Question

n: q1-9mc, q2=-2mc y q3=7me
ermina el valor de la fuera ra

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

$\longrightarrow\underline\pink{\boxed{\bf }}$

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

Where, D refers to discriminant.

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

Solving the polynomial –

$\qquad$$\twoheadrightarrow \sf 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 \sf 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$$\twoheadrightarrow \sf 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$$\twoheadrightarrow \sf x = 0$

$\qquad$______________________

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

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

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

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

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

$\qquad$______________________

Product of roots /zeroes -

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

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

$\qquad$ $\pink{\twoheadrightarrow\bf \alpha\beta = 0}$

Hence proving the product of the roots is c/a

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