The root of X & # 178; + X-1 = 0 is obtained by formula method

The root of X & # 178; + X-1 = 0 is obtained by formula method

a=1,b=1,c=-1
∴x12=[-b±√(b²-4ac)]/2a
=[-1±√(1+4)]/2
∴x12=(-1±√5)/2
solution
x²+x-1=0
△=b²-4ac=1+4=5
Qi
x=(-1+√5)/2
Or x = (- 1 - √ 5) / 2
What is the arithmetical square root of 0.008? What is the value range of √ ~ 10 + 1? If the arithmetical square root of a number √ ~ 5, what is the number?
The arithmetic square root of 0.008 is 0.04 root sign 5
The value range of √ ~ 10 + 1 is greater than 4 and less than 5
If the arithmetic square root of a number is 5, then the number is 5
When m is a real number, the square - (1-m) x + M = 0 of the quadratic equation MX with respect to X has no real root?
Excuse me, everyone
-Find the solution set of inequality by the square of 3x + 5x-4x > 0 x (1-x) > x (2x-3) + 1
The square of MX - (1-m) x + M = 0
The equation has no real roots,
When m ≠ 0,
Discriminant = (1-m) ^ 2-4m * M0
(3m-1)(m+1)>0
-1> M or M > 1 / 3
When m = 0,
The equation becomes - x = 0 and x = 0 has real roots
To sum up:
m> 1 / 3 or MX (2x-3) + 1
x-x^2>2x^2-3x+1
3x^2-4x+1
Using discriminant = (1-m) square - 4m square or M
Solve equation 7200 (1 + x) ² = 12168 with formula method
Divide by 7200
1+2x+x²=1.69
x²+2x-0.69=0
a=1,b=2,c=-0.69
So △ = B & # 178; - 4ac = 6.76
So x = (- 2 ± 2.6) / 2
x1=-2.3,x=0.3
Given 2 / A + 1 / b = 1, find the minimum sum of the arithmetic square roots of (a + b) + (a ^ 2 + B ^ 2)
2/a+1/b=1>=2*sqrt(2/ab)
We can get: sqrt (AB) > = 2 * sqrt (2)
a+b+sqrt(a^2+b^2)>=2*sqrt(ab)+sqrt(2ab)>=4+4*sqrt(2)
If and only if a = B, the equal sign holds
So the minimum value is (4 + 4 * sqrt (2))
Why does the quadratic equation MX2 - (1-m) x + M = 0 have real roots when m is a real number?
In order to make the quadratic equation MX2 - (1-m) x + M = 0 of X have real roots, the discriminant △ = (1-m) 2-4m2 ≥ 0 is sorted out to be - 3m2-2m + 1 ≥ 0, that is, 3M2 + 2m-1 ≤ 0. The solution is - 1 ≤ m ≤ 13 and m ≠ 0. In conclusion, the value range of M is - 1 ≤ m ≤ 13 and m ≠ 0
Using formula method to solve equation (- 2 / 3) x & # 178; + (1 / 2) x + 1 = 0
（-2/3）x²+（1/2）x+1=0
-4x²+3x+6=0
4x²-3x-6=0
x=[3±√(9+96)]/8
=(3±√102)/8
If you don't understand, I wish you a happy study!
X1 = 3 + 3 / radical 17 x2 = 3-3 / radical 17
Does any real number have an arithmetic square root?
In the complex number range, any real number has an arithmetic square root;
In the real range, only nonnegative numbers have arithmetic square roots
No, both negative and positive numbers have square roots.
Upstairs is full of nonsense, negative numbers also have arithmetic square root, and it is imaginary. For example, the arithmetic square root of - 1 is I! I don't have to say more about non negative numbers.
No negative number
If the roots of the following two equations are 2x 2 + 2x 2 = 5x 1-2x 2, respectively
(1) Absolute value of (x1-x2) (2) X1 ^ 3 + x2 ^ 3 (3) 3x1 ^ 2 + x2 ^ 2 + 5x1
According to Weida's theorem: X1 + x2 = - 5 / 2, X1 * x2 = - 3 / 2, (1), (x1-x2) ^ 2 = (x1 + x2) ^ 2-4x1x2 = (- 5 / 2) ^ 2-4 * (- 3 / 2) = 49 / 4, x1-x2 = V (x1-x2) ^ 2 = 7 / 2; (2), X1 ^ 3 + x2 ^ 3 = (x1 + x2) ^ 3-3x1x2 (x1 + x2) = (- 5 / 2) ^ 3-3 * (- 3 / 2) * (- 5 / 2) = - 215 / 8; (3), 3x1 ^ 2
Factorization (2x-1) (x + 3) = 0, X1 = 1 / 2, X2 = - 3, the absolute value of x1-x2 is 7 / 2, 1 / 8 + (- 3) ^ 3 = - 215 / 8. Third, we need to discuss which value is 5x1 after the third question
Using the formula to calculate (X & # 178; - Y & # 178;) / (x + y)
(x²-y²)÷(x+y)
=(x-y)(x+y)÷(x+y)
=x-y