Question 92942
"Solve by using the quadratic formula
x^2-x-2=0"


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-x-2=0}}} ( notice {{{a=1}}}, {{{b=-1}}}, and {{{c=-2}}})


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




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




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




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




{{{x = (1 +- sqrt( 9 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (1 +- 3)/(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 = (1 +- 3)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (1 + 3)/2}}} or {{{x = (1 - 3)/2}}}


Lets look at the first part:


{{{x=(1 + 3)/2}}}


{{{x=4/2}}} Add the terms in the numerator

{{{x=2}}} Divide


So one answer is

{{{x=2}}}




Now lets look at the second part:


{{{x=(1 - 3)/2}}}


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

{{{x=-1}}} Divide


So another answer is

{{{x=-1}}}


So our solutions are:

{{{x=2}}} or {{{x=-1}}}


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


{{{ graph( 500, 500, -11, 12, -11, 12,1*x^2+-1*x+-2) }}}


and we can see that the roots are {{{x=2}}} and {{{x=-1}}}. This verifies our answer



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"Solve by using the quadratic formula
4x^2-3x+3=0 "



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 {{{4*x^2-3*x+3=0}}} ( notice {{{a=4}}}, {{{b=-3}}}, and {{{c=3}}})


{{{x = (--3 +- sqrt( (-3)^2-4*4*3 ))/(2*4)}}} Plug in a=4, b=-3, and c=3




{{{x = (3 +- sqrt( (-3)^2-4*4*3 ))/(2*4)}}} Negate -3 to get 3




{{{x = (3 +- sqrt( 9-4*4*3 ))/(2*4)}}} Square -3 to get 9  (note: remember when you square -3, you must square the negative as well. This is because {{{(-3)^2=-3*-3=9}}}.)




{{{x = (3 +- sqrt( 9+-48 ))/(2*4)}}} Multiply {{{-4*3*4}}} to get {{{-48}}}




{{{x = (3 +- sqrt( -39 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (3 +- i*sqrt(39))/(2*4)}}} 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 = (3 +- i*sqrt(39))/(8)}}} Multiply 2 and 4 to get 8




After simplifying, the quadratic has roots of


{{{x=3/8 + sqrt(39)/8*i}}} or {{{x=3/8 - sqrt(39)/8*i}}}


Notice if we graph the quadratic {{{y=4*x^2-3*x+3}}}, we get


{{{ graph( 500, 500, -14.625, 15.375, -12.5625, 17.4375, 4*x^2-3*x+3) }}} graph of {{{y=4*x^2-3*x+3}}}


And we can see that there are no real roots


To visually verify the answer, check out <a href=http://www.math.hmc.edu/funfacts/ffiles/10005.1.shtml>this page</a> to see a visual representation of imaginary roots