Question 117161
#1



{{{x^2=196}}} Start with the given equation



{{{x^2-196=0}}} Move all of the terms to the left side


Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{x^2-196=0}}} (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like {{{x^2+0*x-196=0}}}  notice {{{a=1}}}, {{{b=0}}}, and {{{c=-196}}})





{{{x = (0 +- sqrt( (0)^2-4*1*-196 ))/(2*1)}}} Plug in a=1, b=0, and c=-196




{{{x = (0 +- sqrt( 0-4*1*-196 ))/(2*1)}}} Square 0 to get 0  




{{{x = (0 +- sqrt( 0+784 ))/(2*1)}}} Multiply {{{-4*-196*1}}} to get {{{784}}}




{{{x = (0 +- sqrt( 784 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-0 +- 28)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (0 +- 28)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (0 + 28)/2}}} or {{{x = (-0 - 28)/2}}}


Lets look at the first part:


{{{x=(0 + 28)/2}}}


{{{x=28/2}}} Add the terms in the numerator

{{{x=14}}} Divide


So one answer is

{{{x=14}}}




Now lets look at the second part:


{{{x=(0 - 28)/2}}}


{{{x=-28/2}}} Subtract the terms in the numerator

{{{x=-14}}} Divide


So another answer is

{{{x=-14}}}


So our solutions are:

{{{x=14}}} or {{{x=-14}}}


Notice when we graph {{{x^2-196}}}, we get:


{{{ graph( 500, 500, -24, 24, -24, 24,1*x^2+0*x+-196) }}}


and we can see that the roots are {{{x=14}}} and {{{x=-14}}}. This verifies our answer



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#2


Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{x^2-4*x+2=0}}} ( notice {{{a=1}}}, {{{b=-4}}}, and {{{c=2}}})





{{{x = (--4 +- sqrt( (-4)^2-4*1*2 ))/(2*1)}}} Plug in a=1, b=-4, and c=2




{{{x = (4 +- sqrt( (-4)^2-4*1*2 ))/(2*1)}}} Negate -4 to get 4




{{{x = (4 +- sqrt( 16-4*1*2 ))/(2*1)}}} Square -4 to get 16  (note: remember when you square -4, you must square the negative as well. This is because {{{(-4)^2=-4*-4=16}}}.)




{{{x = (4 +- sqrt( 16+-8 ))/(2*1)}}} Multiply {{{-4*2*1}}} to get {{{-8}}}




{{{x = (4 +- sqrt( 8 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (4 +- 2*sqrt(2))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (4 +- 2*sqrt(2))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


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



Now break up the fraction



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



Simplify



{{{x=2+sqrt(2)}}} or {{{x=2-sqrt(2)}}}



So these expressions approximate to


{{{x=3.41421356237309}}} or {{{x=0.585786437626905}}}



So our solutions are:

{{{x=3.41421356237309}}} or {{{x=0.585786437626905}}}


Notice when we graph {{{x^2-4*x+2}}}, we get:


{{{ graph( 500, 500, -9.4142135623731, 13.4142135623731, -9.4142135623731, 13.4142135623731,1*x^2+-4*x+2) }}}


when we use the root finder feature on a calculator, we find that {{{x=3.41421356237309}}} and {{{x=0.585786437626905}}}.So this verifies our answer