Is the arithmetic square root of 2 a real number And answer questions A x2 + 4x + 5 = root 2 / 2 has real root Bx2 + 4x + 5 = root 3 / 2 has real root CX2 + 4x + 5 = root 5 / 2 has real root DX2 + 4x + 5 = a (a ≥ 1) has real roots That's right?

Is the arithmetic square root of 2 a real number And answer questions A x2 + 4x + 5 = root 2 / 2 has real root Bx2 + 4x + 5 = root 3 / 2 has real root CX2 + 4x + 5 = root 5 / 2 has real root DX2 + 4x + 5 = a (a ≥ 1) has real roots That's right?

It's a real number
It's the fourth root of two
X1 and X2 are the two parts of the quadratic equation 2x ^ 2 + 5x-2 = 0. Find the following values
1.|X1-X2| 2.1/X1^2+1/X2^2 3.X1^3+X2^3
X1 and X2 are the two parts of the quadratic equation 2x ^ 2 + 5x-2 = 0. Find the following values
x1+x2=-2.5,x1x2=-1
1.|X1-X2|
=Root (6.25 + 4)
=0.5 root 41
2.1/X1^2+1/X2^2
=（6.25+2）/1
=8.25
3.X1^3+X2^3
=(x1+x2)(x1^2-x1x2+x2^2)
=(-2.5)*(6.25+3)
=-23.125
X (x-1) - 1 = 0 formula calculation
Calculation? Should be to solve the equation? If it is: X (x-1) - 1 = 0 (x-1 / 2) ^ 2-1 / 4-1 = 0 (x-1 / 2) ^ 2 = 5 / 4x-1 / 2 = ± √ 5 / 2x = (1 ± √ 5) / 2x1 = (1 + √ 5) / 2, X2 = (1 - √ 5) / 2. If the formula must be used: X (x-1) - 1 = 0 (x-1 / 2) ^ 2-x-1 = 0 {- (- 1) ± √ [(- 1) ^ 2-4 × 1 × (-)
If the sum of squares of an integer 2x-5 is one
1. Find the value of a and X
2. Find the cube root of - A-X + 1
1、a=5²=25
1-2x=-5
X=3
2、-a-x+1
=-25-3+1
=-27
The square root of a = 5, so a = 25
because
Square root has positive and negative
therefore
5=-（1-2x）
X=3
So a = 25 x = 3
Substituting - 25-3 + 1 = - 27
So its cube root = - 3
(1) 5 = 1-2x, x = - 2, so a = 25
(2) -25-(-2)+1=-25+2+1=-22
If X1 and X2 are two of the quadratic equations 2x ^ 2 + 5x-3 = 0 respectively. (1) find the value of | x1-x2 | (2) find the value of 1 / X1 ^ 2 + 1 / x2 ^ 2 (3) find the value of X1 ^ 3 + x2 ^ 3
Solve the equation X1 = - 3; x2 = 1 / 2
(1) lx1-x2l=7/2
(2) 1/x1^2+1/x2^2=37/9
(3) x1^3+x2^3=215/27
(3x + 2) ^ 2 = 4 (x-3) ^ 2
（3x+2)^2=4(x-3)^2
（3x+2)^2-4(x-3)^2=0
(3x+2+2x-6)(3x+2-2x+6)=0
(5x-4)(x+8)=0
X = 4 / 5 or x = - 8
2=4(x-3)^2
（3x+2)^2-4(x-3)^2=0
(3x+2+2x-6)(3x+2-2x+6)=0
(5x-4)(x+8)=0
（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
(5x-4)(x+8)=0
x1=0.8, x2=-8
Find the square root of √ 2x-1 + √ 1-2x + 3
√ 2x-1 + √ 1-2x + 3 is meaningful
Necessary 2x-1 > = 0,1-2x > = 0
==>X=1/2
So the square root of √ 2x-1 + √ 1-2x + 3 = + / - 3
1. The same as above: let X1 and X2 be the two roots of the quadratic equation 2x ^ 2-5x + 1 = 0, solve the equation, and use the relationship between root and coefficient to find the value of (x1) ^ 2 + (x2) ^ 2
2. Proof: no matter what the value of K is, the equation x ^ 2 - (2k-1) x + K-3 / 4 = 0 must have two real roots
(please be clear)
One
