Junior 3 Mathematics Binary Equation Formula Method . Urgently.

First of all, correct, this is a question about a quadratic equation...

x^2+2(k-1)x+k^2-1=0

There are two unequal real roots, then:

4(k-1)^2-4(k^2-1)＞0

4k^2-8k+4-4k^2+4＞0

8k＜8k＜1

x=0. 0+0+k^2-1=0

k 2 = 1k = 1 or k = -1

That is, when k=-1, x=0 is one of the equations.

The equation in this case is:

x^2-4x=0

x(x-4)=0

x=0 or x=4

The other root of the equation is x=4

Solution: (1) Since the equation has two unequal real roots, it is: [2(k-1)] 2-4*(k 2-1)>0

2k-2)^2-4k^2+4>0

4k^2-8k+4-4k^2+4>0

8-8k>0

k<12) substituting x=0 into the equation obtains: k 2-1=0, then k=1, which is inconsistent with the condition that the original equation has two unequal real roots, then x=0 is not the root of the equation.

It's very simple, but the mobile phone is not convenient, roughly speaking, for 1, b squared 4ac is greater than 0, and then the result of the solved inequality is what is sought. 2, first bring in 0, find the k value, and then use the relationship between the root and the coefficient to find the other root.

x^2+2(k-1)x+k^2-1=0

Discriminant = 4(k-1) 2-4(k 2-1)4(k 2-2k+1)-4k 2+4

8-8k>0

Get k<1

Probably, because the product of the two roots is k 2-1

Since 0 is a heel, k -1=0

k<1

So k=-1

Then the equation is. x^2-4x=0

x(x-4)=0

So x=0 or x=4

So the other one follows as 4

The general formula of the binary equation is: ax +bx+c=0, where: a 0 (if the equation a 0 is given, simply multiply -1 on both sides of the equal sign to make a 0). Yes

ax²+bx+c=0

x²+(b/a)x+c/a=0

x²+2×[b/(2a)]x+c/a=0

x²+2×[b/(2a)]x+[b/(2a)]²b/(2a)]²c/a=0

x²+2×[b/(2a)]x+[b/(2a)]²=[b/(2a)]²c/a

x+b/(2a)]²=b²/(2a)²-4ac/(2a)²[x+b/(2a)]²=(b²-4ac)/(2a)²x+b/(2a)=±√[b²-4ac)/(2a)²]x+b/(2a)=±[√b²-4ac)]/(2a)x=-b/(2a)±[b²-4ac)]/(2a)x=[-b±√(b²-4ac)]/(2a)

ax^2+bx+c=0.(a≠0, 2 is squared) divide both sides of the equation by a, get, x 2+bx a+c a=0, shift the term, get:

x 2 + bx a = c a, add half of the square of the primary term coefficient b a on both sides of the equation, that is, add b 2 4a 2 on both sides of the equation, (formula).

x 2+bx a+b 2 4a 2=b 2 4a 2 c a, i.e. (x+b 2a) 2=(b 2-4ac) 4a

x+b/2a=±[√b^2-4ac)]/2a.( indicates the root number) gets:

x=[-b±√(b^2-4ac)]/2a.

The general formula ax 2+bx+c=0 (a, b, c are the real numbers a≠0) For example: x 2+2x+1=0

Matching method: a(x+b 2a) 2=(b 2-4ac) 2a, two-piece a(x-x1)(x-x2)=0

The vertex formula a(x-h) 2+k=0 (h is not equal to x, a is not 0).

x=（-b±√b²-4ac）/2a

where b -4ac 0

a, b, and c are quadratic coefficients, primary coefficients, and constant terms, respectively.

1. (x-m)*(x-2m)=0, x1=m, x2=2m, because x1+x2=3, that is, 3m=3, m=1

2、-3x^2+2x÷5=-1/5*（15x^2-2x)=-1/5[(√15x-1/√15)^2-1/15]

1 5 ( 15x-1 15) 2+1 75, so the maximum unjudged rise value is 1 75

Set the price of x yuan, and sell y down jackets in October.

xy=28000

The solution process is x=1400 y=20

Set to sell x pieces in December.

1400*>9940-20*200-30*100x=49, so at least 49+20+30 pieces.

99 pieces.

First of all, you have a miscalculation: 880 + 220x-20x + 5x 2 = 1600, which should be: 880 + 220x-20x-5x 2 = 1600

Then, 880 + 200 x - 5 x 2 = 1600176 + 40x - x 2 = 320

x^2-40x+144=0

x-36）（x-4）=0

solution, x = 36 or 4

The unary quadratic equation of junior high school mathematics is solved 880+200x+5x 2=1600.

Solution: 5x 2+200x=720

x^2+40x=144

x^2+40x+400=144+400

x+20)^2=544

x+20 = plus or minus (4 roots number 34).

x1=-20+4 root number 34, x2=-20-4 root number 34

The second step was written incorrectly, it should be.

880+220x-20x-5x²=1600-5x²+200x-720=0

x²-40x+144=0

x-4）（x-36）=0

x=4 or x=36