Quadratic equations

$a{{x}^{2}}+bx+c=0, a, b, c\in \mathbb{R}, a\ne 0$

$\Delta ={{b}^{2}}-4ac$

1. Solution formulas for $\Delta >0$

${{x}_{1}}=\frac{-b+\sqrt{\Delta }}{2a}; {{x}_{1}}=\frac{-b-\sqrt{\Delta }}{2a}$

2. Solution formulas for $\Delta =0$

${{x}_{1}}={{x}_{2}}=\frac{-b}{2a}$

In this case the equation $a{{x}^{2}}+bx+c=0$ can also be written: $a{{\left( x+\frac{b}{2a} \right)}^{2}}=0$

3. Solution formulas for $\Delta <0$

${{x}_{1}}=\frac{-b+i\sqrt{\left| \Delta \right|}}{2a}; {{x}_{2}}=\frac{-b-i\sqrt{\left| \Delta \right|}}{2a}$

4. Viète's formulas: $S={{x}_{1}}+{{x}_{2}}=-\frac{b}{a}$; $P={{x}_{1}}\cdot {{x}_{2}}=\frac{c}{a}$

5. Formulas useful in the study of the quadratic equation

${{x}_{1}}^{2}+{{x}_{2}}^{2}={{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}-2{{x}_{1}}{{x}_{2}}={{S}^{2}}-2P$

${{x}_{1}}^{3}+{{x}_{2}}^{3}={{\left( {{x}_{1}}+{{x}_{2}} \right)}^{3}}-3{{x}_{1}}{{x}_{2}}\left( {{x}_{1}}+{{x}_{2}} \right)={{S}^{3}}-3SP$

${{x}_{1}}^{4}+{{x}_{2}}^{4}={{\left( {{x}_{1}}+{{x}_{2}} \right)}^{4}}-6x_{1}^{2}x_{2}^{2}-4{{x}_{1}}{{x}_{2}}(x_{1}^{2}+x_{2}^{2})={{S}^{4}}-4{{S}^{2}}P+2{{P}^{2}}$

The quadratic function

$f:\mathbb{R}\to \mathbb{R}, f(x)=a{{x}^{2}}+bx+c, a, b, c\in \mathbb{R}, a\ne 0$

The graph of a quadratic function is a parabola.

This function can also be written in the form $f(x)=a{{\left( x+\frac{b}{2a} \right)}^{2}}+\frac{-\Delta }{4a}$, called the completed-square (canonical) form. The relation shows that the graph of any quadratic function is obtained from the graph of $f(x)={{x}^{2}}$ by translation along the Ox and Oy axes by ${{x}_{v}}=-\frac{b}{2a}$ and ${{y}_{v}}=-\frac{\Delta }{4a}$ respectively.

Graph of the function $f(x)={{x}^{2}}$

Graph of the function $f(x)=a{{\left( x+\frac{b}{2a} \right)}^{2}}+\frac{-\Delta }{4a}$

There are no three collinear points on the graph of the function $f(x)={{x}^{2}}$.

Three distinct points ${{M}_{0}}\left( {{x}_{0}},{{y}_{0}} \right), {{M}_{1}}\left( {{x}_{1}},{{y}_{1}} \right), {{M}_{2}}\left( {{x}_{2}},{{y}_{2}} \right)$ on the graph of the function are collinear if and only if**:** $\frac{{{y}_{2}}-{{y}_{0}}}{{{x}_{2}}-{{x}_{0}}}=\frac{{{y}_{1}}-{{y}_{0}}}{{{x}_{1}}-{{x}_{0}}}$

Maximum or minimum of a quadratic function

1. If $a>0$, the function $f:\mathbb{R}\to \mathbb{R}, f(x)=a{{x}^{2}}+bx+c, a, b, c\in \mathbb{R}, a\ne 0$ has a minimum equal to $-\frac{\Delta }{4a}$, attained at $x=-\frac{b}{2a}$.

2. If $a<0$, the function $f(x)=a{{x}^{2}}+bx+c$ has a maximum equal to $-\frac{\Delta }{4a}$, attained at $x=-\frac{b}{2a}$.

