Question 414332
one root of the quadratic equation x²-kx+8=0 is negative of the other. the value of k is

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<pre><font face = "consolas" color = "indigo" size = 4><b>

There are two ways to do this:

First way:  Let the roots be +r and -r

Then

x² - kx + 8 = 0

whould have to factor as

(x - r)(x + r) = 0

which multiplies out to

x² + rx - rx + r² = 0

          x² + r² = 0 

So since the terms in x cancel, the coefficient of r
must be zero, which makes k = 0.

However the roots are imaginary since r² must equal to 8

x² + 8 = 0

    x² = -8
           __ 
     x = ±<font face = "symbol">Ö</font>-8
             ___
     x = ± i<font face = "symbol">Ö</font>4*2
             _
     x = ±2i<font face = "symbol">Ö</font>2

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The other way to do it is to use the quadratic formula:


x² - kx + 8 = 0



{{{x = (-(-k) +- sqrt((-k)^2-4*1*8 ))/(2*1) }}}
               
{{{x = (k +- sqrt(k^2-32 ))/2 }}}

The two roots are 

{{{(k + sqrt(k^2-32 ))/2 }}} and {{{(k - sqrt(k^2-32 ))/2 }}}

Setting one equal to the negative or the other:

{{{((k + sqrt(k^2-32 ))/2) = -((k - sqrt(k^2-32 ))/2) }}}

Multiply through by 2

{{{k + sqrt(k^2-32 ) = -(k - sqrt(k^2-32 )) }}}

{{{k + sqrt(k^2-32 ) = -k + sqrt(k^2-32 ) }}}

Subtracting the radical from both sides:

          k = -k

         2k = 0

          k = 0

And the two roots are

{{{x = (k +- sqrt(k^2-32 ))/2 }}}

{{{x = (0 +- sqrt(0^2-32 ))/2 }}}

{{{x = "" +- sqrt(-32)/2 }}}

{{{x = "" +- i*sqrt(32)/2 }}}

{{{x = "" +- i*sqrt(16*2)/2 }}}

{{{x = "" +- 4i*sqrt(2)/2 }}}

{{{x = "" +- 2i*sqrt(2) }}}

Edwin</pre>