Question

The zeros of the polinomial x²-2x-3 are half of the zero of polinomial ax²+bx+c, than find the value of(a-b-c)

Answer 1

▪ Given :-

• Zeros of Polynomial x² - 2x - 3 are Half of zeros of Polynomial ax² + bx + c.

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▪ To Find :-

• Value of (a - b - c)

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▪ Solution :-

We Have 2 Polynomials,

• x² - 2x - 3 - - - - - - - - (1)

• ax² + bx + c - - - - - - - (2)

《Calculating Zeros of Polynomial (1)》

$\sf:\longmapsto \: {x}^{2} - 2x - 3 \\ \\ \sf:\longmapsto \: {x }^{2} - 3x + x - 3 \\ \\ \sf:\longmapsto \: x(x - 3) + 1(x - 3) \\ \\ \sf:\longmapsto \: (x - 3)(x + 1)$

$\underline{ \bf\huge \mathcal{So}} \: ,$

Zeros of Polynomial (1) are 3 And - 1

Now,

According to the Given Condition:

• The zeros of the polinomial x²-2x-3 are half of the zero of polinomial ax²+bx+c

Hence,

Zeros of Polynomial (2) must be 6 And -2

《Finding Polynomial (2)》

We Know That ,

A Quadratic Polynomial having zeros α and β

must be of the form :

x² - ( α + β)x + αβ

Hence,

Polynomial (2) Should be : x² - 4x - 12

Comparing it with ax² + bx + c We Get,

• a = 1
• b = - 4
• c = - 12

Hence ,

$\sf a - b - c = 1 -(-4) -(-12) \\ \\ = 1 + 4 + 12\\\\ = \huge \purple{ \mathfrak{17}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Answer:

Given :-

• The zeros of the polynomial x² - 2x - 3 are half of the zero of polynomial of ax² + bx + c.

To Find :-

• What is the value of (a - b - c).

Solution :-

$\mapsto$ The zeros of the polynomial x² - 2x - 3.

So,

$\bigstar\: \sf\bold{\purple{x^2 - 2x - 3 =\: 0}}$

$\implies \sf x^2 - 2x - 3 =\: 0$

$\implies \sf x^2 - (3 - 1)x - 3 =\: 0$

$\implies \sf x^2 - 3x + x - 3 =\: 0\: \bigg\lgroup \small\sf\bold{\pink{Splitting\: middle\: term}}\bigg\rgroup$

$\implies \sf x(x - 3) + 1(x - 3) =\: 0$

$\implies \sf (x - 3)(x + 1) =\: 0$

$\implies \bf{ (x - 3) =\: 0}$

$\implies \sf x - 3 =\: 0$

$\implies \sf \bold{\green{x =\: 3}}$

$\implies \bf{(x + 1) =\: 0}$

$\implies \sf x + 1 =\: 0$

$\implies \sf\bold{\green{x =\: - 1}}$

Hence, the zeros are 3 and - 1.

Now,

$\mapsto$ The zeros of the polynomial x² - 2x - 3 are half of the zeros of polynomial ax² + bx + c, then,

$\leadsto \sf\bold{The\: zeros\: of\: ax^2 + bx + c\: is\: 6\: and\: -\: 2}$

Hence, the polynomial will be :

$\implies \sf x^2 - (\alpha + \beta)x + (\alpha\beta) =\: 0$

We have :

• α = 6
• β = - 2

$\implies \sf x^2 - \{6 + \{- 2)\}x + \{6 \times (- 2)\} =\: 0$

$\implies \sf x^2 - (6 - 2)x + (- 12) =\: 0$

$\implies \sf x^2 - 4x - 12 =\: 0$

where,

• a = 1
• b = - 4
• c = - 12

Now, we have to find the value of (a - b - c) :

$\leadsto \sf\bold{(a - b - c)}$

$\longrightarrow \sf \{1 - (- 4) - (- 12)\}$

$\longrightarrow \sf (1 + 4 + 12)$

$\longrightarrow \sf (5 + 12)$

$\longrightarrow \sf (17)$

$\longrightarrow \sf\bold{\red{17}}$

${\small{\bold{\underline{\therefore\: The\: value\: of\: (a - b - c)\: is\: 17\: .}}}}$