Question 86122
1.

*[invoke completing_the_square -1, 4, 2]


2.

*[invoke completing_the_square 1, 1, 1]


We can find the x-intercepts by the quadratic formula

1.

Starting with the general quadratic


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


the general form of the quadratic equation is:


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


So lets solve {{{x^2+2*x-8}}}


{{{x = (-2 +- sqrt( (2)^2-4*1*-8 ))/(2*1)}}} Plug in a=1, b=2, and c=-8




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




{{{x = (-2 +- sqrt( 4+32 ))/(2*1)}}} Multiply {{{-4*-8*1}}} to get {{{32}}}




{{{x = (-2 +- sqrt( 36 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-2 +- 6)/(2*1)}}} Simplify the square root




{{{x = (-2 +- 6)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-2 + 6)/2}}} or {{{x = (-2 - 6)/2}}}


Lets look at the first part:


{{{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=-8/2}}} Subtract the terms in the numerator

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our solutions are:

{{{x=2}}} or {{{x=-4}}}


Notice when we graph {{{x^2+2*x-8}}} we get:


{{{ graph( 500, 500, -14, 12, -14, 12,1*x^2+2*x+-8) }}}


and we can see that the roots are {{{x=2}}} and {{{x=-4}}}. This verifies our answer


2.

Starting with the general quadratic


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


the general form of the quadratic equation is:


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


So lets solve {{{x^2-5*x-10}}}


{{{x = (5 +- sqrt( (-5)^2-4*1*-10 ))/(2*1)}}} Plug in a=1, b=-5, and c=-10




{{{x = (5 +- sqrt( 25-4*1*-10 ))/(2*1)}}} Square -5 to get 25




{{{x = (5 +- sqrt( 25+40 ))/(2*1)}}} Multiply {{{-4*-10*1}}} to get {{{40}}}




{{{x = (5 +- sqrt( 65 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- sqrt(65))/(2*1)}}} Simplify the square root




{{{x = (5 +- sqrt(65))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (5 + sqrt(65))/2}}} or {{{x = (5 - sqrt(65))/2}}}



Which approximate to


{{{x=6.53112887414927}}} or {{{x=-1.53112887414927}}}



So our solutions are:

{{{x=6.53112887414927}}} or {{{x=-1.53112887414927}}}


Notice when we graph {{{x^2-5*x-10}}} we get:


{{{ graph( 500, 500, -11.5311288741493, 16.5311288741493, -11.5311288741493, 16.5311288741493,1*x^2+-5*x+-10) }}}


when we use the root finder feature on our calculator, we find that {{{x=6.53112887414927}}} and {{{x=-1.53112887414927}}}.So this verifies our answer