Question

what are the zeros of x^2-3? (A) 3,3 (B)√3,-√3 (C)9,-9 (D)3,-3
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Answer 1

$x^{2} -3 = 0\\(x)^{2} - (\sqrt{3)^{2} } = 0\\( x -\sqrt{3})(x+\sqrt{3} )=0\\=> x = \sqrt{3} , -\sqrt{3}$

Answer 2

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.