Completing the square and quadratic structure
Введено| English | Значение |
|---|---|
| completed square/kəmˈpliːtɪd skweə/ | полный квадрат |
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?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
At x=3, y=18−36+11=−7.
When the coefficient of x² is negative the parabola has a maximum, not a minimum.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
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.