Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.
Step-by-Step Solution
Key Concept: For a quadratic polynomial $ax^2+bx+c$, the zeroes (roots) $\alpha$ and $\beta$ satisfy $\alpha+\beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$. The roots can be obtained using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$.
1. Identify the coefficients\
The given polynomial is $x^2-3=0$.\
Hence $a=1$, $b=0$, $c=-3$.\
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2. Compute the discriminant\
$$\Delta = b^2-4ac = 0^2-4(1)(-3)=12.$$\
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3. Apply the quadratic formula\
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-0 \pm \sqrt{12}}{2\cdot1}=\frac{\pm \sqrt{12}}{2}=\pm \frac{\sqrt{12}}{2}.$$\
Simplify $\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}$, therefore\
$$x = \pm \frac{2\sqrt{3}}{2}=\pm \sqrt{3}.$$\
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4. Zeroes (roots)\
$$\alpha = \sqrt{3}, \qquad \beta = -\sqrt{3}.$$\
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5. Verify the relationships\
- Sum of the zeroes: $\alpha+\beta = \sqrt{3}+(-\sqrt{3}) = 0 = -\dfrac{b}{a}$ (since $b=0$).\
- Product of the zeroes: $\alpha\beta = (\sqrt{3})(-\sqrt{3}) = -3 = \dfrac{c}{a}$ (since $c=-3$).\
Both relationships hold, confirming the theory.
Correct Answer: The zeroes are $\sqrt{3}$ and $-\sqrt{3}$. Their sum is $0$ (equals $-b/a$) and their product is $-3$ (equals $c/a$).