If difference of roots of the equation x2-p x+q=0 is 1 , then p2+4 q2 equals -

Question

If difference of roots of the equation x2-p x+q=0 is 1 , then p2+4 q2 equals - (A) 2 q+3 (B) (1-2 q)2 (C) (1+2 q)2 (D) 2 q-3

Tardigrade Admin · Accepted Answer

x2-p x+q= <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/5747cbc3f897d4dabbbced89e2d0f0e9-.png" /> α-β=1 (α+β)2-4 α β=(1)2 p2-4 q=1........(i) ⇒ from (i) p2=1+4 q 1+4 q+4 q2=(2 q+1)2

Q. If difference of roots of the equation $x^{2} - p x + q = 0$ is 1 , then $p^{2} + 4 q^{2}$ equals -

$2 q + 3$

$(1 - 2 q)^{2}$

$(1 + 2 q)^{2}$

$2 q - 3$

$x^{2} - p x + q =$ $α - β = 1$
$(α + β)^{2} - 4 α β = (1)^{2}$
$p^{2} - 4 q = 1$ ........(i)
$\Rightarrow$ from (i)
$p^{2} = 1 + 4 q$
$1 + 4 q + 4 q^{2} = (2 q + 1)^{2}$

Q. If difference of roots of the equation x2−px+q=0 is 1 , then p2+4q2 equals -

2q+3

(1−2q)2

(1+2q)2

2q−3