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
Answer:
points are very less. ...
Explanation:
increase the points
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
4±
16−12
\begin{gathered}:\implies \sf{x = \dfrac{4 \pm \sqrt{4}}{2 \times 3}} \\ \end{gathered}
:⟹x=
2×3
4±
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+
3×
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.