If the difference of the real roots of equation x2+4 x+(p/q)=0 is 1 (where p and q are coprime), then the value of (p+q) is

Question

If the difference of the real roots of equation x2+4 x+(p/q)=0 is 1 (where p and q are coprime), then the value of (p+q) is (A) 10 (B) 15 (C) 19 (D) 39

Tardigrade Admin · Accepted Answer

Given, |α-β|2=1 ⇒(α+β)2-4 α β=1 ⇒ (15/4)=α β text So, (p/q)=(15/4) ∴ p=15, q=4 text Hence, p+q=19

Q. If the difference of the real roots of equation $x^{2} + 4 x + \frac{p}{q} = 0$ is 1 (where $p$ and $q$ are coprime), then the value of $(p + q)$ is

10

15

19

39

Given, $∣ α - β ∣^{2} = 1 \Rightarrow (α + β)^{2} - 4 α β = 1 \Rightarrow \frac{15}{4} = α β$
$So, \frac{p}{q} = \frac{15}{4}$
$∴ p = 15, q = 4$
$Hence, p + q = 19$

Q. If the difference of the real roots of equation x2+4x+qp​=0 is 1 (where p and q are coprime), then the value of (p+q) is

10

15

19

39

Given, ∣α−β∣2=1⇒(α+β)2−4αβ=1⇒415​=αβ So, qp​=415​ ∴p=15,q=4 Hence, p+q=19

Q. If the difference of the real roots of equation $x^{2} + 4 x + \frac{p}{q} = 0$ is 1 (where $p$ and $q$ are coprime), then the value of $(p + q)$ is

Given, $∣ α - β ∣^{2} = 1 \Rightarrow (α + β)^{2} - 4 α β = 1 \Rightarrow \frac{15}{4} = α β$
$So, \frac{p}{q} = \frac{15}{4}$
$∴ p = 15, q = 4$
$Hence, p + q = 19$