Нелинейные уравнения и смешанные системы
Введено| English | Значение |
|---|---|
| discriminant/dɪˈskrɪmɪnənt/ | The value b squared minus 4ac, which determines the nature of a quadratic equation's roots. |
| extraneous/ekˈstreɪnɪəs/ | Produced by the solving steps but not a true solution of the original equation. |
A root that doesn't work
- Solving $\sqrt{x + 6} = x$ by squaring gives $x^2 - x - 6 = 0$, so $x = 3$ or $x = -2$. But only one of them is a real solution.
- Nonlinear equations can produce candidate roots that fail the original equation. The SAT tests whether you check them.
Quadratic equations
- Factor when you can: $x^2 - 5x + 6 = 0$ gives $(x - 2)(x - 3) = 0$, so $x = 2$ or $3$.
- Otherwise use the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. The discriminant 判别式 $b^2 - 4ac$ tells you how many real solutions: positive two, zero one, negative none.
y = √(x + 6) meets y = x only at x = 3; the candidate x = −2 fails
How many real solutions does $x^2 + 4x + 7 = 0$ have?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
$b^2 - 4ac = 16 - 28 = -12 < 0$.
Check candidates in the original
- For $\sqrt{x + 6} = x$: $x = 3$ gives $\sqrt{9} = 3$ ✓; $x = -2$ gives $\sqrt{4} = 2 \ne -2$ ✗. Only $x = 3$ works.
- Rational equations: reject any value that makes a denominator zero. Absolute value: $|x - 4| = 3$ gives $x - 4 = 3$ or $x - 4 = -3$, so $x = 7$ or $1$.
$\dfrac{x}{x - 2} = \dfrac{2}{x - 2}$ multiplies to $x = 2$ — but $x = 2$ makes the denominator zero, so there is no solution.
Keep or reject the candidate?
Check candidate roots in the original equation.
What is the solution of $\sqrt{2x + 3} = x$?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Squaring: $x^2 - 2x - 3 = 0$, so $x = 3$ или $-1$. $x = -1$ gives $\sqrt{1} = 1 \ne -1$, so only $x = 3$.
What are the solutions of $|x + 1| = 5$?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
$x + 1 = 5$ или $x + 1 = -5$.
A line meets a curve
- Set the expressions equal: $y = x^2$ and $y = x + 2$ meet where $x^2 = x + 2$, so $x^2 - x - 2 = 0$, $x = 2$ or $-1$; points $(2, 4)$ and $(-1, 1)$.
- The number of intersection points equals the number of real solutions of that equation.
The line $y = 3$ and the parabola $y = x^2 - 1$ meet at two points. What is the positive $x$-coordinate?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
$x^2 - 1 = 3$, so $x^2 = 4$, $x = \pm 2$.
Traps
- Keeping extraneous 增根的 roots after squaring or clearing denominators.
- Dividing by the variable: from $x^2 = 5x$, dividing by $x$ loses $x = 0$. Factor instead: $x(x - 5) = 0$.
Dividing both sides of $x^2 = 7x$ by $x$ gives all the solutions.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
It loses $x = 0$. Factor: $x(x - 7) = 0$.
Recap
- Factor or use the formula; the discriminant counts real solutions.
- Check every candidate in the original equation; never divide by an expression that could be zero.
- extraneous: Produced by the solving steps but not a true solution of the original equation.