Question

Find the value of c such that the equation 4x^2-2(c+1)x+(c+4)=0 has real and equal roots.

Answer 1

$\huge\boxed{\texttt{\fcolorbox{Red}{aqua}{Hey Mate!!!}}}$

Here
a = 4
b = -2(c +1)
c=( c+4)
If the equation has real and equal roots, than
$b {}^{2} - 4ac = 0 \\ put \: the \: value \: \: we \: get \\( - 2( c + 1) ){}^{2} - 4 \times 4 \times (c + 4) = 0 \\ 4(c {}^{2} + 1 + 2c) - 16c - 64 = 0 \\ 4c {}^{2} + 4 + 8c - 16c - 64 = 0 \\ 4c {}^{2} - 8c - 60 = 0 \\ 4c {}^{2} - 20c + 12c - 60 = 0 \\ 4c(c - 5) + 12(c - 5) = 0 \\ (c - 5)(4c + 12) = 0 \\ c - 5 = 0 \: \: or \: \: \: 4c + 12 = 0 \\ c = 5 \: \: \: or \: \: \: c = \frac{ - 12}{4} \\ c = 5 \: \: \: \: or \: \: \: - 3$
$\large{\red{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\underline{\underline{Hope\:it\:helps\: you}}}}}}}}}}}}}}$

Answer 2

Hey mate ^_^

Check this attachment

#Be Brainly❤️