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The more application problems of quadratic equation in one variable, the better, and the sooner the better. It is best to sell the problem in detail.
answer

1. solution: let Xiao Peng raise x pigeons. then

(X+ 1)^2=9(X+ 1)

The solution is X=8 or X=- 1 (truncation).

So Xiao Peng raised eight pigeons.

2. Solution: Assuming that the single digit with two digits is X, then

X^2= 10(X-3)+X

The solution is X=6 or X=5.

So these two digits are 36 or 25.

3. Solution: The price of the second year is 25*( 1-0.2)=20W.

Let the depreciation rate of 2.3 years be x.

Then 20 * (1-x) 2 = 16.2.

The solution is X= 10.

So the depreciation rate is 10%.

4. Solution: It makes sense if the selling price is set at X yuan (X is a number greater than 50).

[500-(X-40)* 10]*X=8000

The solution is X= 10 or X=-80 (truncation).

So the price is 60 yuan, and the stock is 500-(60-50)* 10=400.

When an old man dies, he will distribute his inheritance in the following ways: the eldest gets 100 francs and the remaining tenth, the second gets 200 francs and the remaining tenth, the third gets 300 francs and the remaining tenth, and so on. Finally, it is found that all children will get equal inheritance after all inheritance is completed. Seek the total amount of inheritance, the number of children, and the amount of inheritance received by each child.

Let's assume that the total amount of the estate is X francs, the eldest brother gets100+110 * (x-100), and the second son gets 200+10 * (x-/).

Because the eldest brother and the second brother share the same amount, they have to

100+ 1/ 10 *(X- 100)= 200+ 1/ 100 *(X- 100)- 1/ 10 *(X- 100)-200)

The result is X=8 100 (francs).

The number of francs allocated to each child is

100+ 1/ 10 *(X- 100)= 100+ 1/ 10 *(8 100- 100 ...