Question

X2 +k(4x+k-1)+2 find the value of k by which the polynomial hai equal roots

Answer 1

Answer: k=-1 or 2/3

Step-by-step explanation:

x^2 + k(4x+k-1)+2

= x^2 + 4kx + (k^2-k+2)

When the roots are equal b^2=4ac

So (4k)^2=4*1*(k^2-k+2)

16k^2=4k^2-4k+8

12k^2+4k-8=0

3k^2+k-2=0

(3k-2)(k+1)=0

So k=2/3 or -1

So the polynomial is x^2-4x+4=(x-2)^2

Or

x^2+8x/3+16/9=(x+4/3)^2

Answer 2

Answer:

The value of k by which the given polynomial has equal roots will be k=2/3 and k=-1.

Step-by-step explanation:

The given polynomial is   $x^{2}+k(4x+k-1)+2=0$

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

We know that  $D=b^{2}-4ac$

From eq. (1) we get $a=1, b=4k$   and $c=k(k-1)+2$

$D=b^{2}-4ac=(4k)^{2}-4(1)(k^{2}-k+2)$

$D=16k^{2}-4k^{2}+4k+8$

$D=12k^{2}+4k+8$

$D=4(3k^{2}+k+8)$

$D=4(3k(k+1)-2(k+1))\\ D=4(3k-2)(k+1)$

Because the given polynomial has equal roots:

∴    $D=0\\4(3k-2)(k+1)=0$

∴     $3k-2=0\\ k=\frac{2}{3}$        ,          $k+1=0\\k=-1$