Quadratic and Polynomial Relationships - GRE Quantitative Reasoning

Roots sum $=-p$, product $=q$. For monic quadratics, Vieta's formulas simplify to sum of roots $=-p$ and product $=q$.

$\Delta=65$. The discriminant is calculated as $b^2 - 4ac = 9 - 4(2)(-7) = 9 + 56 = 65$.

Sum $=4$, product $=\frac{5}{3}$. Applying Vieta's formulas to $3x^2 - 12x + 5 = 0$ gives sum $=12/3=4$ and product $=5/3$.

$x^2-(r+s)x+rs$. The monic quadratic is constructed using Vieta's formulas with sum $r+s$ and product $rs$.

Sum $=-\frac{b}{a}$, product $=\frac{c}{a}$. Vieta's formulas state that for $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product is $c/a$.

$k=7$. Sum of roots $3+4=7$ equals $k$ for the quadratic $x^2 - kx +12=0$.

$x^2+x-6$. Using sum $2 + (-3) = -1$ and product $2 imes (-3) = -6$ in monic form.

$x=2,3$. Factoring gives $(x-2)(x-3)=0$, so roots are $2$ and $3$.

$(x-5)(x-4)$. Factors are found by pairs that multiply to $20$ and add to $-9$, yielding $ -4$ and $-5$.

$(3,2)$. The vertex is at $x=3$, and $y= (3)^2 -6(3) +11=2$, so $(3,2)$.

Two distinct real roots. A positive discriminant indicates the quadratic crosses the x-axis at two distinct points.

One real double root. A zero discriminant means the quadratic touches the x-axis at exactly one point, a repeated real root.

No real roots (two complex conjugates). A negative discriminant indicates no real solutions, resulting in two complex conjugate roots.

If $a>0$, min $=k$; if $a<0$, max $=k$. In vertex form, $k$ is the extremum value, minimum if $a>0$ and maximum if $a<0$.

Upward if $a>0$; downward if $a<0$. The sign of $a$ determines the parabola's direction: positive $a$ opens upward, negative $a$ opens downward.

$c$. The $y$-intercept occurs at $x=0$, so substituting gives $y=c$.

$a(x-r_1)(x-r_2)$. The factored form arises from the roots being the values that make the quadratic zero, scaled by the leading coefficient $a$.

$y=a(x-h)^2+k$; vertex $=(h,k)$. Vertex form is obtained by completing the square, with $(h, k)$ representing the vertex's coordinates.

$\left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right)$. The vertex coordinates are found by using the axis of symmetry for $x$ and substituting to find $y$, yielding the given expressions in terms of $a$, $b$, $c$.

$x=-\frac{b}{2a}$. The axis of symmetry for a parabola $y = ax^2 + bx + c$ is the vertical line through the vertex at $x = -b/(2a)$.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. The quadratic formula derives from completing the square to solve $ax^2 + bx + c = 0$ for $x$.

$\Delta=b^2-4ac$; sign gives number of real roots. The discriminant $Delta = b^2 - 4ac$ determines the nature of roots: positive for two real, zero for one real, negative for no real roots.