Question 383525
Please take this equation and solve by factoring, completing the square and quadratic equation.  Then leave an equation for your classmates to try.
<pre>
{{{x^2+2x-8=0}}}

To solve by factoring, the easiest way when it works:

{{{(x+4)(x-2)=0}}}

Use the zero factor problem, by setting each factor = 0

{{{matrix(2,5,

x+4=0,"","","",x-2=0,
 x=-4,"","","",x=2)}}}

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To solve by completing the square, the hardest way but it always works.

{{{x^2+2x-8=0}}}

Isolate the terms in x on the left by adding +8 to both sides:

{{{x^2+2x=8}}}

Multiply the coefficient of x, which is 2, by {{{1/2}}}, getting 1, then
square 1, getting {{{red(1)}}}.  Add {{{red(1)}}} to each side:

{{{x^2+2x+red(1)=8+red(1)}}}

Factor the left side as the product of two identical binomials. Combine
the numbers on the right:

{{{(x+1)(x+1)=9}}}

Write the product of two identical binomials as a perfect square:

{{{(x+1)^2=9}}}

Use the principle of square roots:

{{{x+1="" +- sqrt(9)}}}

Since the square root of 9 is 3 we have:

{{{x+1 = "" +- 3}}}

We make two equations, one using the + and the other using the -

{{{matrix(2,5,

x+1=""+3, "","","",x+1=-3,
     x=2, "","","",x=-4)}}}

-------------------------

To solve by using the quadratic formula: (the medium hard way that
always works):

{{{x^2+2x-8=0}}}

There is a 1 coefficient understood before the first term:

{{{1x^2+2x-8=0}}} 

Compare that to 

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

and find that {{{a=1}}}, {{{b=2}}}, {{{c=-8}}}

Then substitute those three values in

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

 {{{x = (-(2) +- sqrt( (2)^2-4*(1)*(-8) ))/(2*(1)) }}}

 {{{x = (-2 +- sqrt( 4+32 ))/2 }}}

 {{{x = (-2 +- sqrt(36))/2 }}}

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

We make two equations, one using the + and the other using the -

{{{matrix(3,5,

x=(-2+6)/2, "","","",x=(-2-6)/2,
x=4/2,      "","","",x=-8/2,
x = 2,      "","","",x=-4)}}}

-----------------------------------------

To make an equation for your classmates to try

Choose two arbitrary numbers of x to be the solutions of the
equation that you're making up.  Let's pick, say -3 and 7

Then reverse the steps in solving by factoring.

Going backward from the zero factor principle:

{{{matrix(2,5,
 x=-3,"","","",x=7,
x+3=0,"","","",x-7=0
)}}}

{{{(x+3)(x-7)=0}}}

Going backward from factoring by multiplying this out using FOIL:

{{{x^2-7x+3x-21=0}}}

{{{x^2-4x-21=0}}}

That's an equation for your classmates to try.

Edwin</pre>