Can anyone help? For what values of K does the equation (2k+1)x^2 +2x= 10x - 6 have two real and equal roots?

For the equation (2k + 1)x² + 2x = 10x - 6 to have two real and equal roots, the value of k = 5/6.

Since the equation is (2k + 1)x² + 2x = 10x - 6, we collect subtract 10x from both sides and add 6 to both sides.

So, we have (2k + 1)x² + 2x - 10x + 6 = 10x - 6 - 10x + 6

(2k + 1)x² - 8x + 6 = 0

For the equation, (2k + 1)x² + 2x = 10x - 6 to have two real and equal roots, this new equation (2k + 1)x² - 8x + 6 = 0 must also have two real and equal roots.

For the equation to have two real and equal roots, its discriminant, D = 0.

D = b² - 4ac where b = -8, a = 2k + 1 and c = 6.

So, D =  b² - 4ac

D =  (-8)² - 4 × (2k + 1) × 6 = 0

64 - 24(2k + 1) = 0

Dividing through by 8, we have

8 - 3(2k + 1) = 0

Expanding the bracket, we have

8 - 6k - 3 = 0

Collecting like terms, we have

-6k + 5 = 0

Subtracting 5 from both sides, we have

-6k = -5

Dividing through by -6, we have

k = -5/-6

k = 5/6

So, for the equation (2k + 1)x² + 2x = 10x - 6 to have two real and equal roots, the value of k = 5/6.

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