Question

please answer this question ............. Find the quadratic equation whose roots are the reciprocals of the roots of x2 + 4x - 10 =0​

Answer 1

Answer:

x2+4x-10=0

x2+4x=10

x(x+4)=10

Answer 2

x^2+4x-10

x=$\frac{-b\pm \sqrt{b^{2} -4ac} }{2a} =-2+\sqrt{14} /-2-\sqrt{14}$

The required quadratic eqn is x^2-(sum of roots)x+(product of roots)

($x^{2}-(\frac{1}{2-\sqrt{14}} + \frac{1}{2+\sqrt{14}})x+(\frac{1}{2-\sqrt{14}} * \frac{1}{2+\sqrt{14}})\\x^{2} -(\frac{4}{2^{2} -\sqrt{14^{2} }} )x+( \frac{1}{2^{2} -\sqrt{14^{2}}})\\x^{2} -(\frac{-4}{10})x+(\frac{-1}{10})\\ x^2+\frac{2}{5} x-\frac{1}{10} \\10x^{2} +4x-1)=0$

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