For the parabola $y=a{x}^{2}+bx+c$

$a{ x }^{ 2 }+bx+c=0$ on the x-axis

$a{ x }^{ 2 }+bx+c>0$ above the x-axis

$a{ x }^{ 2 }+bx+c<0$ below the x-axis

### Examples:

1. Solve ${x}^{2}-3x+2 \ge 0$.

For x-intercepts, y=0a>0 so it is concave upwards.

$0={ x }^{ 2 }-3x+2\\ 0=(x-2)(x-1)\\ x-2=0\qquad or\qquad x-1=0\\ x=2\qquad or\qquad x=1$

$y \ge 0$ on and above the x-axis.

So ${x}^{2}-3x+2 \ge 0$ on and above the x-axis.

$\therefore x \le 1, x \ge 2$

2. Solve $4x-{x}^{2}>0$.

For x-intercepts, y=0. a<0 so it is concave downwards.

$0=4x-{ x }^{ 2 }\\ 0=x(4-x)\\ x=0\qquad or\qquad 4-x=0\\ x=0\qquad or\qquad x=4$

$y>0$ above the x-axis.

So $4x-{x}^{2}>0$ above the x-axis.

$\therefore 0

## The Discriminant

The values of x and y that satisfy a quadratic equation are called the roots of the equation. The roots of $a{x}^{2}+bx+c=0$ are the x-intercepts of the graph $y=a{x}^{2}+bx+c$.

In the quadratic formula $x=\cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$. the expression ${b}^{2}-4ac$ is called the discriminant. It gives us information about the roots of the quadratic equation $a{x}^{2}+bx+c=0$.

Notice that when there are 2 real roots the discriminant ${ b }^{ 2 }-4ac>0$.

When there are 2 equal roots (or just 1 real root), ${ b }^{ 2 }-4ac=0$.

When there are no real roots, ${ b }^{ 2 }-4ac<0$.

We often use $\Delta={ b }^{ 2 }-4ac$.

### Examples:

1. Show that the equation $2{x}^{2}+x+4=0$ has no real roots.

$\Delta ={ b }^{ 2 }-4ac\\ \Delta ={ 1 }^{ 2 }-4(2)(4)\\ \Delta =1-32\\ \Delta =-32\\ \Delta <0$

So the equation has no real roots.

2. Find the values of k for which the quadratic equation $5{x}^{2}-2x+k=0$ has real roots.

$\Delta \ge 0\\ { b }^{ 2 }-4ac\ge 0\\ { (-2) }^{ 2 }-4(5)(k)\ge 0\\ 4-20k\ge 0\\ 4\quad \ge \quad 20k\\ \cfrac { 1 }{ 5 } \ge \quad k$

3. Show that ${ x }^{ 2 }-2x+4>0$ for all x.

$f\quad a>0\quad and\quad \Delta <0,\quad then\quad a{ x }^{ 2 }+bx+c>0\quad for\quad all\quad x.\\ \\ a=1\quad >\quad 0\\ \Delta ={ b }^{ 2 }-4ac\\ \Delta ={ (-2) }^{ 2 }-4(1)(4)\\ \Delta =4-16\\ \Delta =-12\quad <\quad 0\\ \\ Since\quad a>0\quad and\quad \Delta <0,\quad { x }^{ 2 }-2x+4>0\quad for\quad all\quad x.$

As a general rule: if two quadratic expressions are equivalent to each other then the corresponding coefficients must be equal.

If ${ a }_{ 1 }{ x }^{ 2 }+{ b }_{ 1 }x+{ c }_{ 1 }\equiv { a }_{ 2 }{ x }^{ 2 }+{ b }_{ 2 }x+{ c }_{ 2 }$ for all real x then ${ a }_{ 1 }={ a }_{ 2 },\quad { b }_{ 1 }={ b }_{ 2 },\quad { c }_{ 1 }={ c }_{ 2 }$

### Examples:

1. Find values for ab and c if ${ x }^{ 2 }-x\equiv a{ (x+3) }^{ 2 }+bx+c-1$.

2. Find the equation of the parabola that passes through the points (-1, -3), (0, 3) and (2, 21).

## Sum and Product of Roots

The general quadratic equation can be written in the for

${x}^{2}-(\alpha + \beta)x+\alpha \beta = 0$

where $\alpha$ and $\beta$ are the roots of the equation.

