Question

If alpha and beta are zeros of the quadratic polynomial f(x)=x²-3x-2, find a polynomial whose zeros are 2 alpha + 3 beta and 3 alpha + 2 beta​

Answer 1

Answer :-

$\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Quadratic Equations and its zeroes has been used. According to this, the equation which has highest variable degree of 2 is known as Quadratic Equation. This equation when drawn as line on graph, intersects X - axis twice that is it gives the roots of equations as α ans β. Let's do it !!

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★ Formula Used :-

If a Quadratic equation in the form of ax² + bx + c = 0 is given, then

$\qquad \large{\tt{\leadsto \: \: \alpha \: + \: \beta \: = \: \dfrac{(-b)}{a}}}$

$\qquad \large{\tt{\leadsto \: \: \alpha \: \times \: \beta \: = \: \dfrac{c}{a}}}$

where a is the coefficient of x², b is the coefficient of x and c is the constant term.

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★ Correct Question :-

If alpha and beta are zeros of the quadratic polynomial f(x)=x² - 3x + 2, find a polynomial whose zeros are 2α + 3β and 3α + 2β.

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★ Solution :-

Given,

• p(x) = x² - 3x + 2

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

Let the zeroes of this polynomial p(x) be α and β.

Then,

➣ α + β = -(-3) = 3 ... (i)

➣ αβ = 2 ... (ii)

Let the required polynomial be 'ax² + bx + c = 0'.

Then, its zeroes will be '2α + 3β' and '3α + 2β'.

Now let us apply the formula,

❐ Sum of zeroes = -b/a

❐ Product of zeroes = c/a

Then, let's find this out :-

~ Sum of the zeroes :

➺ 2α + 3β + 3α + 2β = -b/a

➺ 5α + 5β = -b/a

➺ 5(α + β) = -b/a ... (iii)

From equation (i) and (iii) , we get

✒ 5(3) = -b/a

$\: \large{\tt{\Longrightarrow \: \: \dfrac{15}{1} \: = \: \dfrac{(-b)}{a}}}$

On comparing LHS and RHS, we get,

• a = 1 , -b = 15

• a = 1 and b = -15

$\: \large{\boxed{\boxed{\sf{a \: = \: 1 \: \: and \: \: b \: = \: -15}}}}$

~ Product of Zeroes :-

➺ (2α + 3β) × (3α + 2β) = c/a

➺ 6α² + 4αβ + 9αβ + 9β² = c/a

➺ 6(α² + β²) + αβ(4 + 9) = c/a

➺ 6[(α + β)² - 2αβ] + 13αβ = c/a ... (iv)

From equation (i), (ii) and (iv), we get,

✒ 6[(3)² - 2(2)] + 13(2) = c/a

✒ 6[9 - 4] + 26 = c/a

✒ (6 × 5) + 26 = c/a

✒ 30 + 26 = c/a

✒ 56/1 = c/a

$\: \large{\tt{\Longrightarrow \: \: \dfrac{56}{1} \: = \: \dfrac{c}{a}}}$

On comparing LHS and RHS, we get,

• c = 56 and a = 1

$\: \large{\boxed{\boxed{\sf{a \: = \: 1 \: \: and \: \: c \: = \: 56}}}}$

By using these values, we get, ;

=> The required equation is : x² - 15x + 56 = 0

$\: \boxed{\large{\rm{Thus, \: the \: required \: polynomial \: is \: : \: \boxed{\bf{x^{2} \: - \: 15x \: + \: 56 \: = 0}}}}}$

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$\: \: \: \huge{\boxed{\tt{\large{More \: to \: know \: :-}}}}$

• Polynomials are the equations formed using constant and variable terms but of many variable degree.

• Linear Polynomial
• Quadratic Polynomial
• Cubic Polynomial
• Bi - Quadratic Polynomial

• Linear Equations are the equations formed using constant and variable terms but of single variable degree.

• Linear Equations in One Variable
• Linear Equations in Two Variables

Answer 2

Correct question :

If alpha and beta are zeros of the quadratic polynomial f(x)= x² - 3x + 2 , find a polynomial whose zeros are 2 alpha + 3 beta and 3 alpha + 2 beta​

Given :

• alpha and beta are zeros of the quadratic polynomial f(x) = x² - 3x + 2

To find :

• Quadratic polynomial which has zeros : (2α + 3β) and (3α + 2β)

Solution :

f(x) = x² - 3x + 2 = 0

⇒  x² - 2x - x + 2 = 0

⇒  x(x - 2) - 1(x - 2) = 0

⇒  (x - 2)(x - 1) = 0

⇒  (x - 2) = 0  or,  (x - 1) = 0

⇒  x = 2  or,  x = 1

∴ Value of α = 2

∴ Value of β = 1

Now zeros of new quadratic polynomial :

1. 2α + 3β = 2(2) + 3(1)

                 = 4 + 3

                 = 7

2. 3α + 2β = 3(2) + 2(1)

                 = 6 + 2

                 = 8

∴ Sum of zeros (α + β) = 7 + 8 = 15

∴ Product of zeros (αβ) = 7 * 8 = 56

Now we know,

⇒  Quadratic polynomial = x² - (α + β)x + αβ

⇒  Quadratic polynomial = x² - (15)x + 56

⇒  Quadratic polynomial = x² - 15x + 56

Therefore,

Required quadratic polynomial = x² - 15x + 56