How do I solve the formula #kx^2 + 8x + 4 = 0# for #x#?

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Answer 1 · 2014-10-19T15:48:13

By Quadratic Formula #x={-bpmsqrt{b^2-4ac}}/{2a}#,

#x={-8pmsqrt{8^2-4cdot k cdot 4}}/{2k}={-8pmsqrt{64-16k}}/{2k}#

by factoring #4# out of the square-root,

#={-8pm4sqrt{4-k}}/{2k}#

by dividing the numerator and the denominator by #2#,

#={-4pm2sqrt{4-k}}/{k}#

I hope that this was helpful.