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The two roots of the quadratic equation x ^ 2-3x + 1 = 0 are X1 and X2 respectively. Then the value of X is a.3b. - 3c.1/3 D. - 1 / 3
If the question is right, there is no answer
The solution of x ^ 2-3x + 1 = 0 is: x = (3 + 5 ^ 0.5) / 2 or (3-5 ^ 0.5) / 2
1. If X1 and X2 are two of the univariate quadratic equations X & # 178; - 3x + 2 = 0, then the value of X1 + X2 is () A. - 2 B.2 C.3 D.1
The solution equation: (1) x & # 178; - 4x-3 = 0 (2) (x-3) &# 178; + 2x (x-3) = 0 (3) 4 (x-1) &# 178; = 36
Choose C to solve the equation: (1) x & # 178; - 4x-3 = 0 & nbsp; X & # 178; - 4x + 4 = 7 (X-2) & # 178; = 7X-2 = ± √ 7x1 = 2 + √ 7 & nbsp; & nbsp; & nbsp; x2 = 2 - √ 7 & nbsp; (2) (x-3) & # 178; + 2x (x-3) = 0 & nbsp; (x-3) (x-3 + 2x) = 0 (x-3) (3x-3) = 0x1 = 3 & nbsp; x2 = 1 & nbsp; (...)
If two of the quadratic equations x ^ 2 + 3x-1 = 0 are X1 and X2 respectively, then X1 + x2 = - 3, x1x2 = - 1, X1 ^ 2 + x2 ^ 2 =?
The quadratic equations of one variable with the reciprocal of two roots of equation 2x2-3x-7 = 0 as roots are
Let 2x2-3x-7 = 0 be x1, x2
Then X1 + x2 = 3 / 2
x1x2=-7/2
therefore
1/x1+1/x2=(x1+x2)/x1x2=-3/7
1/x1*1/x2=1/(x1x2)=-2/7
So it's X & # 178; + 3x / 7-2 / 7 = 0
That is 7x & # 178; + 3x-2 = 0
Univariate quadratic equation: (2x2 + 3x) 2-4 (2x2 + 3x) - 5 = 0
Because (2x2 + 3x) 2-4 (2x2 + 3x) - 5 = 0
Let 2x2 + 3x = a
Then 2a-4a-5 = o
So a = - 2.5
That is, 2x2 + 3x = - 2.5
Just work out X
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