Question 271650
3. By using the Newton-Raphson’s method find the positive root of the quadratic equation 
<pre><font size = 4 color = "indigo"><b>
    {{{5x^2 + 11x – 17 = 0}}} correct to 3 significant figures.

{{{f(x)=5x^2+11x-17}}}
{{{"f'(x)"=10x+11}}}

Choose {{{x[1]=1}}}, 
{{{f(x[1])=f(1)=5(1)^2+11(1)-17=5+11-17=-1}}}
{{{"f'"(x[1])="f'(1)"=10(1)+11=10+11=21}}}


{{{x[2]=x[1]- (f(x[1])) /(  "f'"(x[1]) ) =1- (f(1)) /"f'(1)" =1-(-1/21)=1+1/21=1.047619048 }}}

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{{{x[2]=1.047619048}}}, 
{{{f(x[2])=f(1.047619048)=5(1.047619048)^2+11(1.047619048)-17=.01133786848}}}
{{{"f'"(x[2])="f'(1.047619048)"=10(1.047619048)+11=10.47619048+11=21.47619048}}}


{{{x[3]=x[2]- (f(x[2])) /(  "f'"(x[2]) ) =1.047619048- (f(1.047619048)) /"f'(1.047619048)" =1.047619048-(.01133786848/21.47619048)=1.046562998 }}}

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Checking it by the quadratic formula

{{{x = (-11 +- sqrt( 11^2-4*5*(-17) ))/(2*5) }}}

{{{x = (-11 +- sqrt( 121+340 ))/10 }}}

{{{x = (-11 +- sqrt(461 ))/10 }}}

Use the + to get the positive root:

{{{x=1.047091055}}}

So we only needed 1 iteration of the Newton-Rhaphson method
to get it to three significant figures, for what we had then
would have rounded to 1.05.  By taking it one more step we
have it to 4 significant figures 1.047.

Edwin</pre>