Let the quadratic equation x 2+3 x - k =0 has roots a , b and x 2+3 x -10=0 has roots c , d such that modulus of difference of the roots of the first equation is equal to twice the modulus of the difference of the roots of the second equation. If the value of ' k ' can be expressed as rational number in the lowest form as m / n then find the value of (m+n).

Question

Let the quadratic equation x 2+3 x - k =0 has roots a , b and x 2+3 x -10=0 has roots c , d such that modulus of difference of the roots of the first equation is equal to twice the modulus of the difference of the roots of the second equation. If the value of ' k ' can be expressed as rational number in the lowest form as m / n then find the value of (m+n). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/4bdb37bf4521fae3c41fcffdfb4ba3b1-.png" /> text given | a - b |=2| c - d | (a-b)2=4(c-d)2 (a+b)2-4 a b=4[(c+d)2-4 c d] 9+4 k =4[9+40] 4 k=27+160=187 ⇒ k=(187/4) ⇒ m+n=191

given ∣a−b∣=2∣c−d∣ (a−b)2=4(c−d)2 (a+b)2−4ab=4[(c+d)2−4cd] 9+4k=4[9+40] 4k=27+160=187⇒k=4187​ ⇒m+n=191

$given ∣ a - b ∣ = 2∣ c - d ∣$
$(a - b)^{2} = 4 (c - d)^{2}$
$(a + b)^{2} - 4 ab = 4 [(c + d)^{2} - 4 c d]$
$9 + 4 k = 4 [9 + 40]$
$4 k = 27 + 160 = 187 \Rightarrow k = \frac{187}{4}$
$\Rightarrow m + n = 191$