Question

3. The zeroes of the quadratic polynomial x2 + 99x + 127 are
(A) both positive
(B) both negative
(C) one positive and one negative
(d) both equal

NEED THE ANSWER WITH FULL EXPLANATION​

Answer 1

Answer:

The correct option is (B) both negative.

Step-by-step explanation:

Consider the provided polynomial.

$x^2+99x+127=0$

For a quadratic equation of the form ax²+bx+c=0 the solutions are:

$\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a=1, b=99 and c=127 in above formula.

$x_{1,\:2}=\frac{-99\pm \sqrt{99^2-4\cdot \:1\cdot \:127}}{2\cdot \:1}$

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

$x_{1,\:2}=\frac{-99\pm \sqrt{9293}}{2}$

$x=\frac{-99+\sqrt{9293}}{2},\ or\ \:x=\frac{-99-\sqrt{9293}}{2}$

$x=-1.29....\ or\ x=-97.70...$

Hence, both the roots are negative.

Therefore, the correct option is (B) both negative.

Answer 2

Answer:

plzz mark it as brainliest

correct option is [B]

Step-by-step explanation:

We know that,

In a quadratic expression ax²+ bx + c

If a , b , c will have the same sign

then the both zeroes of the

expression are negative

Compare given expression

x² + 99x + 127 with ax² + bx + c ,

a = 1 , b = 99 ,c = 127

All are positive sign ,

Therefore ,

The expressions both zeroes has negative sign.

Option B ) is correct.

I hope this helps you.

:)