According to scientific analysis, the stage host should stand on the golden section of the front of the stage. Given that the front of the stage is 20 meters, where should the host stand?

According to scientific analysis, the stage host should stand on the golden section of the front of the stage. Given that the front of the stage is 20 meters, where should the host stand?

618, a total of two positions, one on the left and one on the right
And the golden section is more accurate than √ (5) - 2
Using factorization to solve the following equations is a fast and good Bonus!
1.（3x+1）^2=4（x-2）^2
2.（x-1）^2+4（x-1）+4=0
3.（3x+2）^2=4（x-3）^2
^This is the meaning of square, should understand it!
(3x＋1)²－(2x－4)²＝0(3x＋1＋2x－4)(3x＋1－2x＋4)＝0(5x－3)(x＋5)＝0x1＝3/5,x2＝－5(x－1＋2)²＝0(x＋1)²＝0x1＝x2＝－1(3x＋2)²－(2x－6)²＝0(3x＋2＋2x－6)(3x＋2－2x＋6)＝0(...
1..（3x+1）^2=4（x-2）^2
(3x+1-2x+4)(3x+1+2x-2)=0 x1= -5 x2=1/5
2..（x-1）^2+4（x-1）+4=0
(x-1+2)^2=0 x1=x2=-1
3.(3x+2）^2=4（x-3）^2 (3x+2）^2-4（x-3）^2=0
(3x+2-2x+6)(3x+2+2x-6)=0 x1=-8 x2=4/5
The analysis shows that: 4 (X-2) ^ 2 = [2 (X-2)] & sup2; = (2X-4) & sup2;, after the shift, using the square difference formula, 4 (x-3) ^ 2 is equal to (2x-6) & sup2;
One
(3x+1+2x-4)(3x+1-2x+4)=0
(5x-3)(x+5)=0
x1=3/5 x2=-5
2.(x-1+2)²=0
... unfold
The analysis shows that: 4 (X-2) ^ 2 = [2 (X-2)] & sup2; = (2X-4) & sup2;, after the shift, using the square difference formula, 4 (x-3) ^ 2 is equal to (2x-6) & sup2;
One
(3x+1+2x-4)(3x+1-2x+4)=0
(5x-3)(x+5)=0
x1=3/5 x2=-5
2.(x-1+2)²=0
x-1+2=0
x=-1
3.(3x+2+2x-6)(3x+2-2x+6)=0
(5x-4)(x+8)=0
X1 = 4 / 5 x2 = - 8 * fold up
（3x+1）^2=4（x-2）^2
（3x+1）^2-4（x-2）^2=0
[（3x+1）-2（x-2）][（3x+1）+2（x-2）]=0
(3x+1-2x+4)(3x+1+2x-4)=0
(x+5)(5x-3)=0
X = - 5 or x = 3 / 5
（x-1）^2+4（x-1）+4=0
(x-1+2)^2=0
(x+1)^2=0
X =... Expand
（3x+1）^2=4（x-2）^2
（3x+1）^2-4（x-2）^2=0
[（3x+1）-2（x-2）][（3x+1）+2（x-2）]=0
(3x+1-2x+4)(3x+1+2x-4)=0
(x+5)(5x-3)=0
X = - 5 or x = 3 / 5
（x-1）^2+4（x-1）+4=0
(x-1+2)^2=0
(x+1)^2=0
x=-1
（3x+2）^2=4（x-3）^2
（3x+2）^2-4（x-3）^2=0
[（3x+2）-2（x-3）][（3x+2）+2（x-3）]=0
(3x+2-2x+6)(3x+2+2x-6)=0
(x+8)(5x-4)=0
X = - 8 or x = 4 / 5
Given that line AB = 10, point C is the golden section point of line AB, and AC > BC, calculate the length of AC and BC
The ratio of golden section is an irrational number, which is expressed as (√ 5-1) / 2 by fraction. The approximate value of the first three digits is 0.618. This value can be used directly
So AC = 10x0.618 = 6.18, BC = 10-6.18 = 3.82
Solving an equation by factorization
2(x-2)^2=4-x^2
2(x-2)^2=4-x^2
2*(x^2-4x+4)=4-x^2
2x^2-8x+8-4+x^2=0
3x^2-8x+4=0
(3x-2)*(x-2)=0
X = 2 / 3 or x = 2
