if one zero of the polynomial A square + 9 X square + 13x + 6a is reciprocal of the other find the value of A
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let one zero =a
other zero =1/a
product of zeroes =6a/A2 +9
a×1/a =6a/a2 +9
1×a2+9 =6a
a2 +9 =6a
a2 -6a +9 =0
a2 -3a-3a +9=0
a(a-3)-3(a-3)=0
a-3 =0
a=3
other zero =1/a
product of zeroes =6a/A2 +9
a×1/a =6a/a2 +9
1×a2+9 =6a
a2 +9 =6a
a2 -6a +9 =0
a2 -3a-3a +9=0
a(a-3)-3(a-3)=0
a-3 =0
a=3
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