If p - q are odd integers- then the roots of the equation 2 p x2-2 p-q x-q-0 areA- rationalB- irrationalC- non-realD- equal

The correct option is A rational2px^2+(2p+q)x+q=0 ⇒ x= (2p+q)±√((2p+q)^2 8pq)4p ⇒ x= (2p+q)±√((2p q)^2)4p = (2p+q)±(2p q)4p ∴ x= 1, q2p nbsp; From he ...

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Standard XII

Mathematics

Arithmetic Progression

If p , q are ...

Question

If $p, q$ are odd integers, then the roots of the equation
$2 p x^{2} + (2 p + q) x + q = 0$ are

rational

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irrational

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non-real

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equal

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Solution

The correct option is A rational
$2 p x^{2} + (2 p + q) x + q = 0 \Rightarrow x = \frac{- (2 p + q) \pm \sqrt{(2 p + q)^{2} - 8 p q}}{4 p} \Rightarrow x = \frac{- (2 p + q) \pm \sqrt{(2 p - q)^{2}}}{4 p} = \frac{- (2 p + q) \pm (2 p - q)}{4 p} ∴ x = - 1, - \frac{q}{2 p}$
From here, we can say that the roots are not equal but rational.

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If p,q are odd integers, then the roots of the equation 2px2+(2p+q)x+q=0 are

The correct option is A rational2px2+(2p+q)x+q=0⇒x=−(2p+q)±√(2p+q)2−8pq4p⇒x=−(2p+q)±√(2p−q)24p =−(2p+q)±(2p−q)4p∴x=−1,−q2p From here, we can say that the roots are not equal but rational.

If $p, q$ are odd integers, then the roots of the equation
$2 p x^{2} + (2 p + q) x + q = 0$ are