A quadratic polynomial p(x) has 1+√5 and 1−√5 as its zeros and p(1)=2. Then the value of p(0) is
A
−45
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B
85
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C
−85
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D
45
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Solution
The correct option is B85 Let p(x)=a(x−(1+√5))(x−(1−√5)) p(x)=a(x2−x(1+√5+1−√5)+(1−5))p(x)=a(x2−2x−4)
Now, p(1)=2 ⇒2=a(1−2−4)⇒a=−25 ⇒p(x)=−25(x2−2x−4)∴p(0)=85