SOLUTION: Solve each of the following quadratic equations by completing the square.
{{{2x^2+x+11=0}}}
My answer was:x={{{-1/4+(2sqrt(87)/4)*i}}},x={{{-1/4-(2sqrt(87)/4)*i}}}. Is this cor
Algebra.Com
Question 75017This question is from textbook Beginning Algebra
: Solve each of the following quadratic equations by completing the square.
My answer was:x=,x=. Is this correct? Thanks.
This question is from textbook Beginning Algebra
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
I'm not sure where the 2 that multiplies came from. Let's go through the
problem, and maybe you can catch it.
.
Start with:
.
.
You want the coefficient of the term to be 1, so divide all terms on both sides
of the given equation by 2. This results in:
.
.
Take half of the multiplier of the x term. So take half of and you get .
Square that and you get . Add and subtract to the equation and you have:
.
.
The first three terms are now a perfect square. Write them as the square:
.
.
Combine the and by putting the over the common denominator
of 16. Do this by multiplying by to get . Then recognize
that plus is . Substitute this into the equation and
it becomes:
.
.
Subtract from both sides and you have:
.
.
Take the square root of both sides to get:
.
.
Subtract from both sides and you have:
.
.
But the square root of simplifies to which further
simplifies to and
.
Substituting these two results results in the two answers:
.
and
.
In this way, you don't get that multiplier of 2 that you have in your answer.
.
I suspect that you might have forgotten to make the coefficient of the term equal
to 1. Early on you want to divide all the terms on both sides of the quadratic equation by
the coefficient of the term if that multiplier is anything other than 1 ...
but I'm not sure whether or not that was the problem..
.
The rest of it was OK and it appears that, with the exception of that one glitch, you
worked everything correctly.
.
Hope this clarifies things for you.
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