72
(a) The sum of areas of two squares is $2650 \text{ cm}^2$. If the sum of their perimeters is $280 \text{ cm}$, find the sides of the two given squares.
OR
(b) Express the equation $\frac{1}{x} - \frac{1}{x - 2} = 3, (x \neq 0, 2)$ as a quadratic equation in standard form. Hence, find the roots of the quadratic equation so obtained.
OR
(b) Express the equation $\frac{1}{x} - \frac{1}{x - 2} = 3, (x \neq 0, 2)$ as a quadratic equation in standard form. Hence, find the roots of the quadratic equation so obtained.
Show SolutionHide Solution↓
Solution: (a) Let the sides of squares be $x$ and $y$
A.T.Q.
$x^2 + y^2 = 2650$ --------(i) [1 mark]
$4x + 4y = 280 \Rightarrow x + y = 70$ ---(ii) [1 mark]
getting $2x^2 - 140x + 2250 = 0$ or $x^2 - 70x + 1125 = 0$ [1 mark]
$\Rightarrow (x - 25)(x - 45) = 0$
$\Rightarrow x = 25$ and $x = 45$ [1 mark]
$\therefore y = 45$ and $y = 25$
sides of square are $25 \text{ cm}$ and $45 \text{ cm}$. [1 mark]
OR
(b) $\frac{x - 2 - x}{x (x - 2)} = 3$ [1 mark]
$\Rightarrow 3x^2 - 6x + 2 = 0$ [1 mark]
$\text{Discriminant} = 36 - 24 = 12$ [1 mark]
$\text{Roots are } \frac{6 + \sqrt{12}}{6} \text{ and } \frac{6 - \sqrt{12}}{6}$
$\text{or } 1 + \frac{\sqrt{3}}{3} \text{ and } 1 - \frac{\sqrt{3}}{3}$ [1+1 marks]
A.T.Q.
$x^2 + y^2 = 2650$ --------(i) [1 mark]
$4x + 4y = 280 \Rightarrow x + y = 70$ ---(ii) [1 mark]
getting $2x^2 - 140x + 2250 = 0$ or $x^2 - 70x + 1125 = 0$ [1 mark]
$\Rightarrow (x - 25)(x - 45) = 0$
$\Rightarrow x = 25$ and $x = 45$ [1 mark]
$\therefore y = 45$ and $y = 25$
sides of square are $25 \text{ cm}$ and $45 \text{ cm}$. [1 mark]
OR
(b) $\frac{x - 2 - x}{x (x - 2)} = 3$ [1 mark]
$\Rightarrow 3x^2 - 6x + 2 = 0$ [1 mark]
$\text{Discriminant} = 36 - 24 = 12$ [1 mark]
$\text{Roots are } \frac{6 + \sqrt{12}}{6} \text{ and } \frac{6 - \sqrt{12}}{6}$
$\text{or } 1 + \frac{\sqrt{3}}{3} \text{ and } 1 - \frac{\sqrt{3}}{3}$ [1+1 marks]