Question 122759
# 1


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^2-7*x-1=0}}} ( notice {{{a=1}}}, {{{b=-7}}}, and {{{c=-1}}})





{{{x = (--7 +- sqrt( (-7)^2-4*1*-1 ))/(2*1)}}} Plug in a=1, b=-7, and c=-1




{{{x = (7 +- sqrt( (-7)^2-4*1*-1 ))/(2*1)}}} Negate -7 to get 7




{{{x = (7 +- sqrt( 49-4*1*-1 ))/(2*1)}}} Square -7 to get 49  (note: remember when you square -7, you must square the negative as well. This is because {{{(-7)^2=-7*-7=49}}}.)




{{{x = (7 +- sqrt( 49+4 ))/(2*1)}}} Multiply {{{-4*-1*1}}} to get {{{4}}}




{{{x = (7 +- sqrt( 53 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (7 +- sqrt(53))/(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 = (7 +- sqrt(53))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (7 + sqrt(53))/2}}} or {{{x = (7 - sqrt(53))/2}}}



Now break up the fraction



{{{x=+7/2+sqrt(53)/2}}} or {{{x=+7/2-sqrt(53)/2}}}



Simplify



{{{x=7 / 2+sqrt(53)/2}}} or {{{x=7 / 2-sqrt(53)/2}}}



So these expressions approximate to


{{{x=7.14005494464026}}} or {{{x=-0.140054944640259}}}



So our solutions are:

{{{x=7.14005494464026}}} or {{{x=-0.140054944640259}}}


Notice when we graph {{{x^2-7*x-1}}}, we get:


{{{ graph( 500, 500, -10.1400549446403, 17.1400549446403, -10.1400549446403, 17.1400549446403,1*x^2+-7*x+-1) }}}


when we use the root finder feature on a calculator, we find that {{{x=7.14005494464026}}} and {{{x=-0.140054944640259}}}.So 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-4=0}}} ( notice {{{a=1}}}, {{{b=-4}}}, and {{{c=-4}}})





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




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




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




{{{x = (4 +- sqrt( 32 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


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



Now break up the fraction



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



Simplify



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



So these expressions approximate to


{{{x=4.82842712474619}}} or {{{x=-0.82842712474619}}}



So our solutions are:

{{{x=4.82842712474619}}} or {{{x=-0.82842712474619}}}


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


{{{ graph( 500, 500, -10.8284271247462, 14.8284271247462, -10.8284271247462, 14.8284271247462,1*x^2+-4*x+-4) }}}


when we use the root finder feature on a calculator, we find that {{{x=4.82842712474619}}} and {{{x=-0.82842712474619}}}.So this verifies our answer




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# 3



{{{6x^2=9x}}} Start with the given equation




{{{6x^2-9x=0}}} Subtract 9x from both sides




{{{3x(2x-3)=0}}} Factor the left side 




Now set each factor equal to zero:

{{{3x=0}}} or  {{{2x-3=0}}} 


{{{x=0}}} or  {{{x=3/2}}}    Now solve for x in each case



So our answer is 

 {{{x=0}}} or  {{{x=3/2}}} 



Notice if we graph {{{y=6x^2-9x}}}  we can see that the roots are {{{x=0}}} and  {{{x=3/2}}} . So this visually verifies our answer.



{{{ graph(500,500,-10,10,-10,10,0, 6x^2-9x) }}}