Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.
x2 – 2x – 8
Let f(x) = x2 – 2x – 8
To find out zeros of the given polynomial.
We put f(x) = 0
⇒ x2 – 2x – 8 = 0
To find out roots of this polynomial we use splitting of middle term method.
According to this method we need to find two numbers whose sum is – 2 and product is 8.
∴ x2 – (4 – 2)x – 8 = 0
⇒ x2 – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ (x + 2)(x – 4) = 0
∴ x = – 2 and x = 4.
⇒ Our zeros are α = – 2 and β = 4.
⇒ sum of zeros = α + β = – 2 + 4 = 2.
⇒ Product of zeros = αβ = ( – 2) × 4 = – 8.
⇒ Comparing f(x) = x2 – 2x – 8 with standard equation ax2 + bx + c = 0.
We get, a = 1, b = – 2 and c = – 8
We can verify,
⇒ Sum of zeros = ![]()
i.e. α + β = ![]()
∴ α + β = 2
⇒ Product of zeros = ![]()
αβ = ![]()
αβ = – 8.
Hence, relationship between zeros and coefficient is verified.
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