Question

Find the value of p, for which one root of the quadratic equation px²-14x+8=0 is 6 time the other. By quadratic method

Answer 1

$p {x}^{2} - 14x + 8 = 0$

roots are

$( - ( - 14) + \sqrt{ {14}^{2} - (4 \times p \times 8}) )÷2p\: and \\ (- ( - 14) - \sqrt{ {14}^{2} - (4 \times p \times 8} ))$÷2p

$(14 + \sqrt{196 - 32p} )÷2p\: and \\ (14 - \sqrt{196 - 32p} )$÷2p

$14 + \sqrt{196 - 32p} = 6 \times (14 - \sqrt{196 - 32p}$
2p cancelled from both sides so p can not be zero

multiplying both sides by

$14 + \sqrt{196 - 32p}$

$196 + 196 - 32p + 28 \sqrt{196 - 32p} = 6 \times (196 - 196 + 32p) \\ 28 \sqrt{196 - 32p} = 224p \: - 392$

$\sqrt{196 - 32p} = 8p - 14$

squaring both sides

$196 - 32p = 64 {p}^{2} + 196 - 224p \\ 64 {p}^{2} - 192p = 0 \\ {p}^{2} - 3p = 0 \\ p(p - 3) = 0 \\ p = 0 \: and \: p = 3$
p = 0 not possible so p = 3