If one root of the quadratic equation px 2+ qx + r =0( p ≠ 0) is a surd of the form (√ a /√ a +√ a - b ) where p , q , r ∈ Q and a , b are positive integers which are not perfect square of integer, then the other root is

Question

If one root of the quadratic equation px 2+ qx + r =0( p ≠ 0) is a surd of the form (√ a /√ a +√ a - b ) where p , q , r ∈ Q and a , b are positive integers which are not perfect square of integer, then the other root is (A)  (√a/√a-√a-b) (B) a+(√a(a-b)/b) (C) (a+√a(a-b)/b) (D) (√a-√a-b/√b)

Tardigrade Admin · Accepted Answer

α=(√a/√a+√a-b)=(√a(√a-√a-b)/a-(a-b))=(a-√a(a-b)/b) Conjugate of α is (a+√a(a-b)/b) ⇒(C) =(a2-(a(a-b))/b(a-√a(a-b)))=(a/a-√a(a-b))=(√a/√a-√a-b) ⇒ (A)

Q. If one root of the quadratic equation $p x^{2} + q x + r = 0 (p \neq = 0)$ is a surd of the form $\frac{a}{a + a - b}$ where $p, q, r \in Q$ and $a, b$ are positive integers which are not perfect square of integer, then the other root is

$\frac{a}{a - a - b}$

$a + \frac{a ( a - b )}{b}$

$\frac{a + a ( a - b )}{b}$

$\frac{a - a - b}{b}$

$α = \frac{a}{a + a - b} = \frac{a ( a - a - b )}{a - ( a - b )} = \frac{a - a ( a - b )}{b}$
Conjugate of $α$ is $\frac{a + a ( a - b )}{b} \Rightarrow (C)$
$= \frac{a ^{2} - ( a ( a - b ))}{b ( a - a ( a - b ) )} = \frac{a}{a - a ( a - b )} = \frac{a}{a - a - b} \Rightarrow (A)$

Q. If one root of the quadratic equation px2+qx+r=0(p=0) is a surd of the form a​+a−b​a​​ where p,q,r∈Q and a,b are positive integers which are not perfect square of integer, then the other root is

a​−a−b​a​​

a+ba(a−b)​​

ba+a(a−b)​​

b​a​−a−b​​