Four routes to the same roots
Every quadratic equation can be written as $ax^2+bx+c=0$ with $a\neq 0$, and every one of them can be finished with the quadratic formula. But the formula is almost never the fastest route, and on a timed module the difference between fifteen seconds and ninety seconds is worth more than the algebra itself.
- No constant term ($c=0$) $\rightarrow$ factor out $x$.
- Already a square equal to a number, like $(2x-7)^2=25$ $\rightarrow$ square root both sides.
- Integer coefficients that factor cleanly $\rightarrow$ factor.
- $a=1$ and $b$ even $\rightarrow$ complete the square (also the fastest way to the vertex).
- Nothing factors, or the question asks for an exact irrational answer $\rightarrow$ formula.
Solve $\;4x^2=20x$.
Move everything to one side: $4x^2-20x=0$, so $4x(x-5)=0$ and $\boxed{x=0 \text{ or } x=5}$.
Dividing both sides by $4x$ at the start would give only $x=5$ — the root $x=0$ disappears, because dividing by $x$ silently assumes $x\neq 0$.
Quy tắc sống còn: không bao giờ chia hai vế cho một biểu thức chứa ẩn. Chia là vứt mất một nghiệm. Luôn chuyển hết về một vế rồi đặt nhân tử chung. Đề của SAT có hẳn một phương án nhiễu dành riêng cho người vừa chia mất nghiệm $x=0$.
Factoring, including the case $a\neq 1$
When $a=1$, look for two numbers whose product is $c$ and whose sum is $b$. When $a\neq 1$, use the AC method: find two numbers whose product is $ac$ and whose sum is $b$, split the middle term with them, then group.
Solve $\;2x^2+5x-12=0$.
Here $ac=2\cdot(-12)=-24$. Two numbers with product $-24$ and sum $5$: they are $8$ and $-3$.
Split and group: \[ 2x^2+8x-3x-12 \;=\; 2x(x+4)-3(x+4) \;=\; (2x-3)(x+4). \] So $\boxed{x=\tfrac32 \text{ or } x=-4}$.
Solve $\;(x-3)^2-4(x-3)-12=0$.
Let $u=x-3$. Then $u^2-4u-12=0$, which factors as $(u-6)(u+2)=0$, so $u=6$ or $u=-2$.
Convert back: $x-3=6$ gives $x=9$; $x-3=-2$ gives $x=1$. Answer $\boxed{x=9 \text{ or } x=1}$.
Giải xong ẩn phụ $u$ rồi khoanh luôn $u=6$ và $u=-2$. Đề hỏi $x$ chứ không hỏi $u$ — phải đổi ngược lại. Trong bốn phương án luôn có sẵn cặp $\{6,-2\}$ chờ người quên bước cuối.
Square roots and completing the square
If an equation has the shape $(\text{something})^2=k$ with $k>0$, take square roots of both sides and remember the two signs. If it does not have that shape yet, force it.
$(x+p)^2 = x^2+2px+p^2$. So to complete the square on $x^2+bx$, add $\left(\dfrac{b}{2}\right)^2$. When $a\neq 1$, factor $a$ out of the $x^2$ and $x$ terms first.
Solve $\;(2x-7)^2=25$.
Take square roots: $2x-7=5$ or $2x-7=-5$.
First branch: $2x=12$, so $x=6$. Second branch: $2x=2$, so $x=1$. Answer $\boxed{x=6 \text{ or } x=1}$.
Solve $\;x^2-6x-7=0$ by completing the square.
Move the constant: $x^2-6x=7$. Half of $-6$ is $-3$, and $(-3)^2=9$, so add $9$ to both sides: \[ x^2-6x+9=16 \qquad\Longrightarrow\qquad (x-3)^2=16 . \] Then $x-3=\pm 4$, giving $\boxed{x=7 \text{ or } x=-1}$.
Bỏ dấu $\pm$. $\sqrt{16}=4$ là đúng, nhưng $(x-3)^2=16$ cho hai nhánh $x-3=4$ và $x-3=-4$. Mất một nghiệm là mất câu, vì phương án nhiễu luôn có sẵn đáp án “chỉ một nghiệm”.
The quadratic formula
For $ax^2+bx+c=0$ with $a\neq 0$, \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \] Compute $b^2-4ac$ before anything else: it tells you at once whether the answer will be a nice pair of rationals (perfect square), a pair of irrationals, or nothing real at all.
Solve $\;3x^2-5x-1=0$.
Here $a=3$, $b=-5$, $c=-1$, so $b^2-4ac=25-4(3)(-1)=25+12=37$, not a perfect square. \[ x=\frac{5\pm\sqrt{37}}{6}. \qquad \boxed{x=\frac{5\pm\sqrt{37}}{6}} \]
Hai chỗ sai dấu hay gặp nhất, cả hai đều nằm trong $b^2-4ac$: một là quên rằng $-b$ với $b=-5$ thành $+5$; hai là khi $c$ âm thì $-4ac$ trở thành cộng. Viết hẳn $a$, $b$, $c$ kèm dấu ra lề trước khi thay số, đừng nhẩm trong đầu.
Sum and product of the roots
If $r$ and $s$ are the two solutions of $ax^2+bx+c=0$, then $ax^2+bx+c=a(x-r)(x-s)$, and comparing coefficients gives the two relations below. They answer most parameter questions without solving anything.
\[ r+s=-\frac{b}{a}, \qquad\qquad rs=\frac{c}{a}. \]
One solution of $\;3x^2+kx-6=0\;$ is $x=\tfrac23$. Find $k$ and the other solution.
The product of the roots is $\dfrac{c}{a}=\dfrac{-6}{3}=-2$. If one root is $\tfrac23$, the other is $-2\div\tfrac23=-3$.
The sum is then $\tfrac23+(-3)=-\tfrac73$, and the sum must equal $-\dfrac{k}{3}$, so $k=7$.
Check: $3x^2+7x-6=(3x-2)(x+3)$, whose roots are $\tfrac23$ and $-3$. $\boxed{k=7,\; x=-3}$.
Nhìn thấy chữ “sum of the solutions”, “product of the solutions”, “the solutions are $r$ and $2r$”, hay “the solutions differ by $4$” là dừng tay, đừng giải. Ba câu hỏi đó đều trả lời được chỉ bằng $r+s=-b/a$ và $rs=c/a$. Giải ra nghiệm rồi mới cộng lại thường tốn gấp ba thời gian và mở thêm ba chỗ để sai dấu.
Squaring both sides: roots that are not roots
Squaring is not a reversible step. From $A=B$ it always follows that $A^2=B^2$, but the reverse fails: $A^2=B^2$ also holds when $A=-B$. So squaring can manufacture solutions that do not satisfy the original equation. Every solution must be substituted back.
$\sqrt{E}$ is never negative. So $\sqrt{E}=F$ forces $F\geq 0$ — any candidate that makes the right-hand side negative is dead before you even substitute.
Solve $\;\sqrt{3x+4}=x-2$.
Square both sides: $3x+4=x^2-4x+4$, so $0=x^2-7x$ and $x(x-7)=0$, giving $x=0$ or $x=7$.
Check $x=0$: the left side is $\sqrt{4}=2$, the right side is $-2$. Rejected.
Check $x=7$: the left side is $\sqrt{25}=5$, the right side is $5$. Accepted.
Answer: $\boxed{x=7}$ only.
Bình phương xong ra hai nghiệm đẹp, khoanh phương án “$x=-1$ and $x=3$”. Phương án đó luôn có mặt. Hễ đề có dấu căn là bắt buộc thử lại cả hai nghiệm — không có ngoại lệ.