96
(A) Express $\frac{24}{18-x} - \frac{24}{18+x} = 1$ as a quadratic equation in standard form and find the discriminant of the quadratic equation, so obtained. Also, find the roots of the equation.
OR
(B) The sum of squares of two positive numbers is $100$. If one number exceeds the other by $2$, find the numbers.
OR
(B) The sum of squares of two positive numbers is $100$. If one number exceeds the other by $2$, find the numbers.
Show SolutionHide Solution↓
(A) Given equation can be written as
$24(18 + x) – 24(18 – x) = 324 – x^2$ (1 Mark)
i.e., $x^2 + 48x – 324 = 0$ (1 Mark)
$D = 48^2 – 4(-324) = 3600$ (1 Mark)
Roots are $\frac{-48 \pm 60}{2}$ (1 Mark)
i.e., $6, -54$ (1 Mark)
OR
(B) Let the numbers be $x, x + 2$ (1/2 Mark)
$x^2 + (x + 2)^2 = 100$ (1 1/2 Mark)
simplifying we get
$2x^2 + 4x-96 = 0$ or $x^2 + 2x - 48 = 0$ (1 Mark)
which gives $(x + 8) (x – 6) = 0$ (1 Mark)
$x = 6, - 8$ (1/2 Mark)
As $x > 0$ thus, numbers are $6, 8$ (1/2 Mark)
$24(18 + x) – 24(18 – x) = 324 – x^2$ (1 Mark)
i.e., $x^2 + 48x – 324 = 0$ (1 Mark)
$D = 48^2 – 4(-324) = 3600$ (1 Mark)
Roots are $\frac{-48 \pm 60}{2}$ (1 Mark)
i.e., $6, -54$ (1 Mark)
OR
(B) Let the numbers be $x, x + 2$ (1/2 Mark)
$x^2 + (x + 2)^2 = 100$ (1 1/2 Mark)
simplifying we get
$2x^2 + 4x-96 = 0$ or $x^2 + 2x - 48 = 0$ (1 Mark)
which gives $(x + 8) (x – 6) = 0$ (1 Mark)
$x = 6, - 8$ (1/2 Mark)
As $x > 0$ thus, numbers are $6, 8$ (1/2 Mark)