Find zeros of polynomial 5t^2+12t+7 and verify the relationship between zeros and coefficients
Answers
Answered by
205
5t^2 +(5+7)t + 7
5t^2+5t+7t+7
5t(t+1)+7(t+1)
5t+7=0 t+1=0
t=-7/5 t=-1
hence, zeroes are -7/5 and -1
5t^2+5t+7t+7
5t(t+1)+7(t+1)
5t+7=0 t+1=0
t=-7/5 t=-1
hence, zeroes are -7/5 and -1
Answered by
158
P(x)= 5t^2+12t+7
=> 5t^2 +5t+7t+7
=>5t(t+1)+7(t+1)
=>(5t+7),(t+1)
=>t= (-7/5) or t= (-1)
Sum of the zeroes
= alpha + beta = -b/a= -12/5
Product of the zeroes
=alpha *beta = c/a= 7/5
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