If one of the roots of the square + root sign 3x + M = 0 of the quadratic equation x of X is 1-radical 3, then the value of M is The other root is? Two empty,

If one of the roots of the square + root sign 3x + M = 0 of the quadratic equation x of X is 1-radical 3, then the value of M is The other root is? Two empty,

analysis
According to Veda's theorem
x1+x2=-b/a=-√3
So the other root is - √ 3-1 + √ 3 = - 1
x1x2=c/a=m
x1x2=√3+1
So m = √ 3-1

Solving quadratic equation of one variable 3 (X-2) ^ 2-x (X-2) = 0 x ^ 2-2 root sign 3x + 2 = 0

3 (X-2) ^ 2-x (X-2) = 03 (x? - 4x + 4) - x? + 2x = 02x? - 10x + 12 = 02 (x? - 5x + 6) = 02 (X-2) (x-3) = 0x = 2 or x = 3 x ^ 2-2 root number 3x + 2 = 0 (x-radical 3) 2 + 2-3 = 0 (x-radical 3) mm2 = 1 x-radical 3 = ± 1 x = ± 1 + root 3

The quadratic equation with root 2 + 1 and root 3-1 is_____

Root 2 + 1, root 3-1
Then, by the Weida theorem
X1 + x2 = root 2 + root 3
X1. X2 = Radix 2 + 1 times Radix 3-1 = Radix 6-radix 2 + Radix 3-1
So the quadratic equation of one variable is x square - (root 2 + root 3) x + root 6-radical 2 + root 3-1 = 0

A quadratic equation with roots 3 + 2 and 3-2 as roots

Let the quadratic equation of one variable be x 2 + BX + C = 0, and its solution is X1 = √ 3 + 2, X2 = √ 3-2
According to the Weida theorem
x1+x2=-b
x1x2=c
X1 + x2 = √ 3 + 2 + √ 3-2 = 2 √ 3 = - B, then B = - 2 √ 3
X1x2 = (√ 3 + 2) (√ 3-2) = - 1 = C, then C = - 1
So the quadratic equation of one variable is x? - 2 √ 3x-1 = 0

Question: what is a quadratic equation of one variable with (radical 3) + 2 and (radical 3) - 2 as roots?

According to Veda's theorem (if the two roots of the equation about X are x1, X2, then the coefficient of the equation is expressed by roots: x ^ 2 - (x1 + x2) x + X1 * x2 = 0), we can get: x ^ 2 - {[(radical 3) + 2] + [(radical 3) - 2]} x + [(radical 3) + 2] * [(radical 3) - 2] can be simplified to x ^ 2 + 2 times (radical 3) X-1 = 0 if

Write a quadratic equation of one variable about x such that its two roots are root 3, - root 2 As the title

(x-radical 3) * (x + Radix 2) = 0
It can also be disassembled: X squared + (Radix 2-radix 3) x-radix 6 = 0

A quadratic equation with root 5 plus 2 and root 5 minus 2

The answer is: the square of x minus 2 times the root sign 5 plus 1 equals 0
Methods: (x-x1) multiplied by (x-x2) = 0, expand term by term!

Sine cosine function part 1. F (x) = the range of cos2x + SiNx at (- π / 2, π / 6) 2. If f (x) = sin (1 / 2x + π / 3), find the monotone increasing interval, the monotone increasing interval when x belongs to [- 2 π, 2 π], and the value range when x belongs to [0, π / 4]

1.f(x)=sinx+cos2x
=sinx+1-2sinx*sinx
=-2sinx * SiNx + SiNx + 1 x belongs to (- pi / 2, PI / 6)
Let SiNx = t; then t belongs to (- 1,0.5)
F (T) = - 2T * t + T + 1
There is a maximum of 9 / 8 at 1 / 4
Take the minimum value of - 2 at - 1
[the answer can be calculated by yourself, but the lotus root is not sure]
2. This is a simple sine curve
Monotonically increasing: 0.5x + pi / 3 belongs to (- 0.5pi + 2kpi, 0.5pi + 2kpi)
X belongs to (- 5pi / 3 + 4kpi, PI / 3 + 4kpi)
The monotone increasing interval of X when x belongs to [- 2 π, 2 π]: (- 5pi / 3, PI / 3)
The value range of X when it belongs to [0, π / 4]: [radical 3 / 2, sin (11pi / 24)]

Image of sine cosine function in senior one mathematics~ If M (π / 2, m) is on the image of the function y = SiNx, then M is equal to? And why?

On the graph of sin / M (m)
That is, x = π / 2, y = M
So m = sin π / 2 = 1

Sine cosine function of sum and difference of two angles in senior one mathematics How to calculate this? Sin20sin10-cos10sin70 I know that is equal to negative root three divided by two is not? But the calculation process can not work out, please help me

sin20sin10-cos10sin70
=sin20sin10-cos10cos20
=-(cos20cos10-sin20sin10)
=-cos(20+10)
=-cos30
=-√3/2