Question

Determine the value of'm' so that the equation has real and equal roots. x^2-4x+(m-4)=0​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{Roots\;of\;x^2-4x+(m-4)=0\;are\;real\;and\;equal}$

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

$\textsf{The value of 'm'}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{x^2-4x+(m-4)=0}$

$\mathsf{Comparing\;with\;ax^2+bx+c=0\;we\;get}$

$\mathsf{a=1,\;b=-4,\;c=m-4}$

$\textsf{Since the roots are equal, we have}$

$\;\;\;\;\;\;\;\;\;\;\;\bf\;b^2-4ac=0$

$\implies\mathsf{(-4)^2-4(1)(m-4)=0}$

$\implies\mathsf{16-4(m-4)=0}$

$\implies\mathsf{-4(m-4)=-16}$

$\implies\mathsf{m-4=\dfrac{-16}{-4}}$

$\implies\mathsf{m-4=4}$

$\implies\mathsf{m=4+4}$

$\implies\boxed{\bf\,m=8}$

$\underline{\textbf{Answer:}}$

$\textbf{The value of m is 8}$

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Answer 2

Answer:

Ans => m = 8

Step-by-step explanation:

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