Math, asked by supriyaachyutha4321, 1 month ago

if _1,2,Alfa are the roots 2x^3+x^2_7x_6=0 then Alfa is will u say with explanation

Answers

Answered by adilanwar0086
0

Answer:

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Answered by pawanmerijaan
1

How do I solve the equation 2x^3+x^2-7x-6=0 given that the difference between two of its roots is 3?

The given equation is 2x3+x2−7x−6=0.(1)

Let the roots of the equation be α,α+3,β.

Now let us reduce the roots of the equation by 3 .

We know that the equation with roots k less than the roots of equation f(x)=0 is f(x+k)=0.

So let us reduce the roots of the equation by 3 .

The roots of the new equation are α−3,α,β−3.

We can find this equation simply by repeated synthetic division .

Thus we get

f(x)=2(x−3)3+19(x−3)2+53(x−3)+36.

Substituting (x+3),

f(x+3)=2x3+19x2+53x+36..(2)

The equations 1 and

Let the roots of the equation 2x^3+x^2-7x-6=0 are a , (a-3) and b , acordingly:-

a+(a-3)+b= -1/2. or. 2a+b=5/2=> b=(5–4a)/2………….(1)

a.(a-3)+(a-3).b+b.a=-7/2. or. a^2-3a+b(a-3+a)=-7/2.

=>a^2-3a+(2a-3).b=-7/2………………..(2)

a.(a-3).b=6/2. or. a.(a-3).b=3 …………..(3)

putting b=(5–4a)/2 from eqn. (1) in eqn. (2)

a^2-3a+(2a-3).(5–4a)/2=-7/2

=> 2a^2–6a+10a-15–8a^2+12a+7=0

=> -6a^2+16a-8=0

=> 3a^2–8a+4=0

=> (a-2).(3a-2)=0

a=2 , 2/3

But b=(5–4a)/2=(5–8)/2=-3/2. and. (5–8/3)/2=7/6.

or. b= -3/2 , 7/6.

Roots are a , (a-3) and b .

(1). putting a=2 , b=-3/2 roots are 2 , -1 and -3/2. Answer.

(2). putting a=2/3 , b=7/6 roots are 2/3 , -7/3 and 7/6. Answer.

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