if the zero of polynomial f(x)=k^x^-17x+k+2 are reciprocal of each other then the value of k
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Answered by
11
Sol:
Let the roots of (k2 + 4) x2 + 13x + 4k be p and 1/p.
Product of the roots = p x 1/p = (constant term) / coefficient of x2
⇒ (4k) / (k2 + 4) = p x 1/p
⇒ (4k) / (k2 + 4) = 1
⇒ k2 - 4k + 4 = 0
⇒ (k - 2)2 = 0
⇒ k - 2 = 0
⇒ k = 2
Therefore, the value of k is 2.
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Let the roots of (k2 + 4) x2 + 13x + 4k be p and 1/p.
Product of the roots = p x 1/p = (constant term) / coefficient of x2
⇒ (4k) / (k2 + 4) = p x 1/p
⇒ (4k) / (k2 + 4) = 1
⇒ k2 - 4k + 4 = 0
⇒ (k - 2)2 = 0
⇒ k - 2 = 0
⇒ k = 2
Therefore, the value of k is 2.
IF IT HELPED..PLZ MARK AS BRAINLIEST!!
Answered by
2
hey hii .....the value of k here is there....
f(x) = k^x^-17 x+ k +2 .....
in the place of f(x) we can the value of 1 ( like this f(1) ).....
we can place the value of x place that 1 ....
f(1)= k^(1)^ -17 (1) + (1) + 2
= k ^(1)^ -17 +1+2
= k^1-17+1+2
= 16+1+2
= 19
f(x) = k^x^-17 x+ k +2 .....
in the place of f(x) we can the value of 1 ( like this f(1) ).....
we can place the value of x place that 1 ....
f(1)= k^(1)^ -17 (1) + (1) + 2
= k ^(1)^ -17 +1+2
= k^1-17+1+2
= 16+1+2
= 19
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