Question

The roots of the quadratic equation x2-5x+3(k-1) = 0 are such

that − = . Find the value of K.
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Answer 1

Answer:

hope this helps, thank-you

Answer 2

$= abe = \frac{1}{2} \times ar(acf) \\ = \frac{ar(abe)}{ar(acf)} = \frac{ {ab}^{2} }{ {ac}^{2} } (th.6) \\ = \frac{ {ab}^{2} }{ {ab}^{2} + {bc}^{2} } \\ by \: pythagoras \: therom \\ \frac{ {ab}^{2} }{2 {ab}^{2} } \\ ab = ac \: sides \: of \: square \\ = \frac{1 \times {ab}^{2} }{2 \times {ab}^{2} } \\ = \frac{ar(abe)}{ar(acf)} = \frac{1}{2}$