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]
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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]
A similar question is found at https://brainly.com/question/12144265
