Question

The ZEROES of the QUADRATIC polynomial x²+99x+127 are............​

Answer 1

ANSWER:

BOTH NEGATIVE

EXPLANATION:

ROOTS CAN BE CALCULATED AS,

x= -b +/- √b²-4ac/2a

x=-99 +/- √99²-4×1×127/2×1

x=-99 +/- √9293/2

x=-99 +/- 96.4/2

since 96.4<99

HENCE BOTH ZEROES ARE NEGATIVE

Answer 2

QUESTION -

The zeroes of the quadratic polynomial x²+99x+127 are _______

ANSWER -

Let polynomial p(x) = x2+ 99x + 127

Product of the zeroes of p(x) = 127

(product of zeroes = c/a for ax2+ bx + c)

Since the product of its zeroes is positive, we can say that it is only possible when

'both zeroes are positive' or 'both zeroes are negative'.

The zeroes are not equal as discriminant for

p(x) = $\sqrt{99 {}^{2} - 4(127) }$

is definitely not zero.

Also, the sum of the zeroes is –99 (sum of zeroes = –b/a for ax2+ bx + c)

The sum being negative implies that the inference made earlier that 'both zeroes are positive' is not correct.

Hence we conclude that 'both zeroes are negative'

@HarshPratapSingh