Given that the real numbers s - t satisfy 19 s 2-99 s -1-0- t 2-99 t -19-0 and s t-1- then the absolute value of st -4 s -1- t isA- a prime numberB- divisible by 19C- divisible by 5D- divisible by 2

The correct options are A a nbsp;prime number C divisible by 5We have nbsp;19s^2 + 99s + 1 = 0, t^2 + 99t + 19 =0. s, 1/t are roots of the equation 19x^2+9 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

Roots of a Quadratic Equation

Given that th...

Question

Given that the real numbers $s, t$ satisfy $19 s^{2} + 99 s + 1 = 0, t^{2} + 99 t + 19 = 0$ and $s t \neq 1$ , then the absolute value of $\frac{s t + 4 s + 1}{t}$ is

a prime number

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

divisible by

19

No worries! We‘ve got your back. Try BYJU‘S free classes today!

divisible by

5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

divisible by

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A a prime number
C divisible by $5$
We have $19 s^{2} + 99 s + 1 = 0, t^{2} + 99 t + 19 = 0$ .
$s, \frac{1}{t}$ are roots of the equation $19 x^{2} + 99 x + 1 = 0$
$\frac{s}{t} = \frac{1}{19}, s + \frac{1}{t} = - \frac{99}{19}$
$\frac{s t + 1}{t} = - \frac{99}{19}$
So, $∣ ∣ \frac{s t + 4 s + 1}{t} ∣ ∣ = ∣ ∣ - \frac{99}{19} + \frac{4}{19} ∣ ∣ = 5$

Suggest Corrections

Similar questions

Q. Given that

b^{3} + m b^{2} + n b + c

is divisible by (b-s) if

s^{3} + m s^{2} + n s + c = 0.

Also given that

d^{3} + d^{2} + e d + 1

is divisible by (d-1) and

d^{3} - 4 d^{2} + f d - 3

is divisible by (d-3). What is the value of e+f ?

Q. If

x^{100} + 2 x^{99}

+ k is divisible by (x+ 1) = 0, then the value of k is

A box contains 19 balls bearing numbers 1, 2, 3…… 19. A ball is drawn at random from the box. What is the probability that the number on the ball is

a) a prime number

b) divisible by 3 or 5

If $x^{100} + 2 x^{99} + k$ is divisible by $(x + 1)$ , then the value of $k$ is

Q. If

p

is a prime number, and a prime to

p

, and if a square number

c^{2}

can be found such that

c^{2} - a

is divisible by

p

, show that

a^{\frac{1}{2} (p - 1)} - 1

is divisible by

p

Join BYJU'S Learning Program

Given that the real numbers s, t satisfy 19s2+99s+1=0, t2+99t+19=0 and st≠1, then the absolute value of st+4s+1t is

The correct options are A a prime number C divisible by 5We have 19s2+99s+1=0, t2+99t+19=0. s,1t are roots of the equation 19x2+99x+1=0 st=119,s+1t=−9919 st+1t=−9919 So, ∣∣st+4s+1t∣∣=∣∣−9919+419∣∣=5

Given that the real numbers $s, t$ satisfy $19 s^{2} + 99 s + 1 = 0, t^{2} + 99 t + 19 = 0$ and $s t \neq 1$ , then the absolute value of $\frac{s t + 4 s + 1}{t}$ is