Question

The value of K for which (x+1)+4x²+x+1=0 has real and equcal roots

Answer 1

$\bold{Rewrite \: your \: equation:}$

$\bold{We \: have:}$

$\rightarrow \: (k+1)x2+2(1−k)x+1=(ax+b)2$

$\bold{and \: we \: are \: asked \: to \: find \: k. So:}$

$\rightarrow \: a2=k+1a2=k+1$

$\rightarrow \: 2ab=2(1−k)2ab=2(1−k)$

$\rightarrow \: b2=1,b=±1b2=1,b=±1$

$\rightarrow \: ±a=(1−k)±a=(1−k)$

$\rightarrow \: a2=(1−k)2=k+1a2=(1−k)2=k+1$

Solving (1−k)2=k+1(1−k)2=k+1 has

$\diamond \: k2−2k=kk2−2k=k ,$

$\diamond \: k2=3kk2=3k ,$

$\diamond \: k=0k=0 ,$

$\diamond \: k=3k=3$

Plugging in k=0k=0 ,

we have x2+2x+1=(x+1)2x2+2x+1=(x+1)2 .

Plugging in k=3k=3 , we have 4x2−4x+1=(1−2x)2

So the solutions are

$\boxed{k=0,k=3}$

Answer 2

$\huge{ \underline{ \bold{ᴀɴsᴡᴇʀ....{ \heartsuit}}}}$

Rewriteyourequation:

\bold{We \: have:}Wehave:

\rightarrow \: (k+1)x2+2(1−k)x+1=(ax+b)2→(k+1)x2+2(1−k)x+1=(ax+b)2

\bold{and \: we \: are \: asked \: to \: find \: k. So:}andweareaskedtofindk.So:

\rightarrow \: a2=k+1a2=k+1→a2=k+1a2=k+1

\rightarrow \: 2ab=2(1−k)2ab=2(1−k)→2ab=2(1−k)2ab=2(1−k)

\rightarrow \: b2=1,b=±1b2=1,b=±1→b2=1,b=±1b2=1,b=±1

\rightarrow \: ±a=(1−k)±a=(1−k)→±a=(1−k)±a=(1−k)

\rightarrow \: a2=(1−k)2=k+1a2=(1−k)2=k+1→a2=(1−k)2=k+1a2=(1−k)2=k+1

Solving (1−k)2=k+1(1−k)2=k+1 has

\diamond \: k2−2k=kk2−2k=k ,⋄k2−2k=kk2−2k=k,

\diamond \: k2=3kk2=3k ,⋄k2=3kk2=3k,

\diamond \: k=0k=0 ,⋄k=0k=0,

\diamond \: k=3k=3⋄k=3k=3

Plugging in k=0k=0 ,

we have x2+2x+1=(x+1)2x2+2x+1=(x+1)2 .

Plugging in k=3k=3 , we have 4x2−4x+1=(1−2x)2

So the solutions are

\boxed{k=0,k=3}

k=0,k=3