Suppose \(\displaystyle{x}^{{{1}}}\) and \(\displaystyle{x}^{{{2}}}\) are solutions

michiiiiiakqm

michiiiiiakqm

Answered question

2022-03-31

Suppose x1 and x2 are solutions to x2+x+k=0, if x{1}2+x1x2+x22=2k2, find the value of k.

Answer & Explanation

rebecosasny8a

rebecosasny8a

Beginner2022-04-01Added 6 answers

Step 1
(x1+x2)2=x12+2(x1x2)+x22=2k2
The value in the left hand side is not equal to 2k2 due to the present of 2 in the middle term. Here is the complete solution (as per the process you follow).
Step 2
(x1+x2)2=x12+2x1x2+x22
(1)2=(x12+x1x2+x22)+x1x2
1=2k2+k
2k2+k1=0
k=d1±1242(1)4
k=d1±34
k=1, d12

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