1.

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

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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.)



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