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