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

The equation has two real and equal roots for [tex]k = \frac{5}{6}[/tex]

In this question, we use the concept of the solution of a quadratic equation to solve it, considering that a quadratic equation in the format:

[tex]ax^2 + bx + c = 0[/tex]

has two equal solutions if [tex]\Delta = b^2 - 4ac[/tex] is 0.

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In this question:

The equation is:

[tex](2k+1)x^2 + 2x = 10x - 6[/tex]

Placing in the correct format:

[tex](2k+1)x^2 + 2x - 10x + 6 = 0[/tex]

[tex](2k+1)x^2 - 8x + 6 = 0[/tex]

Thus, the coefficients are: [tex]a = 2k + 1, b = -8, c = 6[/tex]

------------------------------------

Delta:

We want it to be positive, so:

[tex]\Delta = b^2 - 4ac[/tex]

[tex]\Delta = 0[/tex]

[tex]b^2 - 4ac = 0[/tex]

[tex](-8)^2 - 4(2k+1)(6) = 0[/tex]

[tex]64 - 48k - 24 = 0[/tex]

[tex]-48k + 40 = 0[/tex]

[tex]-48k = -40[/tex]

[tex]48k = 40[/tex]

[tex]k = \frac{40}{48}[/tex]

[tex]k = \frac{5}{6}[/tex]

The equation has two real and equal roots for [tex]k = \frac{5}{6}[/tex]

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