-12000+2x^2-1260x=01180x−12000+2x

2

−1260x=0

⇒ -80x+2x^2-12000=0−80x+2x

2

−12000=0

⇒ 2x^2-80x-12000=02x

2

−80x−12000=0

⇒ x^2-40x-6000=0x

2

−40x−6000=0

⇒ x^2-100x+60x-6000=0x

2

−100x+60x−6000=0 ( By splitting middle term )

⇒ x(x-100)+60(x-100)=0x(x−100)+60(x−100)=0

⇒ (x+60)(x-100)=0(x+60)(x−100)=0

Hence, x = -60 or x = 100