find the zeroes of q polynomial xsq2-3x-28 and verify the relationship between the zeroes and the coefficients
Answers
Answer:
Roots are -4, and 7.
Step-by-step explanation:
p(x) = x² - 3x - 28
We'll find the roots by Splitting the middle term.
Sum → -3
Product → -28
-7 × 4 will satisfy the above.
[-7 + 4 = -3, and -7 × 4 = -28]
⇒ x² - 3x - 28 = 0
⇒ x² - 7x + 4x - 28 = 0
⇒ x(x - 7) 4(x - 7) = 0
⇒ (x + 4)(x - 7) = 0
⇒ x + 4 = 0
⇒ x = -4
⇒ x - 7 = 0
⇒ x = 7
Hence, the Zeroes are -3, and 7.
Hence, α = -3 and β = 7.
Relationship between co-efficients and zeroes.
- α + β = -b/a
- αβ = c/a
The terms stand for:
- 'b' stands for the co-efficient of x.
- 'c' stands for co-efficient of the constant term.
- 'a' stands for co-efficient of x².
Now, We substitute,
p(x) = x² - 3x - 28
Using Sum of Zeroes formula.
α + β = -b/a
α + β = -(-3)/1
α + β = 3→ Eq(1)
Substituting the zeroes,
⇒ α + β = -4 + 7
⇒ α + β = 3 → Eq(2)
From Eq(1) and Eq(2)
Hence, α = 3 as a root/zero is verified.
Using Product of Zeroes formula.
⇒ αβ = c/a
⇒ αβ = -28/1
⇒ αβ = -28 → Eq(3)
Substituting the zeroes,
⇒ αβ = (-4)(7)
⇒ αβ = -28 → Eq(4)
From Eq(3) and Eq(4)
Hence, β = -28 as a root/zero is verified.