A) True
B) False
C) Can't say
D) Partially True/False
Correct Answer: B
Solution :
False |
\[f\left( \alpha \right)=f\left( \beta \right)=0\] |
[\[\therefore \] \[\alpha \] and \[\beta \]are the zeroes of\[{{x}^{2}}-6x+a\]] |
\[{{\alpha }^{2}}-6\alpha +a=0\] |
\[{{\beta }^{2}}-6\beta +a=0\] |
\[\therefore \,\,\,\alpha +\beta =6\] ...(i) |
\[3\alpha +2\beta =20\] [Given] ...(ii) |
From Eqs. (i) and (ii) we get \[\alpha =8\], \[\beta =-2\] |
\[{{\beta }^{2}}-6\beta +a=0\] |
\[a=6\times \left( -2 \right)-{{\left( -2 \right)}^{2}}\] |
\[=-12-4=-16\] |
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