Question

Find the value of k for which each of the following quadratic equation has equal roots. x^2-4kx + ( k^2-k+2) =0
With steps..​

Answer 1

Answer:

hope it will help you:-)

Answer 2

Given :

• x² - 4kx + (k² - k+2) = 0
• Roots of the equation are real.

To find :

• Value of k.

Solution :

Compare the given quadratic equation with the general form.

General form :

• ax² + bx + c = 0

We get,

• a = 1
• b = 4k
• c = k² - k + 2

Discriminant :

→ $\mathtt{\triangle\:=\:b^2\:-\:4ac}$

→ $\mathtt{\triangle\:=\:(4k)^2\:-\:4(1)\:(k^2\:-\:k+2)}$

→ $\mathtt{\triangle\:=\:16k^2\:-\:4k^2\:+\:4k\:-\:8}$

→ $\mathtt{\triangle\:=\:12k^2\:+4k\:- \:8}$

Roots are equal.

•°• Δ = 0

→ $\mathtt{12k^2\:+4k\:-\:8\:=\:0}$

→ $\mathtt{12k^2\:+12k\:-\:8k\:-\:\:8\:=\:0}$

→ $\mathtt{12k\:(k\:+\:1)\:\:-\:8\:(k\:+\:1)\:=\:0}$

→ $\mathtt{(k+1) \:\:or\:\:(12k\:-\:8)\:=\:0}$

→ $\mathtt{k-1=0\:\:or\:\:12k\:-\:8\:=\:0}$

→ $\mathtt{k=1\:\:\:or\:\:12k\:=\:+ 8}$

→ $\mathtt{k=-1\:\:or\:\:k\:=\:{\dfrac{8}{12}}}$

$\red{\boxed{\mathtt{k=-1\:\:or\:\:k\:=\:{\dfrac{2}{3}}}}}$