Quadratic equations and Quadratic Functions

Many IB mathematics tutors consider quadratic equations as a very important topic of ib maths. There are following ways to solve a quadratic equation

► Factorization method

►complete square method 

► graphical method

► Quadratic formula method

<img src="ib mathematics tutors.png" alt="ib mathematics tutors>

The quadratic formula is the strongest method to solve a quadratic equation. In this article, I will use …… steps to prove the quadratic formula

Given equation: ax²+bx+c=0

Step-1: transfer constant term to right side

ax²+bx=-c

Step-2: divide both sides by coefficient of x²

                                      x²+bx/a=-c/a
Step-3: write (coefficient of x/2)²     that is (b/2a)²=b²/4a²
Step-4: Add this value to both sides
                                x²+bx/a+b²/4a² =-c/a²+b²/4a²
                                (x+b/2a)²=b²-4ac/4a²
now, take square root on both sides

                                  x+b/2=±√b²-4ac/a²

                                   x=-(b/2a)±√b²-4ac/2a

                                   x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

This formula is known as quadratic formula, we can put values of a, b and c  from any equation and find the value of x (the variable) by directly using this formula.

IB Mathematics tutors can also explain the concept of conjugate roots with the help of quadratic formula. In a quadratic equation,
                                                              ax²+bx+c=0
if a, b and c are all rational numbers and one root of the quadratic equation is a+√b then the second root will automatically become a-√b. that can be understood easily as we use one +ve and one -ve sign in quadratic formula.
These types of roots are called Conjugate Roots.
If  a & b  are  the  roots  of  the  quadratic  equation  ax² + bx + c = 0,  then;
(i)  \alpha {\rm{ }} + {\rm{ }}\beta {\rm{ }} = \frac{{--b}}{a}  (ii)  \alpha {\rm{ }}{\rm{.}}\beta {\rm{ }} = \frac{c}{a}     (iii)  \alpha {\rm{ - }}\beta {\rm{ }} = \frac{{\sqrt D }}{a}
Nature  Of  Roots:
(a) Consider the quadratic equation ax² + bx + c = 0  where a, b, c  \in  R & a \ne 0 then
(i) D > 0   \Leftrightarrow  roots  are  real & distinct  (unequal).
(ii) D = 0  \Leftrightarrow  roots  are  real & coincident  (equal).
(iii) D < 0 \Leftrightarrow  roots  are  imaginary
(B) Consider the quadratic equation ax2+ bx + c = 0 where a, b, c  \in  Q & a \ne 0 then
 If  D > 0  &  is a perfect  square , then  roots  are  rational & unequal.
A quadratic  equation  whose  roots  are  a & b  is  (x – a)(x – b) = 0  i.e.   x2 – (a + b) x + a b = 0 i.e.
                          x2 – (sum of  roots) x +  product  of  roots = 0
Consider  the  quadratic  expression , y = ax² + bx + c  , a, b, c  \in  R & a \ne 0  then
(i) The graph between x, y  is always a  parabola.  If a > 0  then the shape of the parabola is concave upwards &  if a < 0  then the shape of the parabola is concave downwards.
Common  Roots  Of  2  Quadratic  Equations  [Only  One  Common  Root]-   Let  \alpha
be  the  common  root  of  ax² + bx + c = 0  &  a’x2 + b’x + c’ = 0
Therefore    \;a{\alpha ^2}{\rm{ }} + {\rm{ }}b\alpha {\rm{ }} + {\rm{ }}c{\rm{ }} = {\rm{ }}0{\rm{ }}\;and{\rm{ }}\;a{\alpha ^2} + {\rm{ }}b\alpha {\rm{ }} + {\rm{ }}c{\rm{ }} = {\rm{ }}0
 If we solve above pair by cramer’s rule we get

                                            \frac{{{\alpha ^2}}}{{bc' - cb'}} = \frac{\alpha }{{ac' - c'a}} = \frac{1}{{ab' - a'b}}

This will give us                    \alpha  = \frac{{bc' - cb'}}{{ac' - c'a}} = \frac{{ac' - c'a}}{{ab' - a'b}}

                                            {(ac' - c'a)^2} = (ab' - a'b)(bc' - cb')

Every pair of the quadratic equation whose coefficients fulfils the above condition will have one root in common.

