Math, asked by sarthak337, 5 months ago

please solve this problem ​

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Answered by Anonymous
0

 \\  \\ \large\underline{ \underline{ \sf{ \red{given:} }}}  \\  \\

  • p(z) = 4z² - 15zπ - 4π²

 \\  \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}  \\  \\

  • Zeroes of polynomial.

 \\  \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}  \\  \\

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 anindyaadhikari13
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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