The root 3 is not a rational number

The root 3 is not a rational number

Suppose that root 3 is a rational number
Because 1

Rational number: - 1 / 2 + 1 / 3 - (- 1 / 9) - 1 / 4=

Original formula = - 3 / 4 + 4 / 9 = - 27 / 36 + 16 / 36 = - 11 / 36

(- 2 / 3) - (+ 1 / 12) - (- 1 / 4) rational number subtraction

(- 2 / 3) - (+ 1 / 12) - (- 1 / 4)
=-2/3-1/12+1/4
=-8/12-1/12+3/12
=-1/2
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-4 and 8 of 7 - (+ 4 and 4 of 1) - (- 5 and 2 of 1) + (- 3 and 8 of 1), rational number subtraction, urgent

It is known that the opposite number of rational number x is 1 and 2 / 3, and the reciprocal of rational number y is - 5 / 2;

We can get x = - 5 / 3, y = - 2 / 5, so x / y = 25 / 6

Given that x 1 and x 2 are two of the equations x 2 + P x + q = 0, and x 1 + x 2 = 6, x 12 + x 22 = 20, find the values of P and Q

∵ X1 and X2 are two of the equations x2 + PX + q = 0. ∵ P = (x1 + x2) = - 6. X1x2 = 12 [(x1 + x2) 2 - (X12 + X22)] = 12 (36-20) = 8. ∵ △ = p2-4q = (- 6) 2-4 × 8 = 4 > 0. The equation has real roots, so p = - 6, q = 8

Junior three mathematics (one variable quadratic equation) speed!
It is known that the ratio of the two quadratic equations ax ^ 2 + BX + C = 0 (a ≠ 0) with respect to X is 2:1. It is proved that 2B ^ 2 = 9ac

2 * X1 = x2
Weida theorem
x1+x2=-b/a
x1*x2=c/a
So 3 * x2 = - B / A
2*x2^2=c/a
LIANLI, eliminate x2
9/2=b^2/ac
That is, 2b ^ 2 = 9ac

Write out the negative integer solution of bivariate linear equation 2x + 3Y = - 18

2x+3y=18
x=0
y=6
x=9
y=0
x=3
y=4
OK~
x=6
y=2

Find the nonnegative integer solution of the quadratic equation 2x + 3Y = 14

X = 1y = 4 and x = 4Y = 2

The nonnegative integer solution of quadratic equation 2x + 3Y = 24 is obtained

2x+3y=24
y=8 x=0
Y = 7, x = not integer
y=6 x=3
Y = 5, x = not integer
y=4 x=6
Y = 3, x = not integer
y=2 x=9
Y = 1, x = not integer
y=0 x=12
So it's a nonnegative integer
x=0 y=8
x=3 y=6
x=6 y=4
x=9 y=2
x=12 y=0