Question

Find a quadratic polynomial, the sum and product of whose zeroes are √3 and1/√3respectively

Answer 1

EXPLANATION.

Quadratic polynomial,

Sum of the zeroes = √3.

Products of its zeroes = 1/√3.

As we know that,

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ α + β = √3.

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ αβ = 1/√3.

As we know that,

Formula of quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (√3)x + (1/√3) = 0.

⇒ √3x² - 3x + 1 = 0.

                                                                                                                       

MORE INFORMATION.

Nature of th roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

$\huge{ \sqrt{3} {x}^{2} - 3x + 1}$

Step-by-step explanation:

Q.

Find a quadratic polynomial, the sum and product of whose zeroes are √3 and 1/√3 respectively

.

Solution -

$sum \: of \: zeroes = - \frac{b}{a} \\ = > \alpha + \beta = \sqrt{3} \\ \\ product \: of \: zeroes = \frac{c}{a} \\ = > \alpha \beta = \frac{1}{ \sqrt{3} }$

Using the quadratic equation formula

${x}^{2} - ( \alpha + \beta )x + ( \alpha \beta ) = 0 \\ \\ = > {x}^{2} - \sqrt{3} x + \frac{1}{ \sqrt{3} } = 0 \\ \\ (multiply \: by \sqrt{3} ) \\ = > \sqrt{3} ( {x}^{2} - \sqrt{3} x + \frac{1}{\sqrt{3} } = 0) \\ \\ = > \underline \red{\sqrt{3} {x}^{2} - 3x + 1 = 0}$

Hence the polynomial is √3x² - 3x + 1 .

hope it helps.