Question 1192071
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{{{x = 4 - sqrt(17)}}} and {{{x = 4 + sqrt(17)}}} are the two roots of the form {{{m-n}}} and {{{m+n}}}


Where,
{{{m = 4}}}
{{{n = sqrt(17)}}}


Use Vieta's formulas to sum the roots:
{{{root1+root2 = (m-n)+(m+n)}}} 
{{{root1+root2 = 2m }}}
{{{root1+root2 = 8}}}
and multiply the roots as well
{{{root1*root2 = (m-n)*(m+n) }}}
{{{root1*root2 = m^2 - n^2}}} Difference of squares rule
{{{root1*root2 = 4^2 - (sqrt(17))^2}}}
{{{root1*root2 = 16 -17}}}
{{{root1*root2 = -1}}}


The negative of the sum of the roots is the value of b, while c represents the product of the roots.
b = -8 and c = -1


Why does this work? Consider the two roots to be r and s
This means {{{x = r}}} and {{{x = s}}} lead to {{{x-r = 0}}} and {{{x - s = 0}}} respectively.
Furthermore, {{{(x-r)(x-s) = x^2-sx-rx+rs = x^2-(s+r)x+rs = 0}}}
The -(s+r)x term matches with bx, showing that b = -(s+r). We add the two roots, then flip the sign to get the value of b.
The rs term is the product of the roots and matches with c.


Answer: {{{x^2 - 8x - 1 = 0}}}
You can confirm this answer by using the quadratic formula to arrive back to the original roots mentioned.


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Another approach:


{{{x = 4 - sqrt(17)}}}
{{{x-4 = -sqrt(17)}}}
{{{(x-4)^2 = (-sqrt(17))^2}}}
{{{(x-4)^2 = 17}}}


Through very nearly identical steps, you should find that {{{x = 4+sqrt(17)}}} also leads to {{{(x-4)^2 = 17}}}
Using the square root method, solving {{{(x-4)^2 = 17}}} leads back to the original roots mentioned.


Now let's expand things out and get everything to one side.
{{{(x-4)^2 = 17}}}
{{{(x-4)(x-4) = 17}}}
{{{x(x-4)-4(x-4) = 17}}}
{{{x^2-4x-4x+16 = 17}}}
{{{x^2-4x-4x+16-17 = 0}}}
{{{x^2-8x-1 = 0}}}
We get the same answer as before.
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