Question

Find the value of k for which the equation (k+1)x²-2kx-2x-1=0 has real and equal roots.

Answer 1

Answer:

k=-1 or-2

Step-by-step explanation:

For equal and real roots, D=0

b^2-4ac=0

a= (k+1)     b= -2(k+1)    c= -1

4k^2 +8k+ 4 - 4(k+1)(-1)=0

4k^2+8k+4+4k+4=0

4k^2 + 12k +8=0

K^2 + 3k+2=0

k^2 +k+2k+2=0

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

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

Answer 2

The values of k are -2 and -1.

Step-by-step explanation:

Given : The quadratic equation $(k+1)x^2-2kx-2x-1=0$ has real roots.

To find : The values of k ?

Solution :

When the quadratic equation $ax^2+bx+c$ has real roots then discriminant is zero.

i.e. $D=b^2-4ac=0$

Write the equation as  $(k+1)x^2-2x(k+1)-1=0$

Here, a=k+1, b=-2(k+1) and c=-1

Substitute the values,

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

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

$4k^2+4+8k+4k+4=0$

$4k^2+12k+8=0$

$k^2+3k+2=0$

$k^2+k+2k+2=0$

$k(k+1 )+ 2 (k+1)=0$

$(k+2)(k+1)=0$

$k=-2,-1$

Therefore, the values of k are -2 and -1.

#Learn more

Which of the following has two distinct real roots? (a) 2x²-3√2x+9/4=0 (b) x²+x-5=0 (c) x²+3x+2√2=0 (d) 5x²-3x+1=0

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