Question

X square +2bx +9b -14 has only negative roots then exhaustive set of Val ue of b is​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{The\;equattion\;x^2+2bx+(9b-14)=0\;has\;only\;negative roots}$

$\underline{\textsf{To find:}}$

$\textsf{Exhaustive set of values of b}$

$\mathsf{}$

$\underline{\textsf{Solution:}}$

$\mathsf{Since\;the\;equation\;x^2+2bx+(9b-14)=0\;has\;only\;negative roots,\;we\;have}$

$\mathsf{Product\;of\;the\;roots\,>\,0}$

$\implies\mathsf{\dfrac{c}{a}\,>\,0}$

$\implies\mathsf{\dfrac{9b-14}{1}\,>\,0}$

$\implies\mathsf{9b-14\,>\,0}$

$\implies\mathsf{9b\,>\,14}$

$\implies\mathsf{b\,>\,\dfrac{14}{9}}$

$\therefore\boxed{\mathsf{b\,\in\,\left(\dfrac{14}{9},\infty\right)}}$

$\underline{\textsf{Find more:}}$

If the roots of x^2 -bx/ax-c = k-1/k+1 are numerically equal and opposite in sign then k=?​

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If x^2-bx/ax-c =m-1/m+1

has roots which are numerically equal but of opposite sign, then the value of'm'must be

(A) a-b/a+b

(B) a+b/a-b

(C) O

(D) 1

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If the roots of the quadratic equation ax2 - 6x + 3 = 0 are equal then findthe value of 2a

(i)3

(ii)6

(iii)2

(iv)4​

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