Question

Un empleado de una ferretería realiza un inventario. Informa que los 3 5 de los clavos son de dos pulgadas; el 15 %, de tres pulgadas, y el resto, de otras medidas. ¿A qué fracción corresponde el 15 % del total? Coloreen en la cuadrícula los clavos de dos y tres pulgadas que hay según el inventario. Luego, comparen los resultados.

Answer 1

General form of quadratic equation is,

ax² + bx + c = 0

In the given equation ( x²-2√2x ), we have –

Coefficient of x², a = 1

Coefficient of x, b =-2√2

Constant term, c = 0

To find the zeros, set the polynimial to zero and solve for x using the quadratic formula.

x²-2√2x = 0

Here, quadratic formula is given by

$\: \: \: \: \: \: \: \: \bigstar{\underline {\boxed{ \pink{\frak {x = \dfrac{-b±√D}{2a}}} }}} \: \: \: \: \:$

Where, D refers to discriminant.

$\: \: \: \: \: \: \: \: \bigstar{\underline {\boxed{ \pink{\frak {D(discriminant) = {b}^{2} - 4ac }} }}} \: \: \: \: \:$

Solving the polynomial –

$\qquad$$\twoheadrightarrow \bf x = \dfrac{-b±√D}{2a}$

$\qquad$$\twoheadrightarrow \sf x = \dfrac{-(-2\sqrt{2}) ± \sqrt{(2\sqrt{2})^2-4\times 1\times 0}}{2\times 1}$

$\qquad$$\twoheadrightarrow \sf x = \dfrac{-(-2\sqrt{2}) ± \sqrt{(2\sqrt{2})^2}}{2\times 1}$

$\qquad$$\twoheadrightarrow \sf x = \dfrac{-(-2\sqrt{2}) ± \sqrt{8}}{2}$

$\qquad$$\twoheadrightarrow \bf x = \dfrac{-(-2\sqrt{2}) ± 2 \sqrt{2}}{2}$

$\qquad$______________________

$\qquad$$\twoheadrightarrow \sf x = \dfrac{2\sqrt{2} + 2\sqrt{2}}{2}$

$\qquad$$\twoheadrightarrow \sf x = \dfrac{4\sqrt{2}}{2}$

$\qquad$$\pink{\twoheadrightarrow \bf x = 2√2}$

Or,

$\qquad$$\twoheadrightarrow \sf x = \dfrac{2\sqrt{2} -2\sqrt{2}}{2}$

$\qquad$$\twoheadrightarrow \sf x = \dfrac{\cancel{2\sqrt{2}} -\cancel{2\sqrt{2}}}{2}$

$\qquad$$\pink{\twoheadrightarrow \bf x = 0}$

Henceforth, zeroes of the polynomial are 2√2 and 0.

$\qquad$______________________

Now, the sum of the zeroes is given by:-

$\qquad$$\twoheadrightarrow\\underline{\boxed{\sf \alpha+\beta=\dfrac{-b}{a} }}$

$\qquad$$\twoheadrightarrow \sf \alpha +\beta =\dfrac{- (-2\sqrt{2})}{1}$

$\qquad$$\twoheadrightarrow \bf \alpha +\beta = 2\sqrt{2}$

Hence proving the sum of the roots is -b/a.

$\qquad$______________________

Product of roots /zeroes -

$\qquad$$\twoheadrightarrow\underline{\boxed{\sf \alpha.\beta=\dfrac{c}{a}  }}$

$\qquad$ $\twoheadrightarrow\bf \alpha\beta = \dfrac{0}{1}$

$\qquad$ $\twoheadrightarrow\bf \alpha\beta = 0$

Hence proving the product of the roots is c/a

Therefore, the relationship between the roots and the coefficients is verified.