Please help quick substitution problem?!?

solve the system of equations.

x^2+y^2=10

y=x^2+5x+2

(x,y)=( , ) ??

Round to the nearest hundredth.

If there is more than one solution, use the "or"

If applicable, "No solution"

2 Answers

Relevance
  • 8 years ago

    To see the solution, download Graph 4.4 from www.padowan.dk for free.

    On "Function I Insert relation" type x^2+y^2=10, then "OK".

    On "Function I Insert relation" type y=x^2+5x+2, choose another color, then "OK".

    To find the solutions, use www.wolframalpha.com, typing solve (x^2+y^2=10, y=x^2+5x+2), then "=".

    Source(s): Graph 4.4 www.wolframalpha.com
  • 8 years ago

    x^2+y^2=10 ................(i)

    y=x^2+5x+2 .............(ii)

    From (i) and (ii)

    x^2 +(x^2+5x+2)^2 =10

    x^2 +(x^4 +25x^2 +4 +10x^3 +4x^2 +20x) =10

    x^4 +10x^3 +30x^2 +20x -6 =0............(iii)

    Possible solutions are (+/-)1,2,3,6

    let f(x) =x^4 +10x^3+30x^2+20x-6

    f(1) = 55

    f(-1) =-5

    f(2) =250

    f(-2) =10

    f(3) =675

    f(-3) =15

    f(6) = 4650

    f(-6) =90

    It means there is a solution from -1 and 1

    f(0.5) =12.8125

    f(0.4) =7.4656

    f(0.3) =2.9781

    f(0.2) =-0.7184

    This shows there may be solution between 0.2 to 0.3

    f(0.25) =1.03515625

    f(0.22) =-0.03917744

    It means value lies between 0.22 to 0.25

    f(0.23) =0.31146841

    It means value lies between 0.22 to 0.23

    f(0.221) =-0.004445946719

    f(0.222) =0.030359392656

    it means value of x lies between 0.221 to 0.222

    f(0.2213) =0.0059878982822161

    f (0.2211) =-0.0009687369332559

    Hence x =0.2211 approximately then find y from (ii0

Still have questions? Get your answers by asking now.