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Hey!!!!!
let p(x) = x³ - 3√5x² + 13k - 3√5
Since x - √5 is a factor
Then x = √5 is a zeros of p(x)
Thus P(√5) = 0
=> (√5)³ - 3√5(√5)² + 13k - 3√5 = 0
=> 5√5 - 3√5(5) + 13k - 3√5 = 0
=> 5√5 - 15√5 + 13k - 3√5 = 0
=> 13k - 13√5 = 0
=> 13(k - √5) = 0
=> k - √5 = 0
=> k = √5
Hope this helps
let p(x) = x³ - 3√5x² + 13k - 3√5
Since x - √5 is a factor
Then x = √5 is a zeros of p(x)
Thus P(√5) = 0
=> (√5)³ - 3√5(√5)² + 13k - 3√5 = 0
=> 5√5 - 3√5(5) + 13k - 3√5 = 0
=> 5√5 - 15√5 + 13k - 3√5 = 0
=> 13k - 13√5 = 0
=> 13(k - √5) = 0
=> k - √5 = 0
=> k = √5
Hope this helps
Answered by
3
hope helped you...........……
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