Question

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Date : 1
I two zeroes of the polynomial
xsquare - 6x² – 26x2 + 138x - 36 ar
2+√3then find other zeroes​

Answer 1

Answer:

${x}^{2} - 6 {x}^{2} - 26 {x}^{2} + 138x - 36 = 0 \\ = > - 31 {x}^{2} + 138x - 36 = 0 \\ = > x = \frac{ - 138 \pm \ \sqrt{138 { }^{2} - 4 \times - 31 \times - 36 } }{2 \times - 31} \\ = > x = \frac{ - 138 \pm \sqrt{19044 - 4464} }{ - 62} \\ = > x = \frac{ - 138 \pm \sqrt{14580} }{ - 62} \\ = > x = \frac{ - 138 \pm120.748}{ - 62}$

so,

$x = \frac{ - 138 + 120.748}{ - 62} = \frac{ - 17.2523}{ - 62} = 0.278$

or,

$x = \frac{ - 138 - 120.748}{ - 62} = \frac{ - 258.748}{ - 62} = 4.173$