Question

find the quadratic equations whose zeros are (3+√3) and (+3-√3)​

Answer 1

Step-by-step explanation:

Given:-

• Zeroes of a polynomial.
• Let the
• $\alpha = 3 + \sqrt{3}$
• And,
• $\beta = 3 - \sqrt{3}$

To find:-

A quadratic equation whose zeroes are alpha and beta which are given.

Solution:-

Sum of zeroes:-

$\alpha + \beta = \frac{ - b}{a } \\ 3 + \sqrt{3} + 3 - \sqrt{3} = \frac{ - b}{a} \\ 6 = \frac{ - b}{a}$

Now, Product of zeroes:-

$\alpha \times \beta = \frac{c}{a} \\ (3 + \sqrt{3} ) \times (3 - \sqrt{3} ) = \frac{c}{a}$

By using (a+b)(a-b) = a²-b²

Where,

• a = 3
• b = √3

$(3 + \sqrt{3} ) \times (3 - \sqrt{3} ) = \frac{c}{a} \\ {3}^{2} - ( { \sqrt{3} }^{2} ) = \frac{c}{a} \\ 9 - 3 = 6 = \frac{c}{a}$

As we know that a quadratic equation can be made by using a formula,

${x}^{2} - ( \alpha + \beta )x + ( \alpha \beta )$

Now substituting the values,

${x}^{2} - (6)x + 6 \\ {x}^{2} - 6x + 6$

Therefore the quadratic equation is x²-6x+6.