If $\alpha$ and $\beta$ are the roots of the quadratic equation $a{x}^{2}+bx+c=0$,

Sum of roots: $\alpha + \beta = -\cfrac{b}{a}$

Product of roots: $\alpha \beta = \cfrac{c}{a}$

### Examples

1. Find the quadratic equation that has roots 6 and -1.

Using the general formula ${ x }^{ 2 }-(\alpha +\beta )x+\alpha \beta =0$,

$\alpha + \beta = 6 + -1 \\ \alpha + \beta = 5 \\ \\ \alpha \beta = 6 \times -1 \\ \alpha \beta = -6$

Therefore, the quadratic equation is: ${x}^{2}-5x-6=0$

2. Find the quadratic equation that has roots $3+\sqrt{2}$ and $3-\sqrt{2}$.

Using the general formula ${ x }^{ 2 }-(\alpha +\beta )x+\alpha \beta =0$,

$\alpha +\beta =3+\sqrt { 2 } +3-\sqrt { 2 } \\ \alpha +\beta =6\\ \\ \alpha \beta =(3+\sqrt { 2 } )\times (3-\sqrt { 2 } )\\ \alpha \beta ={ 3 }^{ 2 }-{ (\sqrt { 2 } ) }^{ 2 }\\ \alpha \beta =9-2\\ \alpha \beta =7$

Therefore, the quadratic equation is: ${x}^{2}-6x+7=0$

3. Find the value of k if one root of $k{x}^{2}-7x+k+1=0$ is -2.

$k{(-2)}^(2)-7(-2)+k+1=0 \\ 4k+14+k+1=0 \\ 5k+15=0 \\ 5k=-15 \\ k=-3$

4. Evaluate p if one root of ${x}^{2}+2x-5p=0$ is double the other root.

If one root is $\alpha$ then the other root is $2\alpha$.

Sum of roots:

$\alpha +\beta =-\cfrac { b }{ a } \\ \alpha +2\alpha =-\cfrac { 2 }{ 1 } \\ 3\alpha =-2\\ \alpha =-\cfrac { 2 }{ 3 }$

Product of roots:

$\alpha \beta =\cfrac { c }{ a } \\ \alpha \times 2\alpha =-\cfrac { p }{ 1 } \\ 2{ \alpha }^{ 2 }=-5p\\ 2{ \left( -\cfrac { 2 }{ 3 } \right) }^{ 2 }=-5p\\ 2\left( \cfrac { 4 }{ 9 } \right) =-5p\\ \cfrac { 8 }{ 9 } =-5p\\ p=-\cfrac { 8 }{ 45 }$

### Extra: Equation Reducible to Quadratics

1. Solve ${ 9 }^{ x }-{ 4.3 }^{ x }+0$.

2. Solve $2\sin ^{ 2 }{ x } +\sin { x } -1=0\quad for\quad 0^{ \circ }\le x\le 360^{ \circ }$.
$Let\quad \sin { x } =u\\ Then\quad 2{ u }^{ 2 }+u-1=0\\ (2u-1)(u+1)=0\\ 2u-1=0\qquad or\qquad u+1=0\\ u=\cfrac { 1 }{ 2 } \qquad or\qquad u=-1\\ \\ But\quad u=\sin { x } \\ So\quad \sin { x } =\cfrac { 1 }{ 2 } \qquad or\qquad \sin { x } =-1\\ x=30^{ \circ },\quad 150^{ \circ }\qquad or\qquad x=270^{ \circ }\quad (based\quad on\quad a\quad sine\quad graph)\\ \\ \therefore \quad x=\quad 30^{ \circ },\quad 150^{ \circ },\quad 270^{ \circ }$