Question

2) Find a quadratic polynomial with the sum and product of zeroes are 1/4

and - 1/4 respectively.​

Answer 1

EXPLANATION.

Quadratic polynomial.

Sum of the zeroes = 1/4.

Products of zeroes = - 1/4.

As we know that,

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ α + β = 1/4. - - - - - (1).

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ αβ = - 1/4. - - - - - (2).

As we know that,

Formula of quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (1/4)x + (-1/4) = 0.

⇒ x² - x/4 - 1/4 = 0.

⇒ 4x² - x - 1 = 0.

                                                                                                                       

MORE INFORMATION.

Nature of 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

Answer:

Given :-

• The sum and product of zeroes are ¼ and - ¼ respectively.

To Find :-

• What is the quadratic polynomial.

Formula Used :-

$\clubsuit$ Quadratic Equation Formula :

$\footnotesize\mapsto \sf\boxed{\bold{\pink{x^2 - (Sum\: of\: roots)x + (Product\: of\: roots) =\: 0}}}$

Solution :-

Given :-

$\bigstar\: \: \bf Sum\: of\: roots\: (\alpha + \beta) =\: \dfrac{1}{4}$

$\bigstar\: \: \bf Product\: of\: roots\: (\alpha\beta) =\: - \dfrac{1}{4}$

According to the question by using the formula we get,

$\footnotesize\longrightarrow \sf\bold{\purple{x^2 - (Sum\: of\: roots)x + (Product\: of\: roots) =\: 0}}$

$\small\longrightarrow \sf x^2 - (\alpha + \beta)x + (\alpha\beta) =\: 0$

$\small\longrightarrow \sf x^2 - \bigg(\dfrac{1}{4}\bigg)x + \bigg(- \dfrac{1}{4}\bigg) =\: 0$

$\small\longrightarrow \sf x^2 - \dfrac{x}{4} - \dfrac{1}{4} =\: 0$

$\small\longrightarrow \sf \dfrac{4x^2 - x - 1}{4} =\: 0$

By doing cross multiplication we get,

$\small\longrightarrow \sf 4x^2 - x - 1 =\: 4(0)$

$\small\longrightarrow \sf 4x^2 - x - 1 =\: 4 \times 0$

$\small\longrightarrow \sf\bold{\red{4x^2 - x - 1 =\: 0}}$

${\small{\bold{\underline{\therefore\: The\: required\: quadratic\: polynomial\: is\: 4x^2 - x - 1 =\: 0\: .}}}}$

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃

★ EXTRA INFORMATION ★

❒ Quadratic Polynomial with one Variable :

✪ The general form of the equation is ax² + bx + c = 0.

[Note : ◆ If a = 0, then the equation becomes to a linear equation.

◆ If b = 0, then the roots of the equation becomes equal but opposite in sign.

◆ If c = 0, then one of the roots is zero. ]

❒ Nature Of Roots :

✪ b² - 4ac is the discriminate of the equation.

Then,

● If b² - 4ac = 0, then the roots are real & equal.

● If b² - 4ac > 0, then the roots are real & unequal.

● If b² - 4ac < 0, then the roots are imaginary & no real roots.