Question

only correct answers pls​

Answer 1

$\Large\red{ \underline{\sfᗝnswer} : }$

• Here, to find the value of k,

$2 \sqrt{2} k {x}^{2} - 4x + k \sqrt{2} = 0$

• Hence,

$k = ( - 2)$

Answer 2

$\sf\huge\bold{Answer:}$

Given :

• 2√2kx²-4x+k√2=0
• Have two equal roots
• k > 0

To find :

• Value of k

Solution :

When two roots are equal, discriminant is 0.

• General form of quadratic equation :

ax²+bx+c=0

• Comparing given equation with general form,,

a = 2√2k , b = -4, c = k√2

$\boxed{\tt\red{ D = {b}^{2} - 4ac}}$

where,

$\boxed{\tt D = 0}$

$\tt\ 0 = ( - 4) {}^{2} - 4(2 \sqrt{2} )(k \sqrt{2} )$

$\tt\ 0 = 16 - 4 \times 2 \times ( \sqrt{2 \times 2} )k$

$\tt\ 0 = 16 - 8 \times 2k$

$\tt\ 0 = 16 - 16k$

$\tt\ - 16 = - 16k$

$\tt\ k = \frac{ - 16}{ - 16}$

$\tt \: k = \cancel \frac{ - 16}{ - 16}$

$\huge \tt\underline \purple{k = 1}$