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4 votes
Solve 2x2 − 4x − 5 = 0 by completing the square.

User Senotrusov
by
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2 Answers

5 votes
2(x^2-2x+1-1)+5=0
2(x-1)^2+3=0
(x-1)^2=-3/2
x-1=+-√-3÷2

\[x=1+√
(-3)/(2)\]
and \[x=1-√-3/2\]
User Najera
by
8.4k points
2 votes
Rewrite the right side of the equation

2x^2-4x-5 = 0 in the following way:


2x^2-4x-5 = 2(x^2-2x)-5=2(x^2-2x+1-1)-5=2((x-1)^2-1)-5=2(x-1)^2-2-5=2(x-2)^2-7.
Then the equation is
2(x-2)^2-7=0 and
2(x-2)^2=7.

So,
(x-2)^2= (7)/(2) and
x-2=\pm \sqrt{ (7)/(2) }
The equation has two solutions:
x_1=2+ \sqrt{ (7)/(2) } and
x_2=2- \sqrt{ (7)/(2) }.




User Tim Kamm
by
8.7k points

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