Quadratics
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| parabola/pəˈræbələ/ | parabola |
| quadratic/kwɒˈdrætɪk/ | biểu thức bậc hai |
| completing the square/kəmˈpliːtɪŋ ðə skweə/ | hoàn thành bình phương |
| vertex/ˈvɜːteks/ | đỉnh |
| discriminant/dɪˈskrɪmɪnənt/ | biệt thức |
Find the peak without a table of guesses
- A simplified path of a thrown ball can be modelled by a parabola 抛物线. A quadratic 二次式 expression $ax^2+bx+c$, with $a\ne0$, gives such a curve.
- Different forms reveal different information: factors reveal roots; a completed square reveals a turning point. Choose the form that answers the question.
Complete the square and interpret the result
- Completing the square 配方法 writes the expression as $a(x+p)^2+q$. First factor $a$ from the quadratic and linear terms; halve the coefficient of $x$ inside the bracket, then compensate for the added square.
- The vertex 顶点 is $(-p,q)$. Since a square is non-negative, $q$ is a minimum if $a>0$ and a maximum if $a<0$. Check by expanding back to the original expression.
For $2x^2+8x+3$, take out $2$ from the first two terms:
The minimum is $-5$ at $x=-2$. The compensation is $-8$, not $-4$, because the outside factor multiplies it.
For $y=(x-3)^2-4$, the vertex is $(3,-4)$. Include intercepts and the opening direction when sketching.
Completing the square, x² − 6 x + 5 = (x − 3)² + q. What is the minimum value q?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
(x − 3)² − 9 + 5 = (x − 3)² − 4, so q = −4 (the minimum value).
Complete the square: 2 x² + 8 x + 3 = 2(x + p)² + q. What is p?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
2(x² + 4 x) + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5, so p = 2.
Complete the square for $y=-3 x^2+18 x-20$. State the full vertex and its nature. Sketch the graph on paper, finding exact roots and the $y$-intercept to support the sketch.
$y=-3(x-3)^2+7$, giving vertex $(3,7)$, a maximum because the squared term has a negative coefficient. Setting $y=0$ gives $(x-3)^2=7/3$, hence $x=3\pm\sqrt{7/3}$. The $y$-intercept is $-20$. The curve opens downward, is symmetric about $x=3$, and crosses the axis at the two stated roots. A teacher must review the actual sketch.
Solve the equation and keep exact answers
- Factorisation uses the zero-product rule: $(x-2)(x-3)=0$ means $x=2$ or $x=3$. Completing a square gives another route: $(x-3)^2=4$ gives $x-3=\pm2$, so $x=1$ or $x=5$.
- When factorisation is not simple, use $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. Substitute signed coefficients carefully and keep a surd answer exact unless a decimal is requested.
For $x^2+2x-1=0$, the coefficients are $a=1$, $b=2$, $c=-1$:
Both roots satisfy the original equation; omitting the minus choice loses one solution.
Solve x² − 5 x + 6 = 0. What is the smaller root?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
x² − 5 x + 6 = (x − 2)(x − 3) = 0, so x = 2 or x = 3; the smaller is 2.
Solve $x^2-8 x+7=0$ and find the vertex of the graph $y=x^2-8 x+7$. Show a method and explain whether the vertex is a maximum or a minimum.
$(x-1)(x-7)=0$ gives $x=1$ or $7$. Completing the square gives $y=(x-4)^2-9$, so the vertex is $(4,-9)$, a minimum because the squared term has a positive coefficient. Substitution or the midpoint of the roots also locates this vertex.
Solve $3 x^2-4 x-1=0$ exactly. Show your chosen method and check that both answers are possible rather than rounding them immediately.
Using $a=3,b=-4,c=-1$, $x=\dfrac{4\pm\sqrt{16+12}}6=\dfrac{2\pm\sqrt 7}3$. Both signs are needed. Substitution into the original polynomial gives zero for each; equivalently their sum is $4/3$ and product $-1/3$, as required by its factorisation.
