Thread: How to solve this equation by completing the square

1. How to solve this equation by completing the square

Hello

I can solve this question by using the quadratic formula but want to see how to solve by completing the square.

Equation is $\frac{-1}{5}x^2 + 2x - \frac{9}{2} = 0$

So my working is:

$\frac{-1}{5}(x^2 - 10x) - \frac{9}{2} = 0$

$\frac{-1}{5}(x^2 - 10x) = \frac{9}{2}$

$\frac{-1}{5}(x - 5)^2 - 25 = \frac{9}{2}$

$\frac{-1}{5}(x - 5)^2 = \frac{50}{2} + \frac{9}{2} = \frac{59}{2}$

But how do I get rid of the $\frac{-1}{5}$. And is the above correct so far.

My problem is that multiplying both sides by -5 means number on rhs will be negative and so no solution. But I know the curve does cross the x axis - so there is a solution. Can anyone show me the final steps to solve this using completing the square?

Angus

2. Re: How to solve this equation by completing the square

You can multiply both sides by -5, yielding x^2 - 10x + \frac{45}{2} = 0. Then:

x^2 - 10x + 25 + \frac{45}(2} = 25

(x - 5)^2 = 25 - \frac{45}{2}

(x - 5)^2 = \frac{5}{2}

(Sorry, my LaTex isn't compiling)

3. Re: How to solve this equation by completing the square

When you add the 25 within the parentheses, you are really subtracting 5 because of the coefficient of -1/5, so you need to add 5 outside.

What you have done is subtract 30 from the left side, while leaving the right side unchanged. You are correct up to:

$-\frac{1}{5}\left(x^2-10x\right)=\frac{9}{2}$

$-\frac{1}{5}\left(x^2-10x+25\right)=\frac{9}{2}-\frac{1}{5}\cdot25$

Can you finish from there?

4. Re: How to solve this equation by completing the square

Originally Posted by angypangy
Hello

I can solve this question by using the quadratic formula but want to see how to solve by completing the square.

Equation is $\frac{-1}{5}x^2 + 2x - \frac{9}{2} = 0$

So my working is:

$\frac{-1}{5}(x^2 - 10x) - \frac{9}{2} = 0$

$\frac{-1}{5}(x^2 - 10x) = \frac{9}{2}$

$\frac{-1}{5}(x - 5)^2 - 25 = \frac{9}{2}$

$\frac{-1}{5}(x - 5)^2 = \frac{50}{2} + \frac{9}{2} = \frac{59}{2}$

But how do I get rid of the $\frac{-1}{5}$. And is the above correct so far.

My problem is that multiplying both sides by -5 means number on rhs will be negative and so no solution. But I know the curve does cross the x axis - so there is a solution. Can anyone show me the final steps to solve this using completing the square?

Angus
It's easier to multiply through by -5 initially to clear that fraction.

$-\dfrac{1}{5}x^2 + 2x - \dfrac{9}{2} = 0 \Leftrightarrow x^2 - 10x + \dfrac{45}{2} = 0$

$(x-5)^2 - 25 + \frac{45}{2} = 0$

$(x-5)^2 - \dfrac{5}{2} = 0$

$(x-5)^2 = \dfrac{5}{2}$

This should now give two real solution

5. Re: How to solve this equation by completing the square

Originally Posted by e^(i*pi)
It's easier to multiply through by -5 initially to clear that fraction.

$-\dfrac{1}{5}x^2 + 2x - \dfrac{9}{2} = 0 \Leftrightarrow x^2 - 10x + \dfrac{45}{2} = 0$

$(x-5)^2 - 25 + \frac{45}{2} = 0$

$(x-5)^2 - \dfrac{5}{2} = 0$

$(x-5)^2 = \dfrac{5}{2}$

This should now give two real solution
Ah, yes that is easier. thanks.