Question 390980

let larger number be {{{y}}} and smaller number {{{x}}}

if the larger of is {{5}}} more than twice the smaller number, we have:

{{{y = 2x + 5}}} ..........(1)

if the square of the smaller is equal to the larger,we have 

{{{y = x^2}}}.........(2)

solve the system:

{{{y = 2x + 5}}} ..........(1)

{{{y = x^2}}}.........(2).......substitute in (1)

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{{{x^2= 2x + 5}}} ..........(1).solve for {{{x}}}


{{{x^2 -5 = 2x }}} 

{{{x^2 -2x  -5 = 0 }}} 



*[invoke quadratic_formula 1, -2, -5, "x"]



{{{x1= 1 +sqrt(6)=1+2.45}}}

{{{x1= 3.45}}}..................................

{{{x2= 1 +sqrt(6)=1-2.45}}}

{{{x2= -1.45}}}...................................

find {{{y}}}

{{{y1 = (3.45)^2 = 11.9}}}

{{{y2 = (-1.45)^2 = 2.1025}}}

check:



{{{11.9= 2( 3.45)+ 5}}} ..........(1)...plug in {{{x1= 3.45}}}and{{{y1 = 11.9}}}

{{{11.9= 6.9 + 5}}} 

{{{11.9= 11.9 }}} 


{{{y= (x)^2}}}.........(2)....plug in {{{x2= -1.45}}}and{{{y2 = 2.1025}}}


{{{2.1025 = (-1.45)^2}}}

{{{2.1025 = 2.1025}}}