Quadratic equations 2 p-1 z2-2 p-1 z-c-0 and q-1 y2-4 q-1 y-3 c-0 have the same pair of roots- Given that -0- what is the value of p-q -A- 4B- 2C- Cannot be determinedD- 3

The correct option is D 3For the equations to have same pair of roots 2p 1/q+1=2p+1/4q+1=c/3c ∴ 3(2p 1)=q+1 ⇒ 6p q=4 and, 3(2p+1)=4q+1 ⇒ 6p 4q= 2 Solving two e ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Quantitative Aptitude

Quadratic Equations

Quadratic equ...

Question

Quadratic equations $(2 p - 1) z^{2} + (2 p + 1) z + c = 0$ and $(q + 1) y^{2} + (4 q + 1) y + 3 c = 0$ have the same pair of roots. Given that $c \neq 0$ , what is the value of (p+q)?

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Cannot be determined

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 3
For the equations to have same pair of roots
$\frac{2 p - 1}{q + 1} = \frac{2 p + 1}{4 q + 1} = \frac{c}{3 c}$
$∴ 3 (2 p - 1) = q + 1 \Rightarrow 6 p - q = 4$
and, $3 (2 p + 1) = 4 q + 1 \Rightarrow 6 p - 4 q = - 2$
Solving two equation q = 2 and p =1
$∴ (p + q) = 3$

Suggest Corrections

Similar questions

Q. Let

m, n

be positive integers and the quadratic equation

4 x^{2} + m x + n = 0

has two distinct real roots

p

and

q

(p \leq q)

. Also, the quadratic equations

x^{2} - p x + 2 q = 0

and

x^{2} - q x + 2 p = 0

have a common root say

α

. If

p

and

q

are rational, then uncommon root of the equation

x^{2} - p x + 2 q = 0

and

x^{2} - q x + 2 p = 0

is equal to

Q. If

p \in Q

and the quadratic equations

x^{2} - 4 x + 1 = 0

and

p x^{2} - (p^{2} + 3) x + 2 p^{2} - p = 0

have a root in common then the value (s) of p are

Q. If

p, q, r

are real and

p \neq q

, then show that the roots of the quadratic equation

(p - q) x^{2} + 5 (p + q) x - 2 (p - q) = 0

are real and unequal.

Q. Find the value of p such that quadratic equation.

(p - 12) x^{2} - 2 (p - 12) x + 2 = 0

has equal roots.

Q. If the roots of the equation

{p x}^{2} + q x + r = 0

, where

2 p, q, 2 r

are in

G . P

are of the

α^{2}, 4 α - 4

than the value of

2 p + 4 q + 7 r

is :

Join BYJU'S Learning Program

Quadratic equations (2p−1)z2+(2p+1)z+c=0 and (q+1)y2+(4q+1)y+3c=0 have the same pair of roots. Given that c≠0, what is the value of (p+q)?

The correct option is A 3For the equations to have same pair of roots 2p−1q+1=2p+14q+1=c3c ∴3(2p−1)=q+1⇒6p−q=4 and, 3(2p+1)=4q+1⇒6p−4q=−2 Solving two equation q = 2 and p =1 ∴(p+q)=3

Quadratic equations $(2 p - 1) z^{2} + (2 p + 1) z + c = 0$ and $(q + 1) y^{2} + (4 q + 1) y + 3 c = 0$ have the same pair of roots. Given that $c \neq 0$ , what is the value of (p+q)?

The correct option is A 3
For the equations to have same pair of roots
$\frac{2 p - 1}{q + 1} = \frac{2 p + 1}{4 q + 1} = \frac{c}{3 c}$
$∴ 3 (2 p - 1) = q + 1 \Rightarrow 6 p - q = 4$
and, $3 (2 p + 1) = 4 q + 1 \Rightarrow 6 p - 4 q = - 2$
Solving two equation q = 2 and p =1
$∴ (p + q) = 3$