please solve this problem
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- p(z) = 4z² - 15zπ - 4π²
- Zeroes of polynomial.
4z² - 15πz - 4π² = 0
By splitting middle term ,
4z² - 16πz + πz - 4π² = 0
4z² - 16πz + πz - 4π² = 0
Taking common factor ,
4z ( z - 4π ) + π ( z - 4π ) = 0
( 4z + π ) ( z - 4π ) = 0
Factors of the polynomial are (4z+π) and (z - 4π).
By zero product rule , either [4z+π=0] or [z-4π=0]
4z + π = 0
z = -π/4
z - 4π = 0
z = 4π
Hence , zeroes of polynomial are -π/4 and 4π.
OPTION (A)ㅤ 4π , -π/4
Answered by
1
Required Answer:-
Given:
- 4z² - 15πz - 4π² = 0
To Find:
- The zeros of equation.
Solution:
We have,
➡ 4z² - 15πz - 4π² = 0
➡ 4z² - 16πz + πz - 4π² = 0
➡ 4z(z - 4π) + π(z - 4π) = 0
➡ (4z + π)(z - 4π) = 0
By zero product rule,
➡ Either (4z + π) = 0 or (z - 4π) = 0
➡ z = -π/4, 4π
★ Hence, the roots of the quadratic equation are 4π and -π/4. Option A is the correct answer.
Answer:
- The roots of the quadratic equation are 4π and -π/4.
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