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It supports mathematical functions (including trigonometric functions).
Current location：Mathematical operation > History of Mathematical Computation > Answer
Overview: 1 questions will be solved this time.Among them
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation （5+2t）（5+2t）-4（4t+6） = 0 .
Question type: Equation
Solution:Original question:
Remove the bracket on the left of the equation:
The equation is transformed into :
After the equation is converted into a general formula, it is converted into:
( 2t + 1 )( 2t + 1 )=0
From
2t + 1 = 0
2t + 1 = 0
it is concluded that:：
There are 2 solution(s).
解一元二次方程的详细方法请参阅：《一元二次方程的解法》
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation （5+2t）（5+2t）-4（4t+6） = 0 .
Question type: Equation
Solution:Original question:
| ( | 5 | + | 2 | t | ) | ( | 5 | + | 2 | t | ) | − | 4 | ( | 4 | t | + | 6 | ) | = | 0 |
| Left side of the equation = | 5 | ( | 5 | + | 2 | t | ) | + | 2 | t | ( | 5 | + | 2 | t | ) | − | 4 | ( | 4 | t | + | 6 | ) |
| = | 5 | × | 5 | + | 5 | × | 2 | t | + | 2 | t | ( | 5 | + | 2 | t | ) | − | 4 | ( | 4 | t | + | 6 | ) |
| = | 25 | + | 10 | t | + | 2 | t | ( | 5 | + | 2 | t | ) | − | 4 | ( | 4 | t | + | 6 | ) |
| = | 25 | + | 10 | t | + | 2 | t | × | 5 | + | 2 | t | × | 2 | t | − | 4 | ( | 4 | t | + | 6 | ) |
| = | 25 | + | 10 | t | + | 10 | t | + | 4 | t | t | − | 4 | ( | 4 | t | + | 6 | ) |
| = | 25 | + | 20 | t | + | 4 | t | t | − | 4 | ( | 4 | t | + | 6 | ) |
| = | 25 | + | 20 | t | + | 4 | t | t | − | 4 | × | 4 | t | − | 4 | × | 6 |
| = | 25 | + | 20 | t | + | 4 | t | t | − | 16 | t | − | 24 |
| = | 1 | + | 4 | t | + | 4 | t | t |
| 1 | + | 4 | t | + | 4 | t | t | = | 0 |
After the equation is converted into a general formula, it is converted into:
( 2t + 1 )( 2t + 1 )=0
From
2t + 1 = 0
2t + 1 = 0
it is concluded that:：
| t1= | - | 1 2 |
| t2= | - | 1 2 |
There are 2 solution(s).
解一元二次方程的详细方法请参阅：《一元二次方程的解法》