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Soloha48 [4]
3 years ago
8

find the positive value of k that would make he left side of the equation a perfect square trinomial x^2-kx+64

Mathematics
1 answer:
Nataliya [291]3 years ago
6 0

Answer: 16

<u>Explanation:</u>

x² - kx + 64

Since we are looking for a perfect square, then we need to take the square root of 64 to find the factors: √64 = 8

So, the factors are: (x - 8)(x - 8)

Foil (or distribute) to get: x² - 16x + 64 <em>The k-value is 16</em>

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a_1=7\\ \\a_1^2=7^2=49\Rightarrow a_2=4+9+1=14\\ \\a_2^2=14^2=196\Rightarrow a_3=1+9+6+1=17\\ \\a_3^2=17^2=289\Rightarrow a_4=2+8+9+1=20\\ \\a_4^2=20^2=400\Rightarrow a_5=4+0+0+1=5\\ \\a_5^2=5^2=25\Rightarrow a_6=2+5+1=8\\ \\a_6^2=8^2=64\Rightarrow a_7=6+4+1=11\\ \\a_7^2=11^2=121\Rightarrow a_8=1+2+1+1=5\\ \\\text{and so on...}

We can see the pattern

a_5=a_8=a_{11}=a_{14}=...=5\\ \\a_6=a_9=a_{12}=a_{15}=...=8\\ \\a_7=a_{10}=a_{13}=a_{16}=...=11

In other words, for all k\ge 2

a_{3k-1}=5\\ \\a_{3k}=8\\ \\a_{3k+1}=11

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