ASSESSMENT ZONE
( 4. Why
CS Doe
6. Den
7. Whe
8. Wh.
A. Fill in the blanks using the words given in the box.
duck
food
claws
water
shelter
African antelopes
F. Answer
1 Wh
2. Wh
3. Wri
4. Wh
5. Des
(6 6. Sna
1. A habitat provides
and
2.
is an aquatic animal.
3. A tiger has sharp
to catch its prey.
4. An earthworm breathes through its
5.
migrate in large numbers to escape the drought.
B. State 'TRUE' or 'FALSE'.
1. Animals that live on trees are known as arboreal animals,
2. Water insects use flippers to swim.
3. Birds migrate to warmer places in winters.
4. Eagles have very poor eyesight.
5. A rat is a rodent.
7. De
8. Ex
( 9 WI
30 Chapter 2 > Animals in Their Surroundings
Answers
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.
Answer:
nice but i can't solve this question