Question

If ∝ and β are the zeroes of the quadratic Polynomial P(x)= ax2 + bx + c, then find
the values of
(a) ∝ + β (b) ∝ β (c) ∝^ 2+ β^2 (d) 1∝+1β

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Solution

\Large{\underline{\mathfrak{\bf{Given}}}}

Given

Distance of object from concave mirror (U) = -10 cm

Radius of curvature of concave mirror ( R ) = -8 cm = 2F

So, F = -8/2 = -4 cm

\Large{\underline{\mathfrak{\bf{Find}}}}

Find

Distance of image (V) .

\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}

Explanation

\Large{\underline{\mathfrak{\bf{Using\:mirror\:formula}}}}

Usingmirrorformula

\red{\boxed{\sf{\orange{\:\dfrac{1}{F}\:=\:\dfrac{1}{U}\:+\:\dfrac{1}{V}}}}}

F

1

=

U

1

+

V

1

Substitute value of U and F in above equation ,

\begin{gathered}:\implies\sf{\:\dfrac{1}{V}\:+\:\dfrac{1}{-10}\:=\:\dfrac{1}{-4}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:-\:\dfrac{1}{4}\:+\:\dfrac{1}{10}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-10+4}{40}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-14}{40}} \\ \\ :\implies\sf{\:V\:=\:\dfrac{-40}{14}} \\ \\ :\implies\sf{\red{\:V\:=\:\dfrac{-20}{7}\:cm}}\end{gathered}

:⟹

V

1

+

−10

1

=

−4

1

:⟹

V

1

=−

4

1

+

10

1

:⟹

V

1

=

40

−10+4

:⟹

V

1

=

40

−14

:⟹V=

14

−40

:⟹V=

7

−20

cm

\Large{\underline{\mathfrak{\bf{Thus}}}}

Thus

Distance of image (V ) = -20/7 cm

When, the image distance is negative , the image is behind the mirror .

So, the image is virtual and upright.