Standard form -> y=ax^2+bx+c→y=ax2+bx+c
Converting the given equation to standard form gives
y=0=5x^2+2x-6y=0=5x2+2x−6
But y=0y=0 is a specific case
So for the general case we have:
y=5x^2+2x-6y=5x2+2x−6
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
the process of completing the square introduces an error. It ads a term to the original equation. Thus this term must be removed. This is achieved by introducing a correction.
Let kk be the correction value.
color(blue)("Step 1")Step 1
write as:" "y=5(x^2+2/5x)-6 y=5(x2+25x)−6
Include the correction
y=5(x^2+2/5x)-6+ky=5(x2+25x)−6+k
At this stage k=0k=0 as we have not changed the overall values.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 2")Step 2
Move the power from x^2x2 to outside the brackets
y=5(x+2/5x)^2-6+k larr" now "k" starts to have a value"y=5(x+25x)2−6+k← now k starts to have a value
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 3")Step 3
Remove the xx from 2/5x25x
y=5(x+2/5)^2-6+k y=5(x+25)2−6+k
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 4")Step 4
Halve the 2/525
y=5(x+2/10)^2-6+k y=5(x+210)2−6+k
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is the point where we find the value of kk
The error comes from: 5(2/10)^25(210)2. This is an added term and needs to be removed.
So 5(2/10)^2+k=05(210)2+k=0
=>color(red)(k=-(5xx4)/100" "=" "-20/100" " =" " -1/5)⇒k=−5×4100 = −20100 = −15
So we now have:
y=5(x+2/10)^2-6color(red)(-1/5) y=5(x+210)2−6−15
color(white)(2/2)22
" "color(blue)(ul(bar(|color(white)(2/2)y=5(x+2/10)^2-31/5color(white)(2/2)|)))
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set y=0 then we have
+31/25=(x+2/10)^2
=>x+2/10=+-sqrt(31/25)
=>x=-2/10+-sqrt(31/25)
=>x=-2/10+-sqrt(31)/5
x=-1.313 to 3 decimal places
x=+0.914 to 3 decimal places
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