Quadratic Polynomial with Irrational Zeroes Explained

 

❓ Question

Find the zeroes of the polynomial:

$$x^2-3$$

and verify the relationship between the zeroes and the coefficients.

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🖼️ Solution Image

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✍️ Short Explanation

To find the zeroes of the polynomial, we factorise it using the identity:

$$a^2-b^2=(a+b)(a-b)$$

Then we verify the relationship between zeroes and coefficients 💯

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🔹 Step 1 — Write the Polynomial

$$f(x)=x^2-3$$

Compare with:

$$ax^2+bx+c$$

So,

$$a=1,\quad b=0,\quad c=-3$$

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🔹 Step 2 — Factorise the Polynomial

$$x^2-3=x^2-(\sqrt{3})^2$$

Using identity:

$$a^2-b^2=(a+b)(a-b)$$
$$x^2-3=(x+\sqrt{3})(x-\sqrt{3})$$

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🔹 Step 3 — Find the Zeroes

$$x-\sqrt{3}=0$$
$$x=\sqrt{3}$$

and

$$x+\sqrt{3}=0$$
$$x=-\sqrt{3}$$

Therefore, the zeroes are:

$$\boxed{\sqrt{3}\text{ and }-\sqrt{3}}$$

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🔹 Step 4 — Verify Relationship Between Zeroes and Coefficients

Let:

$$\alpha=\sqrt{3},\quad \beta=-\sqrt{3}$$

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(i) Sum of Zeroes

$$\alpha+\beta=\sqrt{3}+(-\sqrt{3})=0$$

Also,

$$-\frac{b}{a}=-\frac{0}{1}=0$$

Verified ✅

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(ii) Product of Zeroes

$$\alpha\beta=(\sqrt{3})(-\sqrt{3})=-3$$

Also,

$$\frac{c}{a}=\frac{-3}{1}=-3$$

Verified ✅

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✅ Final Answer

Zeroes of the polynomial are:

$$\boxed{\sqrt{3}\text{ and }-\sqrt{3}}$$

Relationship verified:

$$\alpha+\beta=-\frac{b}{a}$$

and

$$\alpha\beta=\frac{c}{a}$$

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⭐ Key Insight

For quadratic polynomial:

$$ax^2+bx+c$$

• Sum of zeroes:

$$-\frac{b}{a}$$

• Product of zeroes:

$$\frac{c}{a}$$

🧠 Memory Line:

Sum → minus middle coefficient, Product → constant term 

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📚 Related Topics

• Zeroes and Coefficients of Quadratic Polynomial
• Zeroes of Polynomial Using Graph Concept Explained
• Number of Zeroes of a Polynomial from Graphs