Question

find the zeros of the quadratic polynomial 6x^2-13x+6 and verify the relation between the zeros and its coefficients?

Answer 1

Given polynomial,
6x²-13x+6=0
6x²-9x-4x+6=0
3x(2x-3)-2(2x-3)=0
(2x-3)(3x-2)=0
2x-3=0
x=3/2
(or)
3x-2=0
x=2/3
Let
α=3/2
β=2/3
Relation between zeroes and coefficients,
a=6,b= -13,c=6
Sum of zeroes=3/2+2/3=13/6= -(-13)/6= -b/a
Product of zeroes=3/2×2/3=6/6=c/a

hope it helps

Answer 2

Answer:

$\alpha=\frac{2}{3}$ and $\beta=\frac{3}{2}$

Step-by-step explanation:

Given : Quadratic polynomial $6x^2-13x+6$

To find : The zeros of the quadratic polynomial and verify the relation between the zeros and its coefficients?

Solution :

Quadratic polynomial $6x^2-13x+6$

Applying middle split,

$6x^2-9x-4x+6=0$

$3x(2x-3)-2(2x-3)=0$

$(3x-2)(2x-3=0$

$(3x-2)=0,(2x-3)=0$

$x=\frac{2}{3},x=\frac{3}{2}$

Therefore, Zeros of the quadratic polynomial is

$\alpha=\frac{2}{3}$ and $\beta=\frac{3}{2}$

Now, The relation between zeroes and coefficients is

Sum of zeros is $\alpha +\beta=-\frac{b}{a}$

Product of zeros is $\alpha\times\beta=\frac{c}{a}$

Where, a=6,b=-13 and c=6

Now, Substitute the values

$\frac{2}{3}+\frac{3}{2}=-\frac{13}{6}$

$\frac{4+9}{6}=-\frac{-13}{6}$

$\frac{13}{6}=\frac{-13}{6}$

Sum of zeros is Verified.

$\frac{2}{3}\times\frac{3}{2}=\frac{6}{6}$

$\frac{6}{6}=1$

$1=1$

Product of zeros is Verified.

Therefore, The relation is verified.