Nonlinear equations and mixed systems
SAT Topic 11 3:37 English narration · English + 中文 subtitles burned in
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Transcript
A candidate root must satisfy the original equation.
候选根必须满足原方程。
In the square root of x plus 6 equals x, the right side must be nonnegative.
在√(x+6) = x中，等号右边必须非负。
Squaring gives x squared minus x minus 6 equals 0, so candidates are 3 and negative 2.
平方后得到x²−x−6 = 0，因此候选解为3和−2。
Only 3 survives the original check: the square root of 9 equals 3.
只有3通过原方程检验：√9 = 3。
Each task gives its own data.
每个任务都有其自身的数据。
Do not copy the worked example’s numbers.
请勿照搬例题中的数值。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
Require x to be at least 0.
要求 x 大于等于零。
Squaring gives x squared minus 2 x minus 3, which is x minus 3 times x plus 1, equals 0.
两边平方得到 x 平方减二 x 减三，即 x 减三乘 x 加一，等于零。
The candidates are 3 and negative 1.
候选解是三和负一。
The original gives root 9 equals 3 at 3, but root 1 equals 1, not negative 1, at negative 1.
代回原方程，在三处得到根号九等于三，成立；在负一处得到根号一等于一，不等于负一。
Only x equals 3 works.
只有 x 等于三成立。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
Equate the outputs to get x plus 2 equals x squared minus 2 x plus 1, or x squared minus 3 x minus 1 equals 0.
令两个输出相等，得到 x 加二等于 x 平方减二 x 加一，即 x 平方减三 x 减一等于零。
So x equals 3 plus or minus root 13, over 2, and y equals 7 plus or minus root 13, over 2, with matching signs.
所以 x 等于三加减根号十三再除以二，y 等于七加减根号十三再除以二，正负号对应。
The discriminant, 13, is positive, so there are two distinct real intersections; both expressions for y agree by the derived equation.
判别式十三为正，所以有两个不同的实数交点；由推出的方程可知，y 的两个表达式一致。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
The absolute value gives 3 x plus 2 equals 7, or negative 7, so x equals five thirds or negative 3.
由绝对值得到三 x 加二等于七或负七，所以 x 等于三分之五或负三。
The rational equation requires x not equal to 3, and simplifies to x plus 3 equals 6, whose only candidate, 3, is excluded.
分式方程要求 x 不等于三，化简后是 x 加三等于六，唯一的候选解三被排除了。
It has no solution.
所以它没有解。
Return to the task and explain why the method fits.
回到任务中，解释该方法为何适用。
Original practice does not predict an official score.
原始练习无法预测正式成绩。
Complete required live or recorded evidence separately.
请单独完成所需的现场或录制证据。
Pause and test yourself. Cover the worked answer, rebuild the answer to the last example in your own words, and name one condition that would change it before you check.
暂停一下，自我检测：遮住示范答案，用自己的话重做最后一个例子的答案，并在核对之前说出一个会改变结论的条件。