1) 3х²-27=0
3х²=27
х²=9
х¹=3;х²=-3
2) 2х²+3х-5=0
а=2, b=3, c=-5
D=b²-4ac=3²-4×2×(-5)=9+40=49=9²
x½=-b±D/2a=-3±9/4=-3 и 1,25
3)(х+3)²-5х(х-2)=10(2х+1)
х²+6х+9-5х²+10=20х+10
х²+6х+9-5х²+10-20х-10=0
-4х²-14х+9=0/:-1
4х²+14х-9=0
а=4, b=14, c=-9
D=14²-4×4×(-9)=169+144=313=√313
x½=-14±√313/8
4) (x+2)²=43-6x
x²+4x+4-43+6x=0
x²+10x-39=0
a=1, b=10, c=-39
D=10²-4×1×(-39)=100+156=256=16²
x½=-10±16/2=3 и -13
5) (х-2)²+24=(2+3х)²
х²-4х+4+24=4+12х+9х²
х²-4х+4+24-4-12х-9х²=0
-8х²-16х+24=0/:-1
8х²+16х-24=0
a=8, b=16, c=-24
D=16²-4×8×(-24)=256+768=1024=32²
x½=-16±32/16=-3 и 1
The given equation can be re-written as sin
2
4x−2sin4xcos
4
x+cos
2
x=0
Add and subtract cos
8
x
∴(sin4x−cos
4
x)
2
+cos
2
x(1−cos
6
x)=0
Since both the terms are +ive (cos
6
x≤1), above is possible only when each term is zero for the same value of x.
sin4x−cos
4
x=0 .(1)
and cos
2
x(1−cos
6
x)=0 .(2)
From (2) cosx=0 or cos
2
x=1
∵z
3
=1⇒z=1 only
as other values will not be real.
Case I: If cosx=0 i.e., x=(n+
2
1
)π, then from (1)
sin4(n+
2
1
)π+0=0
or sin(4n+2)π=0 which is true.
∴x=(n+
2
1
)π (3)
Case II: When cos
2
x=1 i.e., sinx=0
∴x=rπ then from (1), sin4rπ−1=0 or −1=0 which is not true. Hence the only solution is given by (3).
ответ: 2.