Where the discriminant comes from
The quadratic formula \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] does only one thing that can fail: the square root. Everything about how many real solutions exist is decided by the single number underneath it.
For $ax^2+bx+c=0$ with $a\neq 0$, write $D=b^2-4ac$.
- $D>0$: two distinct real solutions.
- $D=0$: exactly one real solution, $x=-\dfrac{b}{2a}$ (a double root).
- $D<0$: no real solution.
How many real solutions does $\;3x^2-5x+4=0\;$ have?
$D=(-5)^2-4(3)(4)=25-48=-23<0$, so there are $\boxed{\text{no real solutions}}$.
Biệt thức trả lời câu hỏi “bao nhiêu nghiệm”, không trả lời “nghiệm bằng bao nhiêu”. Đề nào chỉ hỏi số nghiệm mà bạn ngồi giải ra nghiệm là tự làm mất một phút. Ngược lại, đề hỏi nghiệm cụ thể mà bạn chỉ tính $D$ rồi dừng thì chưa xong việc.
Reading the three cases
There is a fourth piece of information hidden in $D$ that the SAT uses: when $a$, $b$, $c$ are integers and $D$ is a perfect square, the two roots are rational, which is the same as saying the quadratic factors over the integers. When $D>0$ but is not a perfect square, the roots are irrational and come as a conjugate pair.
Describe the solutions of $\;2x^2+7x-4=0$.
$D=7^2-4(2)(-4)=49+32=81$, a perfect square, so the roots are rational and the expression factors: $2x^2+7x-4=(2x-1)(x+4)$, giving $x=\tfrac12$ and $x=-4$. $\boxed{\text{Two distinct rational solutions}}$.
Chỗ sai nhiều nhất trong cả unit này là dấu của $-4ac$ khi $c$ âm. Ở ví dụ trên $c=-4$, nên $-4ac=-4(2)(-4)=+32$, tức là $D=49+32=81$ chứ không phải $49-32=17$. Hệ quả dùng được ngay: hễ $a$ và $c$ trái dấu thì $ac<0$, nên $-4ac>0$ và $D>0$ — chắc chắn hai nghiệm, khỏi tính. Cùng dấu thì không kết luận được gì, phải tính thật.
Parameter questions: making the discriminant zero
The single most common Advanced Math question on this topic gives one coefficient as an unknown letter and says “exactly one solution”. Translate that phrase to $D=0$ immediately.
For what value of $c$ does $\;x^2-10x+c=0\;$ have exactly one real solution?
$D=(-10)^2-4(1)(c)=100-4c$. Setting $D=0$ gives $4c=100$, so $\boxed{c=25}$.
Check: $x^2-10x+25=(x-5)^2$, whose only root is $x=5$.
The equation $\;kx^2+12x+9=0$, where $k$ is a nonzero constant, has exactly one real solution. What is the value of $k$?
$D=12^2-4(k)(9)=144-36k$. Setting $D=0$ gives $36k=144$, so $\boxed{k=4}$.
Check: $4x^2+12x+9=(2x+3)^2$, whose only root is $x=-\tfrac32$.
Từ khoá cần dịch ngay sang $D$, không suy nghĩ: “exactly one solution”, “a double root”, “the graph touches the $x$-axis at exactly one point”, “the line is tangent to the parabola”, “$ax^2+bx+c$ is a perfect square” — cả năm cách nói đều là $D=0$. “No real solution”, “the graph does not intersect the $x$-axis”, “the system has no real solution” đều là $D<0$.
Boundaries and inequalities
When the question says “two distinct real solutions” or “no real solutions”, the answer is a range, and the endpoint is the value that gives exactly one solution. That endpoint is always one of the answer choices, and it is always wrong.
For which values of $k$ does $\;x^2+kx+9=0\;$ have two distinct real solutions?
$D=k^2-36>0$, so $k^2>36$, which means $\boxed{k<-6 \text{ or } k>6}$.
At $k=\pm 6$ the equation is $(x\pm 3)^2=0$ — one solution, not two. Between $-6$ and $6$ there are none.
The equation $\;3x^2+12x+k=0\;$ has two distinct real solutions. What is the greatest integer value of $k$?
$D=144-4(3)(k)=144-12k>0$, so $12k<144$ and $k<12$. The greatest integer strictly below $12$ is $\boxed{k=11}$.
At $k=12$ we get $3x^2+12x+12=3(x+2)^2$, exactly one solution — so $12$ itself is excluded.
Đề nói “two distinct real solutions” thì bất đẳng thức là ngặt: $D>0$, không phải $D\geq 0$. Lấy luôn giá trị biên là rơi đúng phương án nhiễu được thiết kế cho lỗi này. Cách tự kiểm: thay giá trị biên vào, nếu ra bình phương đúng của một nhị thức thì đó là nghiệm kép, phải loại.
When the equation might not be quadratic at all
$D=b^2-4ac$ is only meaningful when $a\neq 0$. If the coefficient of $x^2$ is itself the unknown, the value $a=0$ has to be examined separately — and it very often produces a legitimate “exactly one solution” that the discriminant never sees.
The equation $\;kx^2+6x+3=0\;$ has exactly one real solution. Find all values of $k$.
Case $k=0$: the equation becomes $6x+3=0$, a linear equation with the single solution $x=-\tfrac12$. That counts.
Case $k\neq 0$: it is genuinely quadratic, so we need $D=36-12k=0$, giving $k=3$.
Answer: $\boxed{k=0 \text{ or } k=3}$.
Đề viết “the quadratic equation $kx^2+\dots$” thì chữ quadratic đã ngầm cho $k\neq 0$, chỉ xét $D=0$. Nhưng đề viết “the equation $kx^2+\dots$” — không có chữ quadratic — thì $k=0$ là một trường hợp hợp lệ và phải xét riêng. Đọc kỹ đúng một chữ đó là chênh nhau một câu.
Lines and parabolas: tangency is $D=0$
To find where a line meets a parabola, set the two expressions equal. The result is a quadratic whose discriminant counts the intersection points: two, one (tangent), or none.
For what value of $b$ is the line $\;y=6x+b\;$ tangent to the parabola $\;y=x^2+10x+22$?
Set them equal: $x^2+10x+22=6x+b$, so $x^2+4x+(22-b)=0$.
Tangency means exactly one intersection point, so \[ D=16-4(1)(22-b)=16-88+4b=4b-72=0 \quad\Longrightarrow\quad \boxed{b=18}. \] Check: with $b=18$ the equation is $x^2+4x+4=(x+2)^2=0$, touching at $x=-2$ only.
Tangent $\Leftrightarrow$ $D=0$. Crosses twice $\Leftrightarrow$ $D>0$. Misses entirely $\Leftrightarrow$ $D<0$. The same three lines cover every “system of one linear and one quadratic equation” question.