English, asked by prajapatisayang08, 4 months ago

You have recently composed an article on the topic ' Our Culture Our Pride'. Send it to a reputed daily of your city and request the Editor to publish it in the newspaper

Answers

Answered by Alina12A
10

Answer:

points are very less. ...

Explanation:

increase the points

Answered by abdulrubfaheemi
1

Answer:

Answer :

Hence the quadratic equation formed is 3x² - 10x + 1.

Given :

Given quadratic equation, 3x² - 4x + 1.

To find :

The quadratic equation whose roots are a²/b and b²/a.

Knowlwdge required :

Quadratic equation formula :

\begin{gathered}\boxed{\sf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}} \\ \\ \end{gathered}

x=

2a

−b±

b

2

−4ac

Formula to form a quadratic equation :

\begin{gathered}\boxed{\therefore \sf{x^{2} - (\alpha + \beta)x + \alpha\beta = 0}} \\ \\ \end{gathered}

∴x

2

−(α+β)x+αβ=0

Solution :

By using the quadratic equation Formula and substituting the values in it, we get :

\begin{gathered}:\implies \sf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}} \\ \end{gathered}

:⟹x=

2a

−b±

b

2

−4ac

\begin{gathered}:\implies \sf{x = \dfrac{-(-4) \pm \sqrt{(-4)^{2} - 4 \times 3 \times 1}}{2 \times 3}} \\ \end{gathered}

:⟹x=

2×3

−(−4)±

(−4)

2

−4×3×1

\begin{gathered}:\implies \sf{x = \dfrac{4 \pm \sqrt{16 - 12}}{2 \times 3}} \\ \end{gathered}

:⟹x=

2×3

16−12

\begin{gathered}:\implies \sf{x = \dfrac{4 \pm \sqrt{4}}{2 \times 3}} \\ \end{gathered}

:⟹x=

2×3

4

\begin{gathered}:\implies \sf{x = \dfrac{4 \pm 2}{2 \times 3}} \\ \end{gathered}

:⟹x=

2×3

4±2

\begin{gathered}:\implies \sf{x = \dfrac{4 \pm 2}{6}} \\ \end{gathered}

:⟹x=

6

4±2

\begin{gathered}:\implies \sf{x = \dfrac{4 + 2}{6}\:;\:x = \dfrac{4 - 2}{6}} \\ \end{gathered}

:⟹x=

6

4+2

;x=

6

4−2

\begin{gathered}:\implies \sf{x = \dfrac{6}{6}\:;\:x = \dfrac{2}{6}} \\ \end{gathered}

:⟹x=

6

6

;x=

6

2

\begin{gathered}:\implies \sf{x = \dfrac{\not{6}}{\not{6}}\:;\:x = \dfrac{\not{2}}{\not{6}}} \\ \end{gathered}

:⟹x=

6

6

;x=

6

2

\begin{gathered}:\implies \sf{x = 1\:;\:x = \dfrac{1}{3}} \\ \end{gathered}

:⟹x=1;x=

3

1

\begin{gathered}\boxed{\therefore \sf{x = 1;\dfrac{1}{3}}} \\ \end{gathered}

∴x=1;

3

1

Thus, the value of x is 1 and ⅓.

From the above equation, we get :

The first root of the equation, a = 1

The second root of the equation, b = ⅓

But we are have to find the quadratic equation whose roots are a²/b and b²/a, So first let's solve them.

\begin{gathered}:\implies \sf{\alpha = \dfrac{a^{2}}{b} \quad | \quad \beta = \dfrac{b^{2}}{a}} \\\end{gathered}

:⟹α=

b

a

2

∣β=

a

b

2

Now by substituting the values of a and b in the equation, we get : [Here, a = 1 and b = ⅓]

\begin{gathered}:\implies \sf{\alpha = \dfrac{1^{2}}{\dfrac{1}{3}} \quad | \quad \beta = \dfrac{\dfrac{1}{3^{2}}}{1}} \\ \end{gathered}

:⟹α=

3

1

1

2

∣β=

1

3

2

1

\begin{gathered}:\implies \sf{\alpha = \dfrac{1}{\dfrac{1}{3}} \quad | \quad \beta = \dfrac{\dfrac{1}{9}}{1}} \\\end{gathered}

:⟹α=

3

1

1

∣β=

1

9

1

\begin{gathered}:\implies \sf{\alpha = 3 \quad | \quad \beta = \dfrac{1}{9}} \\\end{gathered}

:⟹α=3∣β=

9

1

\begin{gathered}\boxed{\therefore \sf{\alpha = 3 \quad | \quad \beta = \dfrac{1}{9}}} \\\end{gathered}

∴α=3∣β=

9

1

Hence the roots of the equation are 3 and ⅑.

Now by using the equation for forming a quadratic equation and substituting the values in it, we get :

\begin{gathered}:\implies \sf{x^{2} - (\alpha + \beta)x + \alpha\beta = 0} \\ \\ \end{gathered}

:⟹x

2

−(α+β)x+αβ=0

\begin{gathered}:\implies \sf{x^{2} - \bigg(3 + \dfrac{1}{9}\bigg)x + 3 \times \dfrac{1}{9} = 0} \\ \\ \end{gathered}

:⟹x

2

−(3+

9

1

)x+3×

9

1

=0

\begin{gathered}:\implies \sf{x^{2} - \bigg(\dfrac{27 + 1}{9}\bigg)x + \not{3} \times \dfrac{1}{\not{9}} = 0} \\ \\ \end{gathered}

:⟹x

2

−(

9

27+1

)x+

9

1

=0

\begin{gathered}:\implies \sf{x^{2} - \dfrac{30x}{9} + \dfrac{1}{3} = 0} \\ \\ \end{gathered}

:⟹x

2

9

30x

+

3

1

=0

\begin{gathered}:\implies \sf{x^{2} - \dfrac{10x}{3} + \dfrac{1}{3} = 0} \\ \\ \end{gathered}

:⟹x

2

3

10x

+

3

1

=0

By multiplying the whole equation by 3, we get :

\begin{gathered}:\implies \sf{\bigg(x^{2} - \dfrac{10x}{3} + \dfrac{1}{3}\bigg) \times 3 = 0 \times 3} \\ \\ \end{gathered}

:⟹(x

2

3

10x

+

3

1

)×3=0×3

\begin{gathered}:\implies \sf{x^{2} \times 3 - \dfrac{10x}{3} \times 3 + \dfrac{1}{3} \times 3 = 0} \\ \\ \end{gathered}

:⟹x

2

×3−

3

10x

×3+

3

1

×3=0

\begin{gathered}:\implies \sf{x^{2} \times 3 - \dfrac{10x}{\not{3}} \times \not{3} + \dfrac{1}{\not{3}} \times \not{3} = 0} \\ \\ \end{gathered}

:⟹x

2

×3−

3

10x

×

3+

3

1

×

3=0

\begin{gathered}:\implies \sf{3x^{2} - 10x+ 1 = 0} \\ \\ \end{gathered}

:⟹3x

2

−10x+1=0

\begin{gathered}\boxed{\therefore \sf{3x^{2} - 10x+ 1 = 0}} \\ \\ \end{gathered}

∴3x

2

−10x+1=0

Hence the quadratic equation formed is 3x² - 10x + 1.

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