**Solving quadratic functions**

**Introduction:**

In mathematics,The quadratic functions are the polynomial functions of the form f(x) = ax^{2}+bx +c,where a` !=`0. the graph of a quadratic function is a parabola whose major axis is parallel to y-axis. The expression ax^{2} + 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 *ax*^{2 }+ *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

*ax*^{2 }+ *bx *+ *c *= `1/a` (*a*^{2}*x*^{2 }+ *abx *+ *ac*)

=`1/a` [*a*^{2}*x*^{2 }+ *a*(*p*+*q*)*x *+ *pq*]= `1/a`[*a*^{2}*x*^{2 }+ *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 – 2 x – x^{2}= 0**

**Sol : **Writing in the standard form,

15 – 2*x *– *x*^{2 }= –*x*^{2 }– 2*x *+ 15

= (–1) (*x*^{2 }+ 2*x *– 15).

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

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

= (–1) (*x *+5)(*x *– 3)

(*x *+ 5) ( 3 – *x*). = 0

X="-5," x = -3

**Ex 2: ****Solve quadratic function: x^{2 }– x – 132 = 0.**

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

Hence we get *x*^{2 }– *x *– 132 = [*x *+ (–12)] (*x *+ 11) = (*x *– 12) (*x *+ 11). = 0

X = 12, x = -11

**Ex 3:**** ****Solve the quadratic function: 2 x^{2}+ 7x + 3 = 0**

**Sol : **Here *a *= coefficient of *x*^{2 }= 2

*b *= coefficient of *x *= 7

*c *= constant term = 3

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

2*x*^{2 }+ 7*x *+ 3 = 21 (2*x *+ 6) (2*x *+ 1) =(*x*+3)(2*x*+1).= 0

X = -3 , x = -1 / 2

**Ex 4:**** ****Solve the quadratic function: 8 a^{2 }+ 2a – 3 = 0.**

**Sol :**

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

By splitting and grouping, we get

8*a*^{2 }+ 2*a *– 3 = 8*a*^{2 }+ 6*a *– 4*a *– 3

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

A = - 3 / 4 , a = 1 / 2

**Ex 5: ****Solve: x^{2 }+ 9x + 18="0**

**Sol : **

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

*x*^{2 }+ 9*x *+ 18 = (*x *+ 3) (*x *+ 6)=0

x = -3 , x = -6

## practice problems on Solving Quadratic Functions :

**Prob 1 :Solve 2 a^{2 }+ 13a + 15 = 0**

**Ans :** a = -5, -2 / 3

**Prob 2 : Solve 4 x^{2 }+ 8x + 3 = 0**

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