if 1 is one of the root of ax2+3x+5=0 then the second root is
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please check the question
in this question the factor is not done
in this question the factor is not done
jaswanth18:
no second root find
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ax2-3x+1=0
One root of this equation is (2+i)
So put x=(2+i)
a(2+i)2–3(2+i)+1=0
a(4+2i)-6-3i+1=0
a(4+2i)-5-3i=0
a=5+3i/4+2i
Rationalise the denominator
a=(5+3i)(4–2i)/(4+2i)(4–2i)
a=20–10i+12i+6/16+4
a=26–2i/20
So the value of a is 13-i/10
One root of this equation is (2+i)
So put x=(2+i)
a(2+i)2–3(2+i)+1=0
a(4+2i)-6-3i+1=0
a(4+2i)-5-3i=0
a=5+3i/4+2i
Rationalise the denominator
a=(5+3i)(4–2i)/(4+2i)(4–2i)
a=20–10i+12i+6/16+4
a=26–2i/20
So the value of a is 13-i/10
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