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# 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

- GuillermoLv 78 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 - Amar SoniLv 78 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