Let X1 and X2 be the two roots of the quadratic equation 2x ^ 2-5x + 1 = 0, and find the value of X1 ^ 3 + x2 ^ 3

Let X1 and X2 be the two roots of the quadratic equation 2x ^ 2-5x + 1 = 0, and find the value of X1 ^ 3 + x2 ^ 3

x1^3+x2^3=（x1+x2)(x1^2-x1x2+x2^2)=（x1+x2)(x1^2+2x1x2+x2^2-3x1x2)=（x1+x2)((x1^+x2^)2-3x1x2))=(-c/a)((-c/a)2-3b/a)=95/8
Because X1 and X2 are the two roots of the quadratic equation 2x ^ 2-5x + 1 = 0, so
x1 + x2 = -(-5) / 2=2.5
x1* x2 =1/2=0.5
x1^3 + x2^3 = (x1 + x2) (x1^2 - x1*x2 + x2 ^2)
=(x1... Unfold)
Because X1 and X2 are the two roots of the quadratic equation 2x ^ 2-5x + 1 = 0, so
x1 + x2 = -(-5) / 2=2.5
x1* x2 =1/2=0.5
x1^3 + x2^3 = (x1 + x2) (x1^2 - x1*x2 + x2 ^2)
=(x1 + x2) (x1^2 +2 x1*x2 + x2 ^2 -3 x1*x2 )
=(x1 + x2) [ (x1+ x2) ^2 -3 x1*x2 ]
=2.5 * [ 0.5^2 - 3* 2.5]
=2.5*(-0.5)
=-Thank you very much..
Math: factorize the equation: 3x ^ 2 + 8x-3 = 0
(x + 3) (3x-1) = 0, so x = - 3 or x = 1 / 3
（3x-1）(x+3)=0
The left formula of the original equation is 3x & # 178; + 9x - (x + 3) = (3x-1) (x + 3), so the solution of the original equation is X1 = - 3, X2 = 1 / 3.
The two square roots of an integer are 3 - A and 2A + 7 respectively
Pay attention to the format. You math experts help
prove:
Because (3-A) + (2a + 7) = 0
3-a+2a+7=0
10+a=0
So a = 0-10
=-10
PS: the sum of the two square roots should be 0, so the first step (3-A) + (2a + 7) = 0 appears
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Because (3-A) + (2a + 7) = 0
3-a+2a+7=0
10+a=0
So = - 10
prove:
Because (3-A) + (2a + 7) = 0
3-a+2a+7=0
10+a=0
So a = 0-10
=-10
PS: the sum of the two square roots should be 0, so the first step (3-A) + (2a + 7) = 0 appears
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Let X1 and X2 be the two roots of the quadratic equation of one variable 2x ^ 2-5x + 1 = 0. Using the relationship between the roots and the coefficients, we can find the values of the following formulas
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
The following equations are solved by factorization: (1) x2 + 16x = 0; (2) 5x2-10x = - 5; (3) x (x-3) + x-3 = 0; (4) 2 (x-3) 2 = 9-x2
(1) The original equation can be transformed as follows: X (x + 16) = 0, x = 0 or x + 16 = 0. X 1 = 0, x 2 = - 16. (2) the original equation can be transformed as x2-2x + 1 = 0, (x-1) 2 = 0. X 1 = x2 = 1. (3) the original equation can be transformed as (x-3) (x + 1) = 0, x-3 = 0 or x + 1 = 0. X 1 = 3, x 2 = - 1. (4) the original equation can be transformed as 2
If the square root of an integer is 2a-1 and - A + 2, then a=______
1 or negative 1
One
Whether - 3,2,2 / 3 is the root of the quadratic equation 2x ^ 2 + 5x-5 = 1-2x-x ^ 2
2x^2+5x-5=1-2x-x^2
3x^2+7x-6=0
（3X-2）（X+3）=0
X = 2 / 3 or x = - 3
So - 3,2 / 3 is the root of the quadratic equation 2x ^ 2 + 5x-5 = 1-2x-x ^ 2
Solving the equation x & sup2; - M (3x-2m + n) - N & sup2; = 0 with formula method
x^2-3mx+2m^2-mn-n^2=0
△=9m^2-4(2m^2-mn-n^2)=(m+2n)^2
So the root is
x1=(3m+m+2n)/2=2m+n
x2=(3m-(m+2n))/2=m-n
If the square roots of a positive number x are 2a-3 and 5-a respectively, then a=______ ，x=______ .
The square roots of ∵ positive number x are 2a-3 and 5-a, respectively, and ∵ 2a-3 + 5-a = 0. The solution is a = - 2, ∵ 5-a = 5 - (- 2) = 7, x = 72 = 49. So the answer is: - 2; 49
For the quadratic equation kx2-6x + 1 = 0 with one variable of X, if there are two unequal real roots, then the value range of K is ()
A. K ≥ 9b. K < 9C. K ≤ 9 and K ≠ 0d. K < 9 and K ≠ 0
According to the meaning of the question, we get (- 6) 2-4k ＞ 0 and K ≠ 0, and the solution is k ＜ 9 and K ≠ 0