Guadratic Equation

(a) Solve the equation y2+3y-10 = 0 .
y2+3y-10 = 0

(x-2)(x+5) = 0

x=2 or -5
(b) Hence solve the equation (x2+x)2+3(x2 +x)-10= 0 .
let y = x2+x
from (a),

x2+x = 2 or x2 +x = -5
x2+x-2=0 or x2+x+5=0
(x-1)(x+2) = 0 or no solution
x = 1 or -2

a)
0=y^2+3y-10
=(y+5)(y-2)
i.e. y=2 or -5

b)By a)
x^2+x =2 or x^2+x=-5
x^2+x-2=0 or x^2+x+5=0
(x+2)(x-1)=0 or no real sol.
x= 1 or -2

a)
y2+3y-10 = 0
(y+5)(y-2) = 0

(y+5) = 0 or (y-2) = 0
y=-5 　　or y = 2

∴y = -5 or 2


b)
(x2+x)2+3(x2+x)-10= 0
y2+3y-10 = 0 (Substitute y = x2 + x)
y=-5 　　　　　　　　　　or y = 2
x2 + x = -5　　　　　　　　or x2 + x= 2
x2 +x +5 = 0　　　　　　　　or x2+x-2=0
x = [-1 (+/-) √(1)2 - 4(1)(5)]/2(1) or (x+2)(x-1)=0
= [-1 (+/-) √(-19)]/2 (rejected) or x+2=0 or x-1=0
　　　　　　　　　　　　　　　x=-2　or x=1

∴x = -2 or 1

a)
y2+3y-10 = 0
(y+5)(y-2) = 0
y = -5 or 2
b)
(x2+x)2+3(x2+x)-10= 0
(x^2+x+5)(x^2+x-2) = 0
(x+2)(x-1)(x^2+x+5) = 0
x = -2 or 1
if x^2+x+5 = 0
DELTA = 1-20=-19<0
無解!
.'.x= -2 or 1