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(A) It is given that $p^2x^2 + (p^2 - q^2)x - q^2 = 0; (p \neq 0)$
(i) Show that the discriminant (D) of above equation is a perfect square.
(ii) Find the roots of the equation.
OR
(B) Three consecutive positive integers are such that the sum of the square of smallest and product of other two is 67. Find the numbers, using quadratic equation.
(i) Show that the discriminant (D) of above equation is a perfect square.
(ii) Find the roots of the equation.
OR
(B) Three consecutive positive integers are such that the sum of the square of smallest and product of other two is 67. Find the numbers, using quadratic equation.
Show SolutionHide Solution↓
(A) (i) Discriminant $= (p^2 - q^2)^2 + 4p^2q^2 = (p^2 + q^2)^2$ [$1 + 1$ mark]
(ii) $\therefore x = \frac{-(p^2 - q^2) \pm \sqrt{(p^2 + q^2)^2}}{2p^2}$ [$1$ mark]
$= \frac{q^2}{p^2}, -1$ [$1 + 1$ mark]
OR
(B) Let the three consecutive positive integers be $x, x + 1, x + 2$ [$1$ mark]
A.T.Q. $x^2 + (x + 1) (x + 2) = 67$ [$1$ mark]
$\Rightarrow 2x^2 + 3x - 65 = 0$ [$1$ mark]
$\Rightarrow (2x + 13) (x - 5) = 0$ [$1$ mark]
$\Rightarrow x = 5, x = -\frac{13}{2}$ (rejected)
So the three consecutive positive integers are 5, 6 and 7 [$1$ mark]
(ii) $\therefore x = \frac{-(p^2 - q^2) \pm \sqrt{(p^2 + q^2)^2}}{2p^2}$ [$1$ mark]
$= \frac{q^2}{p^2}, -1$ [$1 + 1$ mark]
OR
(B) Let the three consecutive positive integers be $x, x + 1, x + 2$ [$1$ mark]
A.T.Q. $x^2 + (x + 1) (x + 2) = 67$ [$1$ mark]
$\Rightarrow 2x^2 + 3x - 65 = 0$ [$1$ mark]
$\Rightarrow (2x + 13) (x - 5) = 0$ [$1$ mark]
$\Rightarrow x = 5, x = -\frac{13}{2}$ (rejected)
So the three consecutive positive integers are 5, 6 and 7 [$1$ mark]