Graph of the function $f(x)=a{{x}^{2}}, a>1$

Graph of the function $f(x)=a{{x}^{2}}, a\in (0,1)$

Graph of the function $f(x)=a{{x}^{2}}, a<-1$

Graph of the function $f(x)=a{{x}^{2}}, a\in (-1,0)$

Sign of the quadratic function

Let $f:\mathbb{R}\to \mathbb{R}, f(x)=a{{x}^{2}}+bx+c, a, b, c\in \mathbb{R}, a\ne 0$

1. If $\Delta ={{b}^{2}}-4ac<0$ we obtain $f(x)$ has the same sign as the real number $a$, for every $x\in \mathbb{R}$, so there are no real roots.

The sign table of the function is:

$x$

$-\infty$

$+\infty$

$f(x)$

sign of $a$

2. If $\Delta ={{b}^{2}}-4ac=0$ we obtain $f(x)=0$ for ${{x}_{1}}=-\frac{b}{2a}$

The sign table of the function is:

$x$	$-\infty$	${{x}_{1}}=-\frac{b}{2a}$	$+\infty$
$f(x)$	sign of $a$	0	sign of $a$

3. If $\Delta ={{b}^{2}}-4ac>0$ we obtain $f(x)=0$ for ${{x}_{1}}=\frac{-b+\sqrt{\Delta }}{2a}$ and ${{x}_{2}}=\frac{-b-\sqrt{\Delta }}{2a}$. Note that $f(x)=a\cdot (x-{{x}_{1}})(x-{{x}_{2}})$.

The sign table of the function is:

$x$	$-\infty$	${{x}_{1}}$		${{x}_{2}}$	$+\infty$
$f(x)$	sign of $a$	0	opposite sign to $a$	0	sign of $a$

Note: the intervals $\left( -\infty , \frac{-b}{2a} \right]\text{ i }\left[ \frac{-b}{2a}, \infty \right)$ are called the intervals of monotonicity of the function.

Sketching the quadratic function

1. Intersections with the coordinate axes

Intersection with the $Ox$ axis amounts to solving the equation $f(x)=0$ for ${{x}_{1}}$ and ${{x}_{2}}$. This gives the points $A({{x}_{1}},0)\text{ i }B\text{(}{{x}_{2}}\text{,0)}$.

Intersection with the $Oy$ axis amounts to computing $f(0)$. This gives the point $C(0,c)$.

2. The vertex of the parabola and the axis of symmetry

The point $V\left( -\frac{b}{2a}\text{, }-\frac{\Delta }{4a} \right)$ lies on the graph of the function and is called the vertex of the parabola.

The axis of symmetry is the line about which every point of the parabola is symmetric; it has the equation $x=\frac{-b}{2a}$.

3. Fill in a table of values, which may include values other than those already computed, so that the curve can be drawn as accurately as possible.

4. Plot the points in the rectangular system $xOy$, then draw the parabola, bearing in mind that no three distinct points on it are collinear.

Factorising the trinomial $f(X)=a{{X}^{2}}+bX+c, a, b, c\in \mathbb{R}, a\ne 0$, with ${{x}_{1}}$ and ${{x}_{2}}$ the roots of the trinomial.

1. $\Delta >0$, $f(X)=a(X-{{x}_{1}})(X-{{x}_{2}})$;

2. $\Delta =0$, $f(X)=a{{(X-{{x}_{1}})}^{2}};$

3. $\Delta <0$, $f(X)$ is irreducible over $\mathbb{R}$, so $f(X)=a{{X}^{2}}+bX+c.$

Constructing a quadratic equation when its sum $S={{x}_{1}}+{{x}_{2}}$ and product $P={{x}_{1}}\cdot {{x}_{2}}$ are known:

${{x}^{2}}-Sx+P=0.$

Try it

The parabola. a decides which way it opens and how narrow it is, b slides the vertex sideways, c lifts the whole curve.

Set a to 0 and it stops being a parabola at all — that is the condition the chapter keeps insisting on.

The discriminant decides how many times the curve meets the x-axis. Move the sliders until it touches at exactly one point — that is Δ = 0.