Question

IF ONE SOLUTUION OF THE EQUATION 3x²=8x+2k+1 IS SEVEN TIMES THE OTHER.THEN FIND THE SOLUTIONS AND THE VALUE OF k.

Answer 1

Explanation:

3x2 - 8x - 2k-1 = 0

suppose one root is b and other is 7b

a = 3 b = 8 c = -2k-1

suppose two root are alpha and bita

alpha + bita = - b /a

a+ 7a = - (-8)/3

8a = 8/3

a = 1/3

Answer 2

$Given \: Equation \: 3x^{2} - 8x -(2k+1) =0$

$Compare \:above\:equation \: with \\ax^{2} + bx + c = 0 , we \:get$

$a = 3 , b = -8 \:and \: c = -(2k+1)$

$Let \: m, 7m \: are \: two \: roots .\:(given)$

$Sum \:of \:the \:roots = \frac{-b}{a}$

$\implies m + 7m = \frac{-(-8)}{3}$

$\implies 8m = \frac{8}{3}$

$\implies m = \frac{1}{3} \: ---(1)$

$Product \:of \:the \:roots = \frac{c}{a}$

$\implies m \times 7m = \frac{-(2k+1)}{3}$

$\implies 7m^{2} = \frac{-(2k+1)}{3}$

$\implies 21m^{2} = -(2k+1)$

$\implies 21 \times \big(\frac{1}{3}\big)^{2} = -(2k+1)$

$\implies \frac{7}{3} = -2k-1$

$\implies 2k = -1 - \frac{7}{3}$

$\implies 2k = \frac{-3-7}{3}$

$\implies k = \frac{-10}{2\times 3}$

$\implies k = \frac{-5}{3}$

Therefore.,

$One \:root (m) = \frac{1}{3} \\Second \:root(7m) = \frac{7}{3}\\Value \:of \:k = \frac{-5}{3}$

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