what are the zeros of x^2-3? (A) 3,3 (B)√3,-√3 (C)9,-9 (D)3,-3
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2 thoughts on “what are the zeros of x^2-3? (A) 3,3 (B)√3,-√3 (C)9,-9 (D)3,-3<br />​”

1. Given:

   The polynomial –

   • x² – 3

   What To Find:

   We have to find –

   • The zeros of the given polynomial.

   Solution:

   Let us take,

   → p(x) = x² – 3

   Substitute 0 in p(x)

   → 0 = x² – 3

   Take – 3 to LHS,

   → 0 + 3 = x²

   Add 0 and 3,

   → 3 = x²

   Take the square (²) to LHS,

   →√3 = x

   It can also be,

   → – √3 = x

   So it will be,

   → ± √3 = x

   Which means,

   → x = √3 and – √3

   Verification:

   • First Method:-

   We know that –

   → Sum of zeros = Coefficient of x ÷ Coffiecient of x²

   Where –

   • The coefficient of x is 0.
   • The coefficient of x² is 1.

   Substitute,

   → √3 + (- √3) = 0 ÷ 1

   Solve the RHS,

   → √3 + (- √3) = 0

   Solve the LHS,

   → √3 -√3 = 0

   Solve the LHS further,

   → 0 = 0

   ∵ LHS = RHS

   ∴ Hence, verified.

   • Second Method:-

   We know that –

   → Products of zeros = Constant term ÷ Coffiecient of x²

   Where –

   • The constant term is – 3.
   • The coefficient of x² is 1.

   Substitute,

   → (√3) × (-√3) = – 3 ÷ 1

   Solve the RHS,

   → (√3) × (- √3) = – 3

   Solve the LHS,

   → – √9 = – 3

   Solve the LHS further,

   → – 3 = – 3

   ∵ LHS = RHS

   ∴ Hence, verified.

   Final Answer:

   ∴ Thus, the zeros of the given polynomial are √3 and – √3 that is Option B.

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2. [tex]x^{2} -3 = 0\\(x)^{2} – (\sqrt{3)^{2} } = 0\\( x -\sqrt{3})(x+\sqrt{3} )=0\\=> x = \sqrt{3} , -\sqrt{3}[/tex]

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