Question

friends please help me
and please don't post irretative answers..​

Answer 1

I don’t know this answer

Answer 2

Answer:

Option c

Step-by-step explanation:

Given :-

α and β are the roots of the equation is (a+1)x²+(2a+3)x+(3a+4) = 0 and αβ = 2

To find :-

Find the value of α+β ?

Solution :-

Given equation is

(a+1)x²+(2a+3)x+(3a+4) = 0

On comparing with the standard quadratic equation ax²+bx+c = 0 then

a = (a+1)

b = (2a+3)

c = (3a+4)

Given roots are α and β then

We know that

Sum of the roots = -b/a

=> α+β = -(2a+3)/(a+1) ----------(1)

and

Product of the roots = c/a

=>αβ = (3a+4)/(a+1) -------------(2)

According to the given problem

αβ = 2 -------------------------------(3)

Now,(2) becomes

=> (3a+4)/(a+1) = 2

=> 3a+4 = 2(a+1)

=> 3a+4 = 2a+2

=> 3a-2a = 2-4

=> a = -2

On substituting the value of a in (1) then

=> α+β = -[2(-2)+3]/[-2+1]

=> α+β = -(-4+3)/(-1)

=> α+β = -(-1)/-1

=> α+β =1/-1

=> α+β = -1

Therefore, α+β = -1

Answer:-

The value of α+β for the given problem is -1

Check:-

If a = -2 then given equation becomes

(-2+1)x²+(2×-2+3)x+(3×-2+4) = 0

=> (-1)x²+(-4+3)x+(-6+4) = 0

=> -x²-x-2 = 0

=> x²+x+2 = 0

On comparing with the standard quadratic equation ax²+bx+c = 0 then

a = 1, b = 1, c = 2

Sum of the zeroes = -b/a

=> -1/1

=> -1

Product of the zeroes = c/a

=> 2/1

=> 2

We have, α+β = -1 and αβ = 2

Verified the given relations in the given problem

Used formulae:-

• The standard quadratic equation is ax²+bx+c = 0
• Sum of the roots = -b/a
• Product of the roots = c/a