Question

for polynomial x^2+99x+127,statement _________ holds. (a) both zero positive (b) both zero negative (c) equal sign​

Answer 1

Answer:

b.both zero negative

Step-by-step explanation:

b. both zero negative

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Answer 2

Step-by-step-Explanation:

Given: $x^2+99x+127$

To find: For polynomial x²+99x+127,statement _________ holds.

(a) both zero positive

(b) both zero negative

(c) equal sign

Solution:

Find zeros of polynomial to answer;

Use quadratic formula:

$\boxed{\bold{x_{1,2}=\frac{-b±\sqrt{b^2-4ac}}{2a}}}$

here,

a=1

b=99

c=127

$x_{1,2}=\frac{-99±\sqrt{(99)^2-4(1)(127)}}{2(1)}$

$x_{1,2}=\frac{-99±\sqrt{9801-508}}{2}$

$x_{1,2}=\frac{-99±\sqrt{9293}}{2}$

$x_{1,2}=\frac{-99±96.4}{2}$

$x_1=\frac{-99+96.4}{2}\\\\ \bold{x_1=-1.29}$

$x_2=\frac{-99-96.4}{2}\\\\ \bold{x_2=-97.7}$

Final answer:

Option B is correct.

for polynomial x²+99x+127,both zero negative.

Zeros are -1.29 and -97.7

Hope it helps you.

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