If 1 is zero of the polynomial f(x)= a2x2-3ax+3x-1, then prove that a=2
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Answered by
20
f(x) = a^2 x^2 - 3ax - 3x - 1
According to question
f( 1) = 0
=> a^2 ( 1)^2 - 3a(1) + 3(1) - 1 = 0
=> a^2 - 3a + 3 - 1 = 0
=> a^2 - 3a + 2 = 0
=> a^2 - a - 2a + 2 = 0
=> a (a - 1) - 2(a - 1) = 0
=> (a - 1)(a - 2) = 0
a = 1 and 2
According to question
f( 1) = 0
=> a^2 ( 1)^2 - 3a(1) + 3(1) - 1 = 0
=> a^2 - 3a + 3 - 1 = 0
=> a^2 - 3a + 2 = 0
=> a^2 - a - 2a + 2 = 0
=> a (a - 1) - 2(a - 1) = 0
=> (a - 1)(a - 2) = 0
a = 1 and 2
Answered by
0
Answer:
a = 2
Step-by-step explanation:
If 1 is the zero of the polynomial f(x)=a
2
x
2
−3ax+3x−1
Then, f(x)=0
=>a
2
(1)
2
−3a(1)+3(1)−1=0
=>a
2
−3a+3−1=0
=>a
2
−3a+2=0
=>a
2
−2a−a+2=0
=>a(a−2)−1(a−2)=0
=>(a−2)(a−1)=0
a=2
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