Question

find the value of m for which the equation x2+2(m-1)x+(m+5)=0 has real and equal roots​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$given \: equation................. {x}^{2} + 2(m - 1)x + (m + 5) = 0$

$\huge\mathfrak\pink{Solution :-}$

$by \: using \: determinant \: form$

$= {b}^{2} - 4ac > 0$

$= ( {2(m - 1)}^{2} - 4(1)(m + 5) > 0$

$= 4 {(m - 1)}^{2} - 4(m + 5) > 0$

$= 4( {m}^{2} + 1 - 2m) - 4m - 20 > 0$

$= 4m^{2} + 4 - 8m - 4m - 20 > 0$

$4 {m}^{2} - 12m - 16 > 0$

$= {m}^{2} - 3m - 4 > 0$

$= {m}^{2} - 4 m + m - 4 > 0$

$= m(m - 4) + 1(m - 4) > 0$

$m - 4 > 0 = m > 4$

$x \: belongs \: to \: (4. \infty )$