Equations quadratic in another expression
Введено| English | Значение |
|---|---|
| substitution/ˌsʌbstɪˈtjuːʃn/ | замена переменной |
Solving for the new variable is only half the job.
- A fourth-degree equation can hide an ordinary quadratic.
- This lesson studies substitution 代换: Replacing a repeated expression with a new variable to reveal a simpler equation.
The method
- Identify a repeated expression and write u for it, including any restrictions.
- Solve the resulting quadratic in u.
- Then solve the original expression equal to each allowed u value.
- Substitute every final candidate into the original equation; a solution for u is not yet a solution for x.
For x⁴-5x²+4=0 with u=x², why must u=-2-style roots be rejected?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The substitution carries its own domain: u≥0. Each valid u then gives TWO x values from the square root; a rejected u gives none. u-values are never themselves the x-solutions.
A worked case
For x⁴−5x²+4=0, set u=x² with u≥0. Then u²−5u+4=(u−1)(u−4)=0, so x²=1 or 4 and x=±1,±2. For x⁴+x²−2=0, the u-equation is (u−1)(u+2)=0: u=−2 is impossible for real x, leaving x=±1. For (x+1)²−5(x+1)+6=0, set u=x+1: u=2 or 3 gives x=1 or 2. For 3^(2x)−4×3^x+3=0, set u=3^x>0: u=1 or 3 gives x=0 or 1.
Equations quadratic in another expression
Identify a repeated expression and write u for it, including any restrictions
Find the first valid or invalid step in a quadratic method.
Find the largest real solution of x⁴−5x²+4=0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The u roots 1 and 4 give x=±1,±2; the largest is 2.
Where it goes wrong
- Do not report u=1 and u=4 as the x solutions of the quartic.
- Taking a square root needs both signs, but a principal square-root expression is nonnegative.
- A quadratic in u can have two real roots while one or both violate u’s domain.
Not true: Two roots in the substituted variable give exactly two real solutions in x. Each root in the new variable must be checked: u=x² with u<0 gives no real x, and u>0 gives two.
Find the smaller solution of (x+1)²−5(x+1)+6=0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
u=x+1 gives u=2 or 3, hence x=1 or 2.
Each root in the new variable must be checked.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. Each root in the new variable must be checked: u=x² with u<0 gives no real x, and u>0 gives two.
Applying it
- Choose a substitution that repeats exactly: 3^(2x)=(3^x)², while 3^(x²) is a different expression.
- Count distinct real x solutions after returning to x, not before.
- Check by direct substitution: x=−2 gives 16−20+4=0 in the first quartic.
How many distinct real solutions does x⁴−5x²+4=0 have?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The four distinct values −2,−1,1,2 all satisfy the original equation.
Match each substitution step to its purpose.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Every candidate returns to the original variable and is checked; the restriction on u comes from what u stands for.
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.
- Substitution: Replacing a repeated expression with a new variable to reveal a simpler equation.
- Identify a repeated expression and write u for it, including any restrictions.
- Watch for: do not report u=1 and u=4 as the x solutions of the quartic.