X1, X2 are the two roots of the quadratic equation 2x ^ 2-5x + 1 = 0
x1+x2=5/2 x1*x2=1/2
(x1)^2+(x2)^2
=(x1+x2)^2-2x1x2
=(5/2)^2-2*1/2
=25/4-1
=21/4
Two
Using discriminant
△=b^2-4ac
=(2k-1)^2-4*1*(k-3/4)
=4k^2-4k+1-4k+3
=4k^2-8k+4
=4(k^2-2k+1)
=4(k-1)^2>=0
So there must be two real roots (including two of the same real roots)
1. From the relationship between root and coefficient, we can get
x1+x2=-b/a =5/2 x1*x2=c/a=1/2
x1*x1+2x*x2=（x1+x2）^2-2x1*x2=25/4-1=21/4
2. The discriminant of equation x ^ 2 - (2k-1) x + K-3 / 4 = 0 is b * b-4ac = (2k-1) ^ 2-4 * (- 3 / 4) * 1 = = (2k-1) ^ 2 + 1 > 0
So no matter what the value of K is, the equation x ^ 2 - (2k-1) x + K-3 / 4 = 0 must have two real roots
1、X1+X2=5/2
X1X2=1/2
∴X1²+X2²
=(X1+X2)²-2X1X1
=(5/2)²-2×1/2
=25/4-1
=21/4
2、△=[-(2K-1)]²-4×1×(K-3/4)
=4K²-4K+1-4K+3
=4K... Unfold
1、X1+X2=5/2
X1X2=1/2
∴X1²+X2²
=(X1+X2)²-2X1X1
=(5/2)²-2×1/2
=25/4-1
=21/4
2、△=[-(2K-1)]²-4×1×(K-3/4)
=4K²-4K+1-4K+3
=4K²-8K+4
=(2K-2)²≥0
Whatever the value of K, the equation x ^ 2 - (2k-1) x + K-3 / 4 = 0 must have two real roots. Put it away
X1, X2 are the two roots of the quadratic equation 2x ^ 2-5x + 1 = 0
x1+x2=5/2 x1*x2=1/2
(x1)^2+(x2)^2
=(x1+x2)^2-2x1x2
=(5/2)^2-2*1/2
=25/4-1
=21/4
Two
Using discriminant
△=b^2-4ac
=(2k-1)^2-4*1*(k-3/4)
=4k^2-4k+1-4k+3
=4k^2-8k+4
=4(k^2-2k+1)
=4(k-1)^2>=0
One
x1+x2=-b/a=5/2
x1x2=1/2
（x1）^2+（x2）^2=(x1+x2)^2-2x1x2=25/4-1=21/4
Two
△=b²-4ac
=(2k-1)^2-4(k-3/4)
=4k^2-4k+1-4k+3
=4(k^2-2k+1)
=4(k-1)^2>=0
When it is greater than 0, there are two different real roots
When equal to 0, there are two identical real roots
1. x1+x2=-(-5)/2=5/2
x1*x2=1/2
(x1)^2+(x2)^2=(x1+x2)^2-2x1*x2=25/4-2*1/2=21/4
2. The discriminant of the equation (- (2k-1)) ^ 2-4 * 1 * (K-3 / 4), as long as it can be proved that the value of the city is not less than or equal to zero.
(- (2k-1)) ^ 2-4 * 1 * (K-3 / 4) = 4 * (k-1) ^ 2... Expansion
1. x1+x2=-(-5)/2=5/2
x1*x2=1/2
(x1)^2+(x2)^2=(x1+x2)^2-2x1*x2=25/4-2*1/2=21/4
2. The discriminant of the equation (- (2k-1)) ^ 2-4 * 1 * (K-3 / 4), as long as it can be proved that the value of the city is not less than or equal to zero.
(-(2k-1))^2-4*1*(k-3/4)=4*（k-1）^2
No matter what value k is, the result of the discriminant is a real number greater than or equal to zero. Put it away
3x (x-1) = 2 (x-1)
I don't concentrate in class
3x(x-1)=2(x-1)
3x(x-1)-2(x-1)=0
(x-1)(3x-2)=0
X-1 = 0 or 3x-2 = 0
x1=1 x2=2/3
3x(x-1)=2(x-1)
3x=2
x=3/2
Transfer term: 3x (x-1) - 2 (x-1) = 0, put forward the common factor: (x-1) (3x-2) = 0, and get x = 1 or 2 / 3
If the square root of an integer is 2a-1 and - A + 2, then a =, the integer is
It's a positive number
2a－1=－(－a+2)
a=－1
The positive number is nine
If two values are equal
2a-1=-a+2
A=1
The positive number is one
2a－1=－(－a+2)
a=－1
The whole number is 25