Question 922274
7 and 10; 10 is 4 less than twice 7.
A quadratic equation may be expected, but it is not the most efficient way to solve the problem, and if you are going to solve the quadratic equation by factoring, why not just factor to get to the solution directly?
After all we need a solution, so there is no need to make sure we find every possible solution.
 
Listing the pairs of factors is easy enough:
{{{1*70=70}}}
{{{2*35=70}}}
{{{5*14=70}}} and
{{{7*10=70}}} .
The only pair where one factor is less than twice the other is 7 and 10.
 
NOTE:
Finding the right pair of factors seems easy, and it is, because 70 is a nice number for this problem.
The prime factoring of {{{70}}} is {{{70=2*5*7}}} .
That is a product of 3 prime factors, each with an exponent of 1.
That means that {{{70}}} has {{{(1+1)*(1+1)*(1+1)=2*3*3=8}}} factors in 4 pairs.
Listing the pairs of factors is easy enough:
{{{1*70=70}}}
{{{2*35=70}}}
{{{5*14=70}}} and
{{{7*10=70}}} .
The only pair where one factor is less than twice the other is 7 and 10.
 
WITH THE QUADRATIC EQUATION, we can also list the fancier answer (-5 and -14):
{{{x}}}= one factor
{{{2x-4)=2(x-2)}}}= the other factor (4 less than twice {{{x}}} )
At this point, we can write our quadratic equation as
{{{x(2x-4)=70}}}--->{{2x^2-4x=70}}}--->{{{2x^2-4x-70=0}}} ,
or as
{{{2(x-2)x=70}}}--->{{{x(x-2)=35}}}<--->{{{x^2-2x=35}}}<--->{{{x^2-2x-35=0}}} .
 
You can apply the quadratic formula to {{{2x^2-4x-70=0}}} or to {{{x^2-2x-35=0}}} .
You can solve {{{x^2-2x-35=0}}} by factoring or by completing the square.
You can also realize that {{{x(x-2)=35}}} screams {{{x=7}}} .
 
Completing the square:
{{{x^2-2x=35}}}--->{{{x^2-2x+1=35+1}}}--->{{{(x-1)^2=36}}}--->{{{system(x-1=6,"or",x-1=-6)}}}--->{{{system(x=6+1=7,"or",x=-6+1=-5)}}}
With {{{x=7}}} , the other number is {{{2x-4=2*7-4=10}}} .
If we want to be fancy, we say that {{{x=-5}}} is an integer, and {{{2x-4=2*(-5)-4=-10-4=-14}}} is the other integer.
 
Solving by factoring:
{{{x^2-2x-35=0}}}--->{{{(x-7)(x+5)=0}}}--->{{{system(x-7=0,"or",x+5=0)}}}--->{{{system(x=7,"or",x=-5)}}}
 
Applying the quadratic formula to {{{2x^2-4x-70=0}}} we get
{{{a=2}}} , {{{b=-4}}} , {{{c=-70}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} turns into
{{{x = (-(-4) +- sqrt((-4)^2-4*2*(-70)))/(2*2) }}}
{{{x = (4 +- sqrt(16+560))/4 }}}
{{{x = (4 +- sqrt(576))/4 }}}
{{{x = (4 +- 24)/4 }}}--->{{{system(x=(4+24)/4=28/4=7,"or",x=(4-24)/4=(-20)/4=-5)}}}