1) Find the zeroes of the Quadratic Polynomial
(i) $6x^2+x-15$

Solution
$6x^2+x-15$ =0 
$6x^2+10x-9x-15$ =0 
$2x(3x+5)-3(3x+5)$ =0
$(3x+5)(2x-3)=0$ 

$x=-5/3$ and $x=3/2$
Ans: Zeroes of the polynomial $6x^2+x-15$ are $x=-5/3$ and $x=3/2$ 

2) If the zeroes of the quadratic equation are  $1/3$ and $9/4$. Find the quadratic equations. 
Solution

If $1/3$ and $9/4$ are the zeroes of quadratic equations, then the quadratic equation is of the form:
$(x-1/3)(x-9/4)=0 $
$(3x-1)(4x-9)=0 $
$12x^2-27x-4x+9 =0 $
$12x^2-31x+9 =0 $

Ans: The quadratic equation is $12x^2-31x+9$ 

3) Find the quadratic polynomial, the sum and product of whose roots are $31/12$ and $3/4$.

Solution
 Let the quadratic polynomial be $ax^2+bx+c$, and its roots be α and β. We have
 α + β = $31/12$ = $-b/a$
 and αβ = $3/4$ = $c/a$
 
 If $a=1$, $b=-31/12$ and $c=3/4$
 $ax^2+bx+c$ = $x^2-31/12x+3/4$
 
 $12x^2-31x+9$
 Ans: The required polynomial is $12x^2-31x+9$ 
 
 4) Find the value of $k$ such that the polynomial $x^2-(k+6)x+2(2k-1)$ has sum of its zeroes
 equal to half of their product.
 
 Solution
 For the given polynomial, $x^2-(k+6)x+2(2k-1)$ 
 $a=1$ ; $b=-(k+6)$ and $c=2(2k-1)$
 If α and β are the roots of the polynomial,
Therefore
 α+β = $-b/a$ 
 α+β = $-(-(k+6))$=> $(k+6)$ 
 
 and αβ = $c/a$
        = $2(2k-1)$
 Given $α+β$ = $1/2(αβ)$
 Hence,$(k+6)$=$1/2(2(2k-1))$
        $k+6=2k-1$
		$k=7$
 
5)Find all the zeroes of the polynomials $2x^4-9x^3+5x^2+3x-1$ , if two of its
zeroes are (2+√3) and (2-√3)

 Solution
 Since (2+√3) and (2-√3) are the zeroes of the polynomial,
 therefore $(x-(2+√3))(x-(2-√3))$ is factors of the polynomial.
=> $(x-2-√3)(x-2+√3)$
=> $(x-2)^2-(√3)^2$  (Hint: $(a^2-b^2)=(a-b)(a+b)$)
=> $x^2-4x+4 - 3$
=> $x^2-4x+1$ is factor of polynomial $2x^4-9x^3+5x^2+3x-1$
Divide $2x^4-9x^3+5x^2+3x-1$ with $x^2-4x+1$


 
 $2x^4-9x^3+5x^2+3x-1$
${x^2-4x+1}$

$2x^2$

$- x$
$- 1$
$2x^4-8x^3+2x^2$
($-$)
($+$) ($-$)


$-x^3+3x^2+3x-1 $

$-x^3+4x^2-x$
($+$)
$(-)$  ($+$)

$-x^2+4x-1$
$-x^2+4x-1$
($+$)($-$)($+$)


$0$  $0$  $0$