Question

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The value of k = 0.5.

Step-by-step explanation:

Given: Quadratic Equation, $2kx^2-2(1+2k)x+(3+2k)=0$

           Roots are equal.

To find: Value of k.

To find value of k we use Discriminant condition of equal roots.

Discriminant, D = 0 for equal roots

⇒ $\sqrt{b^2-4ac}=0$ ......(1)

where, a = Coefficient of $x^2$ term

           b = Coefficient of x term

          c = Constant term

From given Quadratic equation we get,

a = 2k  , b = -2( 1+2k )   , c = 3+2k

Putting these value in eqn 1 , we get

$\sqrt{(-2(1+2k))^2-4\times2k\times(3+2k)} = 0$

$\sqrt{(-2)^2(1+2k))^2-8k\times(3+2k)} = 0$

Square Both sides of equation to remove square root, we get

$(-2)^2(1+2k)^2-8k\times(3+2k) = 0$

$4\times(1^2+(2k)^2+2\times2k\times1)-8k\times3-8k\times2k = 0$

$4\times(1+4k^2+4k)-24k-16k^2 = 0$

$4+16k^2+16k-24k-16k^2 = 0$

$4+16k^2-16k^2+16k-24k = 0$

$4+(16-16)k^2+(16-24)k = 0$

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

$4+-8k = 0$

$4 = 8k$

$k = \frac{4}{8}$

$k = \frac{1}{2}$

k = 0.5

Therefore, The value of k = 0.5.