Math, asked by arshdeep96, 1 year ago

If the squared difference of the zeros of the polynomial f(x)=x^2+px+45 is equal to 144,find the value of p.

Answers

Answered by sahuraj457

please follow me on brainly

Attachments:

arshdeep96: i am seening this formula first time (a+b)^2-(a-b)^2=4ab.

amitnrw: (a+b)^2 =a^2 + b^2 + 2ab Eq 1 & (a-b)^2 = a^2 + b^2 - 2ab Now Eq1 - Eq 2 you will get (a+b)^2-(a-b)^2= (2ab -(-2ab) = 4ab

arshdeep96: you have done it in wrong way.you have take (a-b)2=144 it is wrong ,i think. here will be (a^2-b^2)=144.

Answered by amitnrw

Answer:

p = ±16.12 , ±8.93i

Step-by-step explanation:

Let say a & b are zeros of the polynomial f(x)=x²+px+45

Then a + b = -p

ab = 45

I a² - b² I = 144 given

I (a+b)(a-b) I = 144

a + b = -p

Squaring both sides

a² + b² + 2ab = p²

=> a² + b² + 2×45 = p²

=> a² + b² = p² - 90

(a-b)² = a² + b² - 2ab

=> (a-b)² = p² - 90 - 90

=> (a-b)² = p² - 180

=> a-b = ±√(p² - 180)

I (a+b)(a-b) I = 144

I (-p)(±√(p² - 180)) I = 144

Squaring both sides

p²(p² - 180) = 144²

=> p²(p² - 180) = 20736

let say p² = y

y ( y-180) = 20736

=> y² - 180y - 20736 = 0

=> y = {180 ± √((180)² - 4×(-20736))}/2

=> y = (180 ± 339.62)/2

=> y = 259.81 , -79.81

p² = y

p = ±√y

p = ±16.12 , ±8.93i

arshdeep96: your answer is wrong

amitnrw: i took a^2 - b^2 = 144 (difference of squares of Zeroes) while it was to be taken as (a-b)^2 (square of difference of Zeroes) for your expected answer . squared difference of the zeros created confusion.

arshdeep96: no problem

Previous Question

Next Question

If the squared difference of the zeros of the polynomial f(x)=x^2+px+45 is equal to 144,find the value of p.​

Answers

If the squared difference of the zeros of the polynomial f(x)=x^2+px+45 is equal to 144,find the value of p.