Question

kx2+kx+1=-4x2-x when D =0(Real roots).

Answer 1

Answer:

hey frnd here is your answer

Step-by-step explanation:

• kx²+kx+1=4x²-x

• kx²-4x²+kx+x+1=0

• (k-4)x²+(k+1)x+1=0.

• Now D=0 (D=b²-4ac).

• b²-4ac=0

• (k+1)²-4(k-4)(1)=0

• (k²+2k+1)-4k+16=0

• k²+2k+1-4k+16=0

• k²-2k+17=0.

so your final answer is-

By solving we get k²-2k+17=0

Note- Here further k²-2k+17 can't be solved bcoz it has imaginary roots.

i hope you get the right answer

thanks for asking

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Answer 2

Step-by-step explanation:

The given equation kx2 + kx + 1 = -4x2 – x

This can be rewritten as,

(k + 4)x2 + (k + 1)x + 1 = 0

Now, this in the form of ax2 + bx + c = 0

Where a = (k +4), b = (k + 1), c = 1

For the equation to have real and equal roots, the condition is

D = b2 – 4ac = 0

⇒ (k + 1)2 – 4(k +4)(1) = 0

⇒ (k +1)2 – 4k – 16 = 0

⇒ k2 + 2k + 1 – 4k – 16 = 0

⇒ k2 – 2k – 15 = 0

Now, solving for k by factorization we have

⇒ k2 – 5k + 3k – 15 = 0

⇒ k(k – 5) + 3(k – 5) = 0

⇒ (k + 3)(k – 5) = 0,

k = -3 and k = 5,

So, the value of k can either be -3 or 5.

• I hope it's help you...☺