Question

One of the roots of the equation
2x2 + 6x = 2k – 1 is twice the other. Find
the value of k and the roots of the equation.

Answer 1

can anybody help me?

Answer 2

working on it. i think there is a snap way, but i don't see it yet. just grinding it out using quadratic formula

Answer 3

but i am sure there is a simple way

Answer 4

i've done it, but i want some answers to make sure that mine is correct

Answer 5

use quadratic you get the two roots in terms of x, equate them solve for k so that 1 is double

Answer 6

did you use the quadratic formula or did you use the sum and product of the roots?
if you have an answer it is easy to check

Answer 7

uhm, i have no answer for that to check it so.

Answer 8

i use sum and product of the roots.

Answer 9

it supposed to be 2x^2 , the question is a bit wrong

Answer 10

i understand the question. you can grind it out by starting with
\[2x^2+6x-2k+1=0\] then
\[x=\frac{-6\pm\sqrt{36-8(-2k+1)}}{4}\] and work it that way. but i think it is silly. instead forget about the 2k-1 and just solve the problem with
\[2x^2+6x+c=0\]

Answer 11

so how to find the k and the other root?

Answer 12

i must be doing something wrong, because if i try to solve it this way i get no solution

Answer 13

are you sure it is
\[2x^2+6x\] or might it be
\[2x^2-6x\]?

Answer 14

it is a plus

Answer 15

ok let me try this
\[2x^2+6x+c=0\]
\[x=\frac{-6\pm\sqrt{36-8c}}{4}\]
\[x=\frac{-6\pm2\sqrt{9-2c}}{2}\]
\[x=\frac{-3\pm\sqrt{9-2c}}{2}\] so two roots are
\[\frac{-3-\sqrt{9-2c}}{2}, \frac{-3+\sqrt{9-2c}}{2}\] and since one is twice the other you can write
\[-\frac{3}{2}-\frac{\sqrt{9-2c}}{2}=-3+\sqrt{9-2c}\] or better still
\[-3-\sqrt{9-2c}=-6+2\sqrt{9-2c}\] then
\[3=3\sqrt{9-2c}\]
\[\sqrt{9-2c}=1\]

Answer 16

wow thats kind of hard way

Answer 17

yes i agree, but i got
\[c=\frac{9}{2}\] and let me check that it works

Answer 18

nope it is wrong. damn

Answer 19

how about the answer for k is 7/2 ?