Math, asked by prianshukabire, 4 months ago

find the zeros of Polinomial =3x^2-8 and verify reaction between zeros and co efecient​

Answers

Answered by TheProphet
9

S O L U T I O N :

We have quadratic polynomial p(x) = 3x² - 8 & zeroes of the polynomial p(x) = 0

Using by Factorisation Method :

➤ 3x² - 8 = 0

➤ 3x² = 8

➤ x² = 8/3

➤ x = ±√8/3

➤ x = √8/3 Or x = -√8/3

Therefore, α = √8/3 & β = -√8/3 are two zeroes of the given polynomial.

As we know that given quadratic polynomial compared with ax² + bx + c;

  • a = 3
  • b = 0
  • c = -8

Now,

Sum of the zeroes :

➤ α + β = -b/a = coefficient of x/coefficient of x²

➤ √8/3 + (-√8/3) = -0/3

➤ √8/3 - √8/3 = 0

0 = 0

Product of the zeroes :

➤ α × β = c/a = constant term/coefficient of x²

➤ √8/3 × (-√8/3) = -8/3

➤ -√8 × 8/√3 × 3 = -8/3

-8/3 = -8/3

Thus,

The relationship between zeroes & coefficient is verified .

Answered by kiran01486
1

Given, quadratic polynomial is √3x2 - 8x + 4√3

= √3x2 - 6x - 2x + 4√3

= √3x2 - √3 * √3 * 2 * x - 2x + 4√3

= √3x(x - 2√3) - 2(x - 2√3)

= (x - 2√3)(√3x - 2)

The value of √3x2 - 8x + 4√3 is zero if x - 2√3 = 0 or √3x - 2 = 0

So, x = 2√3, 2/√3

Therefore, zeroes of x2 – 2x – 8 are 2√3 and 2/√3

Now, Sum of zeroes = 2√3 + 2/√3

= (2√3 * √3 + 2)/√3

= (2 * 3 + 2)/√3

= (6 + 2)/√3

= 8/√3

= -(-8)/√3

= -(Coefficient of x)/ (Coefficient of x2)

Product of zeroes = 2√3 * (2/√3)

= (2√3 * 2)/√3

= 4√3/√3

= Constant term/ (Coefficient of x2)

Hence, the relation between its zero and coefficients is verified.

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