Quadratic Function Properties

Quadratic Inequalities

 

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.

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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.

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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<x<4

 

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.

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\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.

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\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.

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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.

 

Quadratic Identities

 

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.

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

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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.

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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}.

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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.

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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.

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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.

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2. Solve 2\sin ^{ 2 }{ x } +\sin { x } -1=0\quad for\quad 0^{ \circ  }\le x\le 360^{ \circ  }.

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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  }