**Question 1 :**

If α and β are the roots of

2x^{2}-3x-5 = 0

form a quadratic equation whose roots are α^{2} and β^{2}

**Solution :**

By comparing the given equation with general form of quadratic equation we get,

a = 2, b = -3 and c = -5

Sum of the roots : α+β = -b/a α+β = 3/2 |
Product of roots : α β = c/a αβ = -5/2 |

here α = α^{2 } and β = β^{2}

General form of quadratic equation whose roots are α^{2} and β^{2}

x^{2}-(α^{2}+β^{2})x+α^{2}β^{2} = 0 ---(1)

x^{2}-(α^{2}+β^{2}) x+(αβ)^{2} = 0

α^{2}+β^{2} = (α+β)^{2}-2αβ

= (3/2)^{2} - 2 (-5/2)

= (9/4) + 5

α^{2}+β^{2 }= 29/4

By applying the values in (1), we get

x^{2}-(29/4)x+(-5/2)^{2} = 0

4x^{2}-29x+25 = 0

Therefore the required quadratic equation is

4x^{2}-29x+25 = 0

**Question 2 :**

If α and β are the roots of

x^{2}-3x+2 = 0

form a quadratic equation whose roots are -α and -β.

**Solution :**

By comparing the given equation with general form of quadratic equation we get,

a = 1, b = -3 and c = 2

Sum of the roots : α+β = -b/a α+β = 3 |
Product of roots : αβ = c/a αβ = 2 |

Here α = -α and β = -β

General form of quadratic equation whose roots are α^{2} and β^{2}

x^{2}-(-α-β)x+(-α)(-β) = 0

x^{2}+(α+β)x+αβ = 0 -----(1)

By applying the appropriate values in (1), we get

x^{2}-3x+2 = 0

Therefore the required quadratic equations is

x^{2}-3x+2 = 0

**Question 3 :**

If α and β are the roots of

x^{2}-3x-1 = 0

then form a quadratic equation whose roots are 1/α^{2} and 1/β^{2}.

**Solution :**

x^{2}-3x-1 = 0

a = 1, b = -3 and c = -1

Sum of the roots : α+β = -b/a = 3 |
Product of roots : αβ = c/a = -1 |

Here α = 1/α^{2}and β = 1/β^{2}

By applying the above roots in the general form of a quadratic equation, we get

x^{2}-(1/α^{2}+1/β^{2})x+(1/α^{2}) (1/β^{2}) = 0

x^{2}-(α^{2}+β^{2}/α^{2}β^{2})x+[(1/α)(1/β)]^{2 }= 0

x^{2}-[(α^{2}+β^{2})/(αβ)^{2}]x+(1/αβ)^{2} = 0 ----(1)

α^{2}+β^{2} = (α+β)^{2}-2αβ

= 3^{2}-2(-1)

= 9+2

α^{2}+β^{2 }= 11

By applying the value of α^{2}+β^{2 }and αβ in (1), we get

x^{2}-11x+1 = 0

So, the required quadratic equation is x^{2}-11x+1 = 0.

**Question 4 :**

If α and β are the roots of the equation

3x^{2}-6x+1 = 0

form an equation whose roots are

(i) 1/α , 1/β (ii) α^{2}β , β^{2}α (iii) 2α+β, 2β+α

**Solution :**

From the given quadratic equation

a = 3, b = -6 and c = 1

Sum of the roots : α+β = -b/a ==> 6/3 α+β = 2 |
Product of roots αβ = c/a αβ = 1/3 |

Here α = 1/α and β = 1/β

x^{2} - (1/α+1/β)x+(1/α) (1/β) = 0

x^{2}-[(α + β)/αβ]x+[(1/αβ)] = 0 ----(1)

By applying the appropriate values in (1), we get

x^{2}-6x+(2/3) = 0

3x^{2}-18x+2 = 0

So, the required quadratic equation is

3x^{2}-18x+2 = 0

(ii) α^{2}β , β^{2}α

**Solution :**

Here α = α^{2}β and β = β^{2}α

Applying the given roots instead of α and β in the general form.

x^{2}-(α^{2}β+β^{2}α)x+(α^{2}β)(β^{2}α) = 0

x^{2}-αβ(α+β)x+(α^{3}β^{3}) = 0

x^{2}-αβ(α+β)x+(αβ)^{3} = 0 ----(1)

By applying the appropriate values in (1), we get

x^{2}-(2x/3)+(1/27) = 0

27x^{2}-18x+1 = 0

So, the required quadratic equation is

27x^{2}-18x+1 = 0

(iii) 2α+β, 2β+α

**Solution :**

Here α = 2α+β and β = 2β+α

α+β = 2α+β+2β+α = 3α+3β α+β = 3(α+β) |
αβ = (2α+β)(2β+α) = 4αβ+2α²+2β²+αβ αβ = 5αβ+2(α |

x^{2}-3(α+β)x+5αβ+2(α^{2}+β^{2}) = 0 ----(1)

α^{2}+β^{2} = (α+β)^{2}-2αβ

= 2^{2} - 2 (1/3)

= 4-2/3

α^{2}+β^{2 }= 10/3

By applying the values of α^{2}+β^{2}, α+β and αβ in (1), we get

x^{2}-6x+(5/3)+(20/3) = 0

x^{2}-6x+(25/3) = 0

3x^{2}-18x+25 = 0

So, the required quadratic equation is

3x^{2}-18x+25 = 0

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