Question

if the zeros of polynomial p(X)=X2 + 4x + 2a are a and 2/a find value of a​

Answer 1

Solution :-

comparing the given quadratic polynomial p(x) = x² + 4x + 2a with ax² + bx + c , we get,

→ a = 1

→ b = 4

→ c = 2a

Now, we know that, sum of zeros of quadratic polynomial ax² + bx + c is (-b/a) .

and, we have given that, zeros of given polynomial are a and (2/a) .

So,

→ (a + 2/a) = (-b/a)

→ (a² + 2)/a = (-4/1)

→ a² + 2 = - 4a

→ a² + 4a + 2 = 0

Now ,using Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 we get :-

• x = [ -b±√(b²-4ac) / 2a ]
• x = [ - b ± √D /2a ] where D(Discriminant)= b²- 4ac.

Now, we have :-

• a = 1
• b = 4
• c = 2

So,

→ D = b² - 4ac

→ D = 4² - 4*1*2

→ D = 16 - 8

→ D = 8

Therefore,

→ a = [ - b ± √D /2a ]

→ a = [ - 4 ± √8 ] / 2*1

→ a = ( - 4 ± 2√2 ) /2

→ a = (- 4 + 2√2)/2 or , (-4 - 2√2)/2

→ a = (-2 + √2) or, (-2 - √2) (Ans.)

Answer 2

Answer:

- 2 ± √2

Step-by-step explanation:

Polynomials written in form of x² - Sx + P represent S as sum of roots and P as product of roots.

So, here, product of roots is 2a.

Roots are a and 2/a.

= > Their Product = 2a

= > a*(2/a) = 2a

= > a = ± 1

But it doesnt satisfy the conditions as according to this, sum of roots should be 1 + 2 = 3,

But here it is - 4.

So, using different concept:

= > a + (2/a) = - 4

= > a² + 4a + 2 = 0

Using quadratic equation:

= > a = 2(-2±√2)/2 = - 2 ± √2

But this also doesnt satisfy the conditions, as when a = - 2 ± √2, product of roots is not 2(- 2±√2).

Therefore I think your equation is x² + 4x + 2.

That's how:

product = 2

→ a + (2/a) = - 4

→ a² + 4a + 2 = 0

→ a = - 2 ± √2