Question

Find the value of k for which the roots of the following equation are real and equal-:


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



Answer it,m having boards tomorrow!​

Answer 1

$\huge\underline\bold\color{red}\mathfrak{Answer}$

The equation has equal and real roots, so the determinant is equal to zero.

b² - 4ac = 0

[(-2(2k-1)]² - 4 (k²) (4) = 0

(-4k + 2)² - 16k² = 0

16k² - 16k + 4 - 16k² = 0

16k = 4

k = 4/16

k = 1/4

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Answer 2

Answer:

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