Question

For what values of k will the equation y^2-15=k(2y-8) have equal roots?

Answer 1

hey, mate here is ur Ans

for any quadratic equation having equal roots Decrement must be =0

y²-2ky+8k-15=0

D=√{(2k)²-4(1)(8k-15)}

D=0

4k²-32k+60=0

k²-8k+15=0
k=5 or k=3

Answer 2

${y}^{2} - 15 = k(2y - 8) \\ = > {y}^{2} - 15 = 2ky - 8k \\ = > {y}^{2} - 2ky - 15 + 8k = 0 \\ \\ \\ a = 1 \: b = - 2k \: c = 8k - 15$

For equal roots,
${b}^{2} - 4ac = 0 \\ = > {( - 2k)}^{2} - 4 \times 1 \times (8k - 15) = 0 \\ = > 4 {k}^{2} - 32k + 60 = 0 \\ = > 4 {k}^{2} - (12 + 20)k + 60 = 0 \\ = > 4 {k}^{2} - 12k - 20k + 60 = 0 \\ = > 4k(k - 3) - 20(k - 3) \\ = > (k - 3)(4k - 20)$

$\therefore \: k - 3 = 0 \\ = > k = 3 \\ and \\ 4k - 20 = 0 \\ = > 4k = 20 \\ = > k = \frac{20}{4} \\ = > k = 5$

Hope it'll help...