Question

write the zeros of a polynomial 4 x square - 8 x + 3 and verify the relation between the zeros and coefficients of polynomial

Answer 1

Step-by-step explanation:

Given -

• p(x) = 4x² - 8x + 3

To Find -

• Zeroes of the polynomial
• Verify the relationship between the zeroes and the coefficient.

Now,

By using quadratic formula :-

• x = -b ± √b² - 4ac/2a

→ -(-8) ± √(-8)² - 4×(4)×(3)/2(4)

→ 8 ± √64 - 48/8

→ 8 ± √16/8

→ 8 ± 4/8

Zeroes are -

→ 8 + 4/8

→ 12/8

→ 3/2

And

→ 8 - 4/8

→ 4/8

→ 1/2

Hence,

Zeroes of the polynomial are 3/2 and 1/2.

Verification :-

As we know that :-

• α + β = -b/a

→ 3/2 + 1/2 = -(-8)/4

→ 3 + 1/2 = 8/4

→ 4/2 = 2

→ 2 = 2

LHS = RHS

And

• αβ = c/a

→ 3/2 × 1/2 = 3/4

→ 3/4 = 3/4

LHS = RHS

Hence,

Verified...

Answer 2

◙ Given ◙

A quadratic polynomial, f(x) = 4x² - 8x + 3.

◙ Objective ◙

To find the zeros of f(x) and verify the relation between the zeros and coefficients of polynomial.

◙ Solution ◙

Finding the zeros of f(x) :-

$\rm{ 4x^2-8x+3=0}\\\\\rm{\dashrightarrow 4x^2-6x-2x+3=0}\\\\\rm{\dashrightarrow 2x(2x-3)-1(2x-3)=0}\\\\\rm{\dashrightarrow (2x-3)(2x-1)=0}\\\\\rm{\dashrightarrow (2x-3)=0\:\: or \:\: (2x-1)=0}\\\\\rm{\dashrightarrow x=\dfrac{3}{2}\:\:or\:\:x=\dfrac{1}{2} }$

Now, assume :-

$\rm{Let\ \alpha=\dfrac{3}{2} ,\ and\ \beta=\dfrac{1}{2} .}$

In the given polynomial f(x), a = 4, b = -8 & c = 3.

$\rule{170}2$

For sum,

$\rm{\alpha+\beta=\dfrac{3}{2}+\dfrac{1}{2} }\\\\\rm{\dashrightarrow \alpha+\beta=\dfrac{4}{2}}\\\\\rm{\dashrightarrow \alpha+\beta=2}$

$\rule{100}1$

For coefficients,

$\rm{\dfrac{-b}{a}=\dfrac{-(-8)}{4} }\\\\\rm{\dashrightarrow \dfrac{-b}{a}=\dfrac{8}{4} }\\\\\rm{\dashrightarrow \dfrac{-b}{a}=2 }$

Verified ✅

$\rule{170}2$

For product,

$\rm{\alpha\beta=\dfrac{3}{2}\times\dfrac{1}{2} }\\\\\rm{\dashrightarrow \alpha\beta=\dfrac{3}{4}}$

$\rule{100}1$

For coefficients,

$\rm{\dfrac{c}{a}=\dfrac{3}{4} }$

Verified ✅

$\rule{170}2$