Let X1 and X2 be the two roots of the equation 2x + 4x-3, then (x1 + 1) (x2 + 1) and X1 + x2

Let X1 and X2 be the two roots of the equation 2x + 4x-3, X1 + x2 = - 4 / 2 = - 2x1x2 = - 3 / 2 (x1 + 1) (x2 + 1) = x1x2 + X1 + x2 + 1 = - 3 / 2-2 + 1 = - 5 / 2x12 + X22 = (x1 + x2) 2-2x1x2 = (- 2) 2 + 2x3 / 2 = 4 + 3 = 7

RELATED INFORMATIONS

1. Weida theorem | x1-x2| X1 and X2 are two steps of the quadratic equation 2x ^ + 5x-3 = 0 with one variable. The process of finding the value of | x1-x2 | is written by WIDA's theorem
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5. If x = 1, the value of cube + 3bx + 4 of algebraic formula 2aX is 5, then what is the value of cube + 3bx + 4 of algebraic formula 2aX when x = - 1?
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11. If the square of the quadratic equation 2x minus 4x plus 3 = 0 is x1, X2, then x2 divided by X1 plus X1 divided by x2 =? .
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17. The absolute value of solving equation - 12 + 3x = 18 + 2x
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20. The sign | - 2 |, denotes that the absolute value of - 2 is 2, and | + 2 |, denotes that the absolute value of + 2 is 2. If | - x | = 2, then x = 2 or x = - 2. If the equation | - X-1 | = 2 is solved, then X-1 = 2 or X-1 = - 2 can be obtained, and x = 3 or x = - 1 can be obtained by solving the equation respectively. Using the above knowledge, the equation can be solved as follows: | - 2x-1-7 = 0