Answers
Required Answer:-
Given:
- 2x⁴ - 11x³ + 7x² + 13x - 7 = 0
To Find:
- All the zeros if two of its zeros are 3 + √2 and 3 - √2
Solution:
Let f(x) = 2x⁴ - 11x³ + 7x² + 13x - 7
It's given that,
⟹ Two of its zeros are 3 + √2 and 3 - √2
Therefore,
⟹ f(3 + √2) = 0 and f(3 - √2) = 0
So,
⟹ (x - 3 - √2) and (x - 3 + √2) are the factors of the given polynomial. (By Factor Theorem)
So,
(x - 3 - √2)(x - 3 + √2)
= (x - 3)² - (√2)²
= x² - 6x + 9 - 2
= x² - 6x + 7
Therefore, x² - 6x + 7 is a factor of the polynomial.
Let g(x) = x² - 6x + 7
Let's divide f(x) by g(x)
----------------------------------------------------------------------------------------
x² - 6x + 7 ) 2x⁴ - 11x³ + 7x² + 13x - 7 ( 2x² + x - 1
2x⁴ - 12x³ + 14x²
(-) (+) (-)
--------------------------------------------
x³ - 7x² + 13x
x³ - 6x² + 7x
(-) (+) (-)
--------------------------------------------
-x² + 6x - 7
-x² + 6x - 7
(+) (-) (+)
--------------------------------------------
0
----------------------------------------------------------------------------------------
Therefore,
⟹ Quotient = 2x² + x - 1
⟹ Remainder = 0
So,
(2x⁴ - 11x³ + 7x² + 13x - 7)
= (x² - 6x + 7)(2x² + x - 1)
⟹ (x² - 6x + 7)(2x² + x - 1) = 0
⟹ Either (x² - 6x + 7) = 0 or (2x² + x - 1) = 0
It's already given that roots of x² - 6x + 7 = 0 are 3 + √2 and 3 - √2. So, solve for the other part.
⟹ 2x² + x - 1 = 0
⟹ 2x² + 2x - x - 1 = 0
⟹ 2x(x + 1) - 1(x + 1) = 0
⟹ (2x - 1)(x + 1) = 0
⟹ Either (2x - 1) = 0 or (x + 1) = 0
⟹ x = 1/2, -1
Hence, the other two roots are 1/2 and -1.
So,
⟹ 2x⁴ - 11x³ + 7x² + 13x - 7 = 0
⟹ x₁,₂,₃,₄ = 3 + √2, 3 - √2, 1/2, -1
Answer:
- The roots of the given quartic equation are (3 + √2), (3 - √2), 1/2 and -1.