Use the discriminant 判别式 for a real-root condition
- The discriminant is $\Delta=b^2-4ac$. For a genuine quadratic, $\Delta>0$ gives two distinct real roots, $\Delta=0$ one repeated real root, and $\Delta<0$ no real roots.
- To compare a line and a quadratic curve, equate their $y$-expressions first. Apply the discriminant to that new equation: its real roots correspond to intersection points.
For an upward-opening parabola with fixed $a$ and $b$, shifting it up increases $c$ and reduces $b^2-4ac$. A zero discriminant means a repeated root, not no root.
Explore the parabola
y = ax² + bx + c
Change a, b and c to reshape the curve — and watch the vertex, roots and intercepts move.
For x² + 4 x + 1 = 0, what is the discriminant b² − 4 ac?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
b² − 4 ac = 4² − 4(1)(1) = 16 − 4 = 12 (> 0, so two distinct real roots).
If b² − 4 ac < 0, the quadratic equation has:
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
A negative discriminant means the parabola never crosses the x-axis — no real roots.
The equation x² + 2 x + 5 = 0 has two distinct real roots.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Discriminant = 2² − 4(1)(5) = 4 − 20 = −16 < 0, so no real roots.
Solve an inequality using signs between roots
- Put all terms on one side, find its roots, and test the sign on each interval. For an upward-opening quadratic with two distinct roots, values are negative between them and positive outside; a negative leading coefficient reverses those signs.
- Include a root for $\leq$ or $\geq$, and exclude it for $<$ or $>$. When multiplying an inequality by a negative number, reverse its direction.
Solve $-x^2+4x-3\geq0$. Multiplying by $-1$ gives $x^2-4x+3\leq0$, hence $(x-1)(x-3)\leq0$. The upward-opening quadratic is non-positive between its roots, so $1\leq x\leq3$. Testing $x=2$ in the original expression gives a positive value, confirming the interval.
Solve $-2 x^2+10 x-8>0$. Explain the interval and the treatment of the boundary values; do not give only the roots.
$-2(x-1)(x-4)>0$. Dividing by $-2$ reverses the inequality: $(x-1)(x-4)<0$. The factors have opposite signs between the roots, so $1<x<4$. Neither endpoint is included because the original inequality is strict. Testing $x=2$ gives $4>0$; a point outside, such as $x=0$, gives $-8$.
Substitute before solving simultaneous equations
- Use the linear equation to express one variable in terms of the other; substitute into the quadratic equation. Solve the resulting quadratic, then recover the matching value of the other variable for each root.
- Give coordinate pairs, not two unrelated lists. Check each pair in both original equations; substitutions involving a product such as $xy$ still need the full bracket.
For $x+2y=7$ and $x^2+4y^2=25$, use $x=7-2y$:
Then $(2y-3)(y-2)=0$, so $y=\dfrac32$ or $2$. The corresponding pairs are $(4,\dfrac32)$ and $(3,2)$; both satisfy both original equations.
Solve $x+y=4$ and $2 xy+y^2=12$ simultaneously. Give all coordinate pairs and check each in both original equations.
Use $x=4-y$: $2(4-y)y+y^2=12$, hence $y^2-8 y+12=(y-2)(y-6)=0$. Thus $y=2$ gives $x=2$, and $y=6$ gives $x=-2$. The pairs are $(2,2)$ and $(-2,6)$. They both give $x+y=4$; the second expression gives $8+4=12$ and $-24+36=12$.
For the curve $y=x^2-2 x+3$ and line $y=x+k$, find the values of $k$ giving no intersection, one repeated intersection and two distinct intersections. Locate the touching point in the boundary case.
Equating gives $x^2-3 x+3-k=0$ with discriminant $4 k-3$. There is no intersection for $k<3/4$, a repeated intersection for $k=3/4$, and two for $k>3/4$. At equality $x=3/2$ and $y=9/4$, so the touching point is $(3/2,9/4)$. The discriminant belongs to the intersection equation, not the curve considered alone.