The condition that a quadratic function-  
                                           f(x , y) = ax² + 2 hxy + by² + 2 gx + 2 fy + c  may be  resolved  into  two  linear  factors  is  that     abc{\rm{ }} + {\rm{ }}2{\rm{ }}fgh{\rm{ }} - a{f^2} - b{g^2} - c{h^2}{\rm{ }} = {\rm{ }}0{\rm{ }}\; or
                                                                            \left| {\begin{array}{ccccccccccccccc}
a&h&g\\
h&b&f\\
g&f&c
\end{array}} \right| = 0
Reducible Quadratic Equations-These are the equations which are not quadratic in their initial condition but after some calculations, we can reduce them into quadratic equations
(i) If the power of the second term is exactly half to the power of the first term and the third term is a constant, these types of equations can be reduced to quadratic equations.
i.e. x + \sqrt x  - 6 = 0 , {x^6} + {x^3} - 6 = 0 , {e^{2x}} + {e^x} - 6 = 0 ,{a^{2x}} + {a^x} - 6 = 0

{\left( {2x + \frac{3}{{2x}}} \right)^2} + \left( {2x + \frac{3}{{2x}}} \right) - 6 = 0  all these equations can easily be reduced into quadratic equations by applying the method of substitution.

Example-Solve this equation and find x  {\left( {2x + \frac{3}{{2x}}} \right)^2} + \left( {2x + \frac{3}{{2x}}} \right) - 6 = 0 

Ans:

let y=  \left( {2x + \frac{3}{{2x}}} \right) then given equation will become

 

                                   {y^2} + y - 6 = 0  it’s a simple quadratic equation we can be easily factorised it and solve  so y=–3,2

so  \left( {2x + \frac{3}{{2x}}} \right)=-3

                                    \frac{{4{x^2} + 3}}{{2x}} =  - 3

                               \begin{array}{l}
4{x^2} + 3 =  - 6x\\
4{x^2} - 6x + 3 = 0
\end{array}

the final equation can be solved using Quadratic formula and the same process can be repeated for  y=2

(ii) If a variable is added with its own reciprocal, then we get a quadratic equation i.e x + \frac{1}{x} - 6 = 0,   \frac{{2x + 3}}{{x - 2}} + \frac{{x - 2}}{{2x + 3}} = 0 {e^{2x}} + \frac{1}{{{e^x}}} - 6 = 0  {a^{2x}} + \frac{1}{{{a^x}}} - 6 = 0 all these equations can be reduced into quadratic by replacing one term by any other variable.

 

Standard Form of a Quadratic Function-A quadratic function y=ax2+ bx + c can be

reduced into standard form    y = a{(x - h)^2} + k  by method of completing the square. If we

draw the graph of this function we shall get a parabola with vertex (h,k). The parabola will be upward for a>0 and downward for a<0

 

Maximum and Minimum value of a quadratic function- If the function is in the form

y = a{(x - h)^2} + k Then ‘h’ is the input value of the function while ‘k’ is its output.

(i) If a>0 (in case of upward parabola) the minimum value of f is f(h)=k

(ii) If a<0 (in case of downward parabola)the maximum value of f is f(h)=k
If our function is in the form of  y=ax² + bx + c then vertex of the parabola V = \left( { - \frac{b}{{2a}},\frac{D}{{4a}}} \right)
The line passing through vertex and parallel to the y-axis is called the axis of symmetry.
The parabolic graph of a quadratic function is symmetrical about axis of symmetry.

 \Rightarrow  f(x) has a minimum value at vertex if a>0 and {f_{\min }} =  - \frac{D}{{4a}}  at   x =  - \frac{b}{{2a}}

 

 \Rightarrow   f(x) has a maximum value at vertex if a<0 and   {f_{\min }} =  - \frac{D}{{4a}}  at   x =  - \frac{b}{{2a}}

 

In the next post about quadratics, I shall discuss discriminant, nature of roots, relationships between the roots. In the meantime, you can download the pdf and solve practice questions.

 quadratic equation 200 questions.pdf

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 quadratic equation and function.pdf

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