Question

find a quadratic polynomial each with the given number as sum and product of its zero respectively -2√3,-9​

Answer 1

EXPLANATION.

Quadratic polynomial,

Sum of zeroes of quadratic polynomial = -2√3.

Products of zeroes of quadratic polynomial = -9.

As we know that,

General equation of polynomial = ax² + bx + c.

Sum of zeroes of quadratic polynomial.

⇒ α + β = -b/a.

⇒ α + β = -2√3.

Products of zeroes of quadratic polynomial.

⇒ αβ = c/a.

⇒ αβ = -9.

Quadratic polynomial,

⇒ x² - (α + β)x + αβ.

Put the value in equation, we get.

⇒ x² - (-2√3)x + (-9) = 0.

⇒ x² + 2√3x - 9 = 0.

                                                                                                                   

MORE INFORMATION.

Nature of the factors of the quadratic expression.

(1) = Real and different, 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

Answer:

Given :-

• The sum and product of its zero is - 2√3 , - 9 respectively.

To Find :-

• What is the quadratic polynomial of each of the given numbers.

Formula Used :-

❶ To find sum of zeros,

${\red{\boxed{\small{\bold{\alpha +\: \beta =\: \: - \dfrac{b}{a}}}}}}$

❷ To find product of zeros,

${\red{\boxed{\small{\bold{\alpha\beta =\: \dfrac{c}{a}}}}}}$

where,

• α = Alpha
• β = Beta

❸ To find quadratic polynomial,

${\red{\boxed{\small{\bold{{x}^{2} -\: (Sum\: of\: zeros)x +\: (Product\: of\: zeros\:) =\: 0}}}}}$

Solution :-

❶ First we have to find the sum of zeros,

According to the question by using the formula we get,

$\\ \sf \implies \alpha +\: \beta =\: -\dfrac{b}{a}$

$\\ \sf \implies \bold{\alpha +\: \beta =\: {- 2\sqrt{3}}}$

❷ Again we have to find the product of zeros,

According to the question by using the formula we get,

$\\ \sf \implies \alpha\beta =\: \dfrac{c}{a}$

$\\ \sf \implies \bold{\alpha\beta =\: - 9}$

❸ Now, we have to find the quadratic polynomial,

Given :

• Sum of zeros (α + β) = - 2√3
• Product of zeros (αβ) = - 9

According to the question by using the formula we get,

$\\ \sf \implies {x}^{2} -\: (Sum\: of\: zeros)x +\: (Product\: of\: zeros) =\: 0$

$\\ \sf \implies {x}^{2} -\: (\alpha +\: \beta)x +\: (\alpha\beta) =\: 0$

$\\ \sf \implies {x}^{2} -\: (- 2\sqrt{3})x +\: (- 9) =\: 0$

$\\ \sf \implies \bold{\purple{{x}^{2} +\: 2\sqrt{3}x -\: 9}}$

$\therefore$ The quadratic polynomial is ${\underline{\boxed{\small{\bf{{x}^{2} +\: 2\sqrt{3}x -\: 9}}}}}$