Question

for what calue of k will the equation (k+4)x²+(k+1)x+1=0 have real and equal roots​

Answer 1

$\large\red{ANSWER}$

$(k + 4) {x}^{2} + (k + 1)x + 1 = 0$

$</p><p>D = {b}^{2} - 4ac \\ ={ (k + 1) }^{2} - 4 \times (k + 4) \times 1 \\ = {k}^{2} + 1 + 2k - 4k + 16 \\ = {k}^{2} - 2k + 17$

$ACCORDING \: \: TO \: \: QUESTION, \\ equation \: has \: real \: and \: equal \: roots \\ </p><p>=>D=0$

$= > {k }^{2} - 2k + 17 = 0$

$D = {b}^{2} - 4ac \\ = {( - 2)}^{2} - 4 \times 1 \times 17 \\ = 4 - 68 \\ = - 64 < 0 \\ = > no \: real \: roots \: exist \: or \: imaginary \: roots$

$x = \frac{ - b + - \sqrt{D} }{2a} \\ x = \frac{ - ( - 2) + - \sqrt{ - 64} }{2 \times 1} \\ = \frac{2 + - \sqrt{ {8}^{2} {i}^{2} } }{2} \\ = \frac{2 + - 8i}{2} \\ = 1 + - 4i$

$= > x = 1 + 4i \: or \: x = 1 - 4i$

Here this is the value of k.

=> k=1+4i

or k=1-4i

where i(iorta)=√-1

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Answer 2

Answer:

Step-by-step explanation:

b=k+1

a=k+4

c=1

b^2-4ac=0 to be real and equal

(k+1)^2-4(k+4)=0

k^2+1+2k-4k-4=0

k^2-2k-3=0

k^2-3k+k-3=0

k(k-3)+1(k-3)=0

k=3,-1