Question

if 4+√3 is a root of ax2+CX+b and 5+√6 is a root of x2-dx+c then the value of b+c/ade is​

Answer 1

Given :- 4+√3 is a root of ax²+cx + b = 0 and 5+√6 is a root of x²- dx + e = 0 then the value of b+c/ade is ?

a. 29/5

b. 5/29

c. 1/38

d. 1/58

Solution :-

since 4 + √3 is a root of ax²+cx + b = 0 . so, other root will bs 4 - √3 .

so,

→ sum of roots = 4 + √3 + 4 - √3 = 8

→ product of roots = (4 + √3)(4 - √3) = 16 - 3 = 13

then,

→ Required quadratic equation is = x² - (sum of roots)x + product of roots = 0 => x² - 8x + 13 = 0

comparing this with ax²+cx + b = 0 we get,

• a = 1
• c = (-8)
• b = 13

similarly, since 5 + √6 is a root of x² -dx + e = 0 . so, other root will bs 5 - √6 .

so,

→ sum of roots = 5 + √6 + 5 - √6 = 10

→ product of roots = (5 + √6)(5 - √6) = 25 - 6 = 19

then,

→ Required quadratic equation is = x² - (sum of roots)x + product of roots = 0 => x² - 10x + 19 = 0

comparing this with x² -dx + e we get,

• d = 10
• e = 19

therefore,

→ (b + c)/ade

→ (13 - 8)/1 * 10 * 19

→ 5/190

→ 1/38 (Ans.)

Learn more :-

JEE mains Question :-

https://brainly.in/question/22246812

. Find all the zeroes of the polynomial x4

– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.

https://brainly.in/question/39026698

Answer 2

Step-by-step explanation:

answer is 1/58.

hope you understood this ☺️