SAT quadratics: forms, roots, vertex, and solution counts

01Know what each form reveals

A quadratic function can appear in several equivalent forms. Rewriting is useful when it exposes the information the question asks for. It is not necessary to expand every expression immediately.

02Connect roots, intercepts, and the vertex

The three examples in the table describe the same parabola. Its roots are 2 and 4, so it crosses the horizontal axis at $(2,0)$ and $(4,0)$. Its vertex is $(3,-1)$, midway between the roots horizontally. At $x=0$, it has value 8, giving the vertical intercept $(0,8)$.

The sign in vertex form needs care. In $y=2(x+5{)}^2-7$, the vertex is $(-5,-7)$ because $x+5=x-(-5)$. The coefficient 2 makes the parabola open upward, so the vertex gives a minimum value of $-7$.

The vertex's $x$-coordinate and the minimum value are different answers. If the question asks when a modeled quantity is minimized, it may want $x=-5$. If it asks for the minimum quantity, it wants $y=-7$, subject to the model's domain.

03Solve roots exactly when you can

Set the function equal to zero to find roots. For $x^2-6x+8=0$, find two numbers whose product is 8 and sum is $-6$: $-2$ and $-4$. Then $(x-2)(x-4)=0$, so $x=2$ or $x=4$.

The zero-product rule applies when the product equals zero. You cannot set each factor to zero in $(x-2)(x-4)=3$. First move 3 to the left and simplify, or solve the original equation with another valid method.

If factoring is awkward, use $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. For $2x^2+x-3=0$, the discriminant is $1-4(2)(-3)=25$. The roots are $\frac{-1+5}{4}=1$ and $\frac{-1-5}{4}=-\frac{3}{2}$. Check them in the original equation.

04Count real solutions with the discriminant

For $a{x}^2+bx+c=0$ with $a\ne 0$, the discriminant is $D=b^2-4ac$. If $D>0$, there are two distinct real solutions. If $D=0$, there is one distinct real solution, a repeated root. If $D<0$, there are no real solutions.

For $x^2-4x+k=0$, one distinct real solution requires $16-4k=0$, so $k=4$. Two distinct real solutions require $k<4$; no real solutions require $k>4$.

If the coefficient of $x^2$ contains a parameter, check whether it can be zero. At that value the equation may become linear, so the quadratic discriminant rule no longer settles the question. Also read whether the prompt asks for real solutions or another condition.

05Use completing the square and root relationships

To rewrite $x^2+6x+5$, add and subtract 9: $x^2+6x+9-9+5=(x+3{)}^2-4$. This makes the vertex $(-3,-4)$ visible. With a leading coefficient other than 1, factor that coefficient from the quadratic and linear terms before completing the square.

For roots $r_1$ and $r_2$ of $a{x}^2+bx+c=0$, their sum is $r_1+r_2=-\frac{b}{a}$ and product is $r_1{r}_2=\frac{c}{a}$. A question asking for the sum of roots may not require solving either root.

For $3x^2-12x+7=0$, the root sum is 4 and product is $\frac{7}{3}$. These relationships come from expanding the factored form. They are useful only after the equation is written with zero on one side and the coefficients correctly identified.

06Graph the right comparison and keep the domain

In Desmos, graph the quadratic and inspect roots or the vertex. If the equation is $x^2-6x+8=5$, graph the parabola and $y=5$, or subtract 5 and find the new expression's zeros. The original parabola's x-intercepts do not solve that equation.

A decimal display may need an exact algebraic check. Adjust the window before concluding that a parabola has no intersection. In a height model, reject a negative time if the question restricts time to values after launch. In an area problem, a negative length is not a usable solution even if it satisfies the algebra.

Practice quadratics in the SAT question bank. For each problem, name the requested feature before selecting a method. Use Desmos practice to verify graphs, and the SAT math formulas guide for a broader review. A correct quadratic calculation still needs the right root, coordinate, or value for the question.