Math Guide

Solving quadratic functions

Introduction:


                In mathematics,The quadratic functions are the polynomial functions of the form f(x) = ax2+bx +c,where a` !=`0. the graph of a quadratic function is a parabola whose major axis is parallel to y-axis. The expression ax2 + bx + c in the definition of a quadratic function is a polynomial of degree 2 (or) 2nd degree polynomial, because the highest exponent of a is 2, or highest order of x is 2. If the quadratic function is set equalto zero(i.e., euated to 0), then the resultant is a quadratic equation.The solutions to the equation are called as the roots of the equation.


Factoring quadratic Equation:


we consider the quadratic polynomial ax2 + bx + c where a, b, c are integers and a ≠0,1. If we are able to find two integers p and q such that pq = ac and p + q = b. Then

ax2 + bx + c = `1/a` (a2x2 + abx + ac)

=`1/a` [a2x2 + a(p+q)x + pq]= `1/a`[a2x2 + apx + aqx + pq]= `1/a`[ax (ax + p) + q(ax + p)]

= `1/a`(ax + p) (ax + q)


Thus, we are able to factorize the expression


Example problems for Solving quadratic functions :


Ex 1: Solve quadratic function 15 – 2x x2= 0


Sol : Writing in the standard form,

15 – 2x x2 = –x2 – 2x + 15

                 = (–1) (x2 + 2x – 15).

Here, we find –15 = 5 × –3, 5 + (–3) = 2

Hence, we get 15 – 2x x2 = (–1) [(x+5) {x + (–3)}]

                                        = (–1) (x +5)(x – 3)

                   (x + 5) ( 3 – x). = 0

                               X="-5," x = -3


Ex 2: Solve quadratic function: x2 x – 132 = 0.

Sol : We find –132 = (–12) × (11), (–12) + 11 = –1.

Hence we get x2 x – 132 = [x + (–12)] (x + 11) = (x – 12) (x + 11). = 0

X = 12, x = -11


Ex 3: Solve the quadratic function:  2x2+ 7x + 3 = 0

Sol : Here a = coefficient of x2 = 2

b = coefficient of x = 7

c = constant term = 3

We find a × c = 2 × 3 = 6 = 6 × 1, 6 + 1 = 7 = b. Hence

2x2 + 7x + 3 = 21 (2x + 6) (2x + 1) =(x+3)(2x+1).= 0

X = -3 , x = -1 / 2


Ex 4: Solve the quadratic function: 8a2 + 2a – 3 = 0.

Sol :

 Here, we find 8 ×– 3 = –24 = 6 × – 4, 6+(–4) = 2

By splitting and grouping, we get

8a2 + 2a – 3 = 8a2 + 6a – 4a – 3

= 2a (4a + 3) – (1)(4a+3) = (4a + 3) (2a – 1) = 0

  A = - 3 / 4 , a =  1 / 2


Ex 5: Solve:x2 + 9x + 18="0

Sol :

 We find that the sum of the factors 3 and 6 is the coefficient of x. Hence the factorization is

x2 + 9x + 18 = (x + 3) (x + 6)=0

x  = -3 , x  = -6


practice problems on Solving Quadratic Functions :


Prob 1 :Solve  2a2 + 13a + 15 = 0

Ans : a = -5, -2 / 3

Prob 2 : Solve  4x2 + 8x + 3 = 0

Ans : x = -1 / 2, -3 / 2