Recognise a quadratic in another expression
- In $x^4-5x^2+4=0$, let $u=x^2$. Solve $u^2-5u+4=(u-1)(u-4)=0$, then return to $x$: the real solutions are $\pm1$ and $\pm2$.
- Write the restriction on the substitute. If $u=\sqrt x$, then $u\geq0$; a negative root for $u$ cannot represent that square root, even if squaring it gives a positive number.
For $6x+\sqrt x-1=0$, set $u=\sqrt x\geq0$. Then $6u^2+u-1=(3u-1)(2u+1)=0$. Reject $u=-\dfrac12$, keep $u=\dfrac13$, and return to $x=u^2=\dfrac19$. Check this value in the original equation.
Solve both real equations: (a) $x^4-3 x^2-4=0$; (b) $2 x+\sqrt x-3=0$. State the allowed values of each substitute and check any rejected root before returning to $x$.
(a) Let $u=x^2\geq 0$. Then $(u-4)(u+1)=0$; reject $u=-1$ and use $u=4$ to obtain $x=\pm 2$. (b) Let $v=\sqrt x\geq 0$. Then $2 v^2+v-3=(2 v+3)(v-1)=0$. Reject $v=-3/2$, keep $v=1$, and obtain $x=1$. Squaring a rejected negative square-root value does not make it a valid solution of the original equation.
Check whether a parameter removes the quadratic term
- For $kx^2+2x+1=0$ to have two distinct real roots, both $k\ne0$ and $4-4k>0$ are needed. Thus $k<1$ with $k\ne0$; the positive discriminant at $k=0$ does not make a linear equation quadratic.
- At $k=0$ the equation is $2x+1=0$, with one solution. At $k=1$ it is $(x+1)^2=0$, with a repeated root. State which case the question asks for.
Find all real values of $k$ for which $kx^2+4 x+1=0$ has two distinct real roots. Then describe what happens at $k=0$ and at the repeated-root boundary.
It must remain quadratic, so $k\ne 0$. The discriminant condition is $16-4 k>0$, giving $k<4$ with $k\ne 0$. At $k=0$ it is the linear equation $4 x+1=0$, with root $-1/4$. At $k=4$ it is $(2 x+1)^2=0$, with repeated root $-1/2$.
Use given roots to find unknown constants
- A stated root must give zero when substituted into the polynomial. Use each root to form an equation in the unknown constants; alternatively expand $a(x-\alpha)(x-\beta)$ and compare coefficients with $ax^2+bx+c$.
- Expansion gives $\alpha+\beta=-\dfrac ba$ and $\alpha\beta=\dfrac ca$, provided $a\ne0$. Use the given root expressions in these equations and check the resulting constants in the original polynomial.
Suppose $px^2+6x-8$ has roots $2m$ and $-4m$, with $p,m\ne0$. Expanding its root form gives
Matching coefficients gives $pm=3$ and $pm^2=1$. Dividing the second equation by the first gives $m=\dfrac13$, hence $p=9$. The roots are $\dfrac23$ and $-\dfrac43$, and each makes $9x^2+6x-8$ equal to zero.
The polynomial $ax^2+8x-6$ has roots $3t$ and $-2t$, where $a$ and $t$ are non-zero real constants. Find $a$, $t$ and the two actual roots. Show how the root information produces equations and check the answers.
The root form is $a(x-3t)(x+2t)$. The linear and constant coefficients determine the two non-zero constants.
You've got it
- Choose factors, a completed square or the formula for the information required.
- For inequalities, justify intervals and endpoint inclusion; for simultaneous equations, check paired solutions.
- Check substitute restrictions and a non-zero leading coefficient before interpreting roots.
Which list gives all roots of $x^2-10 x+21=0$?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$(x-3)(x-7)=0$, so the roots are 3 and 7. They add to 10 and multiply to 21.