Completing the square and quadratic structure
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| completed square/kəmˈpliːtɪd skweə/ | dạng bình phương hoàn chỉnh |
Which form tells us where it happens without drawing every point?
- A quadratic model reaches a lowest value.
- This lesson studies completed square 配方形式: A quadratic written as a multiple of a squared bracket plus a constant.
The method
- For x²+bx, add and subtract (b/2)² to obtain (x+b/2)²−(b/2)².
- If the leading coefficient a is not 1, factor a from the x² and x terms first.
- The form a(x−h)²+k gives vertex (h,k), axis x=h, and a minimum k if a>0 or a maximum k if a<0.
For 2x²-12x+11, why is the minimum -7 at x=3 rather than at x=6?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Halving the coefficient must happen INSIDE the factored bracket: x²-6x halves to (x-3)². The square is nonnegative, so the minimum is the constant -7 at x=3.
A worked case
2x²−12x+11=2(x²−6x)+11=2[(x−3)²−9]+11=2(x−3)²−7. Its minimum is −7 at x=3. To solve 2x²−12x+11=0, obtain (x−3)²=7/2, so x=3±√(7/2). The roots lie equally far from x=3. In general ax²+bx+c=0 becomes (x+b/(2a))²=(b²−4ac)/(4a²), for a≠0, leading to x=(−b±√(b²−4ac))/(2a). A negative discriminant means no real roots; zero gives one repeated root.
Completing the square and quadratic structure
For x²+bx, add and subtract (b/2)² to obtain (x+b/2)²−(b/2)²
Find the first valid or invalid step in a quadratic method.
Find the x-coordinate of the minimum of 2x²−12x+11.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The vertex is x=−b/(2a)=12/4=3.
Where it goes wrong
- Half the coefficient inside the factored bracket, not the original coefficient of x.
- Keep the outside multiplier when subtracting the added square.
- A positive squared bracket does not imply the whole quadratic is positive: its constant can be negative.
Not true: A completed-square expression represents a quadratic with a minimum. When the coefficient of x² is negative the parabola has a maximum, not a minimum.
Find that minimum value.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At x=3, y=18−36+11=−7.
When the coefficient of x² is negative the parabola has a maximum, not a minimum.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. When the coefficient of x² is negative the parabola has a maximum, not a minimum.
Applying it
- For the same model, y≤1 becomes 2(x−3)²≤8, giving 1≤x≤5.
- Check both endpoints.
- A contextual restriction such as x≥0 must be combined with this interval.
- For y=−2(x−3)²+7, the same vertex x-coordinate gives a maximum instead of a minimum.
Find the larger solution of 2x²−12x+11=11.
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Set y=11: 2x(x−6)=0, so the larger root is 6.
Match each completed-square feature of a(x-h)²+k to its meaning.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Completing the square exposes the vertex directly and leads to the quadratic formula when set to zero.
In your course
- AQA A-level Mathematics (7357), specification section B.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Completed square: A quadratic written as a multiple of a squared bracket plus a constant.
- For x²+bx, add and subtract (b/2)² to obtain (x+b/2)²−(b/2)².
- Watch for: half the coefficient inside the factored bracket, not the original coefficient of x.