Question

गई समीकरण में k का मान ज्ञात करें जिस के लिये समीकरण के मूल बराबर हों: kx²+ 6x + 1 = 0 *


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Answer 1

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

$\mathsf{Roots\;of\;the\;equation\;kx^2+6x+1=0\;are\;equal}$

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

$\textsf{The value of k}$

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

$\mathsf{Since\;the\;roots\;of\;the\;equation\;kx^2+6x+1=0\;are\;equal,we\;have}$

$\mathsf{b^2-4ac=0}$

$\implies\mathsf{6^2-4{\times}k{\times}1=0}$

$\implies\mathsf{36-4k=0}$

$\implies\mathsf{-4k=-36}$

$\implies\mathsf{k=\dfrac{-36}{-4}}$

$\implies\boxed{\mathsf{k=9}}$

$\underline{\textsf{Answer:}}$

$\textsf{The value of k is 9}$

$\textbf{Find more:}$

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

soneone please answer this

https://brainly.in/question/9525569

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

https://brainly.in/question/16638285

Answer 2

$Compare \: given \: Quadratic \: equation \\kx^{2} + 6x + 1 = 0 \: with \: ax^{2}+bx + c = 0, \\we \:get$

$a = k , \: b = 6 \: and \: c = 1$

$\pink{ Discreminant ( \triangle ) = 0 }$

$\blue{ ( Given \: roots \: are \: equal . )}$

$\implies b^{2} - 4ac = 0$

$\implies 6^{2} - 4\times k \times 1 = 0$

$\implies -4k =- 6^{2}$

$\implies k = \frac{ -36}{-4}$

$\implies k = 9$

Therefore.,

$\red{ Value \: of \: k } \green { = 9 }$

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