Math, asked by Vinodshrma67vs12, 5 months ago

find the all zeroes of the polynomial x²-12​

Answers

Answered by Anonymous
3

Answer:

Question:

If tan θ = ¹/√7 then , show that \sf \dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}=\dfrac{3}{4}

csc

2

θ+sec

2

θ

csc

2

θ−sec

2

θ

=

4

3

\huge\bold{Solution :}Solution:

★══════════════════════★

\sf tan\ \theta=\dfrac{1}{\sqrt{7}}tan θ=

7

1

:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}:→tan

2

θ=

(

7

)

2

1

:\to \sf \textsf{\textbf{\pink{tan$^\text{2} \boldsymbol \theta\ $ =\ $\dfrac{\text{1}}{\text{7}}$}}}\ \; \bigstar:→tan

2

θ =

7

1

\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}

cot θ

1

=

7

1

:\to \sf cot\ \theta=\sqrt{7}:→cot θ=

7

:\to \sf cot^2\theta=(\sqrt{7})^2:→cot

2

θ=(

7

)

2

:\to \sf \textsf{\textbf{\green{cot$^\text{2}\ \boldsymbol \theta $\ =\ 7}}}\ \; \bigstar:→cot

2

θ = 7 ★

★══════════════════════★

LHS

:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}:→

csc

2

θ+sec

2

θ

csc

2

θ−sec

2

θ

From Trigonometric identities ,

csc²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}:→

(1+cot

2

θ)+(1+tan

2

θ)

(1+cot

2

θ)−(1+tan

2

θ)

tan²θ = ¹/₇

cot²θ = 7

:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}:→

(1+7)+(1+

7

1

)

(1+7)−(1+

7

1

)

:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}:→

8+

7

8

8−

7

8

:\to\ \sf \dfrac{48}{64}:→

64

48

:\to\ \textsf{\textbf{\orange{$\dfrac{\text{3}}{\text{4}}$}}}\ \; \bigstar:→

4

3

Answered by vashushubu77
11

Hi friend!!!

Given, f(x)=x²-√2x-12

→we have to find the values of x such that f(x)=0

→x²-√2x-12=0

→x²-3√2x+2√2x-12=0

→x(x-3√2)+2√2(x-3√2)=0

→(x+2√2)(x-3√2)=0

so, -2√2 and 3√2 are the zeros of the given polynomial.

I hope this will help u ;)

Happy learning

Hope it helps you...

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