Solve the equation x 3 squares X 4 squares x 5 squares 17X 24 30

x-3) +x-4) -x-5) 17x+24 Solution: x -6x+9+x -8x+16-x +10x-25 17x+24 Shift: x -4x-17x 24 x -21x 24 x -21x-24 0 According to the formula method, the discriminant formula:

x -b b -4ac 2a , discriminant b -4ac discriminant a 1, b 21, c -24 gives b -4ac (-21) -4 1 (-24 ) 0 so x -b b -4ac 2a (-21) 21) +96 2 441 537 2, there are two inequality equation solutions: x 441 + 537 2, x 441- 537 2 . OK Check it out for yourself, right.

Convert the equation into a general form: x 2-21x-24=0, where a=1, b=-21, c=-24, and substitute the root finding formula to calculate.

Left = 3x 2-4x = Right 17x + 24

So 3x 2-21x-24 = 0

i.e. x 2-7x-8 = 0

x-8)(x+1）=0

So x=8 or -1

Hope it helps!

1）（x-3）^2=225/4

x-3=+/25/2

x = 31 2 or x = -19 2

2) Decompose: x(3x-6)=0

x=0 or 3x-6=0

Solution: x=0 or x=2

3) Simplification to get x 2 + x - 3x - 3 = 0

Simplified: x 2-2x-3 = 0

Obtain: (x-3)(x+1)=0 ==x-3=0 or x+1=0

Both: x=3 or x=-1 choose c,7,c,2,the first question x-3= 15 2,the second question x(3x-6)=0,the third problem only c,solve the equation can be solved,2,(1)4(x-3)square=225

x-3)²=15/2）²

x-3=15 2=, or x-3=-15 2=

So x= or x=

2) 3x squared - 6x = 0, circle sum 3x (x-2) = 0, x = 0, or x = 2

2: Case of the solution of equation x(x+1)=3(x+1).

x(x+1)-3(x+1)=0

x-3)(x+1)=0

x=3 or x=-1

So choose C and have fun, 1, 1(1) (x-3) = 225 4, x = 21 2 or x = -9 2

2) 3 (x-2) x = 0, so x = 0 or x = 2

Shift x -2x-3 = 0

i.e. (x-3)(x+1)=0

So x=3 or the maese cavity state x=-1 choose c,0,1: solve equation (1)4(x-3) square=225 (2)3xsquared-6x=0

2: The source of the solution of the equation x(x+1)=3(x+1) () a:x=-1 b:x=3 c:x1=-1,x2=3 d: none of the above.

The answer is 0, and the common factor is 2x 2x=0, which is the same as the previous one.

x square + 3) squared - 4 (x square + 3) = 0

x square + 3) (x square + 3-4) = 0

x square + 3 = 0 (no real number book Kai solution) is medium.

x square + 3-4 = 0

x squared = 1x = 1 x = -1

25 (2x-1) to the power of 2 - 16 = 0

2x-1) = 16 25

2x-1=±4/5

x=9/10 x=-1/10

The square of x + the radical discriminant of x=3x=3 δ=b 2-4ac=33 spike 0, there is a real solution, using the formula for finding the root, x=[-b

b 2-4ac) (1 2)] 2a, the equation is solved as x1==[3 + root number 33] 4, x2==[3 - root number.

33] 4,2, method 1 is the same, method 2 uses the cross multiplication method x squared -2x+1=25, both x 2-2x-24=(x-6)(x+4)=0 x1=6,x2=-4;

3, the same method, method two with ten stoves to sell the word multiplication, the original sorting to obtain 2x 2-9x+10 = (2x-5) (x-2) = 0 both x1 = 2, x2 = 2 5

The square of x + the radical discriminant of x=3x=3 δ=b 2-4ac=33 0, there is a real number solution, using the formula for finding the root, x=[-b

b 2-4ac) (1 2)] 2a, the equation is solved as x1==[3 + root number 33] 4, x2==[3 - root number.

33] 4,2, method 1 is the same, method 2 uses the cross multiplication method x squared -2x+1=25, both x 2-2x-24=(x-6)(x+4)=0 x1=6, x2=-4;

3. Method 1 is the same, method 2 uses cross multiplication, and the original formula is sorted out to obtain 2x 2-9x+10=(2x-5)(x-2)=0 x1=2, x2=2 5

1. Move the term to get 2x square + 3x-3=0

Because =b 2-4ac=9+24=33 0, x=-b root 2a=-3 root 33 4 x1= =-3+root 33 4, x2=-3-root 33 42, and the square of x is -2x-24=0

Factoring yields (x+4)(x-6)=0

x+4=0 or x-6=0

x1=-4,x2=6

3. Move the item to get x(2x-5)-4x+10=0 and simplify it to get 2x 2-9x+10=0

Factoring yields (2x-5)(x-2)=0

2x-5=0 or x-2=0

x1=2/5,x2=2

Type it out word by word, give it points, thank you.

x square + 8x + 16 - x square - 10x - 25 + x square - 6x + 9 = 24 + 4x; x squared - 8x = 24 + 4x; x squared - 12x - 24 = 0;

1）（x-3）^2=225/4

x-3=+/25/2

x = 31 2 or x = -19 2

2) Decompose: x(3x-6)=0

x=0 or 3x-6=0

Solution: x=0 or x=2

3) Simplification to get x 2 + x - 3x - 3 = 0

Simplified: x 2-2x-3 = 0

Obtain: (x-3)(x+1)=0 ==x-3=0 or x+1=0 both: x=3 or x=-1 choose c

1) 4 (x-3) square = 225

x-3)²=15/2）²

x-3 = 15 2=, or x-3 = -15 2 = so x= or x=

2) 3x squared - 6x = 0, 3x (x-2) = 0, x = 0, or x = 22: the case of the solution of the equation x(x(x+1)=3(x+1)=0

x-3)(x+1)=0

x=3 or x=-1

So choose C and wish you a happy day.

1.(1) (x-3) = 225 4, x = 21 2 or x = -9 22) 3 (x-2) x = 0, so x = 0 or x = 2 shift x -2x-3 = 0

i.e. (x-3)(x+1)=0

So x=3 or x=-1 choose c

The first problem x-3 = 15 2, the second problem x(3x-6) = 0, the third problem only has c, and the equation can be solved.