Given that line AB = 10, C and D are two golden section points on AB, the length of line CD is calculated
AC=0.618*AB=6.18 BD=0.618*AB=6.18 BC=AD=10-AC=3.82 CD=AB-AD-BC=10-3.82-3.82=2.36
Solve the following equation (process) by factorization
3X（2X+1）=4X+2 （X-4）²=（5-2X）² 3X（2X+1）=4X+2 （X-1）²=（5-2X）²
2X²+1=2√3 X X（2X-3）=（3X+2）（2X-3） （X-1）²（X²-1）=0 X²-6X+9=（5-2X）²
3x (2x + 1) = 4x + 2 6x ^ 2 + 3x = 4x + 26x ^ 2-x-2 = 0 (3x-2) (2x + 1) = 0 x = 2 / 3 or x = - 1 / 2 (x-4) & # 178; = (5-2x) & # 178; X ^ 2-8x + 16 = 4x ^ 2-20x + 25X ^ 2-4x + 3 = 0 (x-3) (x-1) = 0 x = 3 or x = 1 (x-1) & # 178; = (5-2x) & # 178; X ^ 2-2x + 1 = 4x ^ 2-20x + 25X ^ 2-6x +
If AB = 2, point C is the golden section of line AB, and CD = 1, then ab=
There are two possibilities for the title to be correct~
AB = root 2
LZ, there is something wrong with your topic. There's a D spot out of thin air, and its location is not known at all.
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Check whether the title is wrong, or make the position clear. Ask me again. I will answer for you.
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Two
The following equations are solved by factorization
1 ：（X+3)(2X-5)=0 2 :X²+9X=0 3：（2X+3）²=2X+3
4：X²-4X=4=2（X-2）
Junlang lieying team answers for you
1 ：（X+3)(2X-5)=0
X + 3 = 0 or 2x-5 = 0
X1=-3,X2=5/2；
2 : X²+9X=0
X(X+9)=0
X1=0,X2=-9
3： （2X+3）²=2X+3
(2X+3)［(2X+3)-1］=0
X1=-3/2,X2=-1
4： X²-4X=4=2（X-2）
X(X-2)-2(X-2)=0
(X-2)^2=0
X1=X2=2.
Points c and D are the two golden section points of line ab. if CD = 5, then ab=_____
The answer is 5 × 5,
Let AC be x, because C and D are the golden section points of AB, so CA = BD, X / (x + 5) = (5-1) / 2 under the root, then we can solve x, ab = 5 × 5 under the root
5 times root 5 + 10
5 / [(root 5-1) / 2 - (3-root 5) / 2] = 5 times root 5 + 10
Because: the longer segment is (root 5-1) / 2 of ab
... unfold
5 times root 5 + 10
5 / [(root 5-1) / 2 - (3-root 5) / 2] = 5 times root 5 + 10
Because: the longer segment is (root 5-1) / 2 of ab
The shorter segment is (3-radical 5) / 2 of ab
So CD is ab [(radical 5-1) / 2 - (3-radical 5) / 2]
Use factorization to solve the following equations
⑴ 2x²+5x=0 ⑵(y-1)(y+3)=-4
⑶x(x-2)=4-x² ⑷√3x(x-1)-√2(x-1)=0
⑸4(x-1)²=25(x+1)² ⑹√2x(x+1)=x+1
⑺x²-9 =1/2(x²-6x+9）
1) Left = x (2x + 5) = 0, x = 0 or 2x + 5 = 0, x = - 5 / 22) left right = y ^ 2 + 2Y + 1 = (y + 1) ^ 2 = 0, y = - 13) right = (2 + x) (2-x), x = 2 is a solution, or both sides are approximately divided, - x = 2 + X, x = - 14) left = (x-1) (√ 3x - √ 2) = 0, x = 1, or x = √ 6 / 35) left right square difference = (7x + 3) (- 3x