Question

Find the zeros of following quadratic polynomial and verify the zeroes and cofficient
1:-2x^2-3x+1

Answer 1

Given:-

๛Polynomial:-

➜2x²-3x+1

$\rule{200}{1}$

To Find:-

✪The Zeros of the Polynomial and Verify the zeros & Coefficients?

$\rule{200}{1}$

AnsWer:-

☞Zeros are:- x=½ and x=1

$\rule{200}{1}$

✪Using Splitting the Middle Term✪

↝2x²-3x+1=0

S=-3[Sum of Equation]

P=2[Product of Equation]

★-2+(-1)=-3→S

★-2×-1=2→P

↝2x²-2x-x+1=0

↝2x(x-1)-1(x-1)=0

$\rule{200}{1}$

✪Taking 2x-1

➜2x-1=0

➜2x=1

➜x=$\frac{1}{2}$⇄α

$\rule{200}{1}$

✪Taking x-1

➜x-1=0

➜x=1⇄β

$\rule{200}{1}$

•α+β=$\frac{-b}{a}$

→½+1=$\frac{-(-3)}{2}$

→$\frac{3}{2}$=$\frac{3}{2}$

$\rule{200}{1}$

•αβ=$\frac{c}{a}$

→½×1=$\frac{1}{2}$

→$\frac{1}{2}$=$\frac{1}{2}$

✓erified!

$\rule{200}{2}$

Answer 2

$\underline{\rule{269}{2}}$

GIVEN QUADRATIC POLYNOMIAL :-

2x² - 3x + 1

$\underline{\rule{269}{2}}$

TO FIND :-

The zeros of the polynomial.

$\underline{\rule{269}{2}}$

TO DETERMINE :-

The relationship between the zeros and the coefficients.

$\underline{\rule{269}{2}}$

SOLUTION :-

Solving for the zeros of the polynomial :-

2x² - 3x + 1 = 0

⇒2x² - x - 2x + 1 = 0

⇒x(2x - 1) - 1(2x - 1) = 0

⇒(2x - 1)(x - 1) = 0

⇒(2x - 1) = 0    or    (x - 1) = 0

⇒x = 1/2    or    x = 1

◙ Let α = 1/2 and β = 1.

Sum of the zeros = α + β = -b/a

⇒S = -(-3)/2

⇒S = 3/2                ...(i)

Product of the zeros = αβ = c/a

⇒P = 1/2                ...(ii)

Now,

• Sum of the zeros = 1/2 + 1 = 2/3              ...(iii)

• Product of the zeros = 1/2 × 1 = 1/2          ...(iv)

As eq.(i) = eq.(iii) and eq.(ii) = eq.(iv), so the RELATION IS VERIFIED.

$\underline{\rule{269}{2}}$

# MrNameless