Question

if zeroes of polynomial X2 -7x+k are such that alpha -beta =1 then value of k​

Answer 1

Solution :

The given polynomial :

> x² - 7x + k .

Let the two roots of this polynomial be alpha and beta.

According to the questión ,

Alpha - Beta = 1.

We have to find the value of k .

Let us proceed .

Alpha - Beta = 1

Alpha + Beta = (-b)/a = 7

Adding these two equations -

> 2 Alpha = 8

> Alpha = 4

> Beta = 3

Product of roots > (c/a) > k

> k = 4 × 3 = 12

The value of k is 12 and the polynomial becomes x² - 7x + 12

Verifying the roots -

x² - 7x + 12

> x² - 4x - 3x + 12

> x( x - 4) - 3( x - 4)

> ( x - 3)( x - 4)

The roots are 3 and 4 .

This is the the required answer.

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Answer 2

Step-by-step explanation:

$\sf\begin{gathered} Compare \: given \: polynomial \\x^{2} - 7x + k\: with \:ax^{2} + bx + c , we \:get \end{gathered}$

$\sf a = 1 , \: b = -7 \: and \: c = ka=1,b=−7andc=k$

$\sf\begin{gathered} \alpha \: and \:\beta \: are \: zeroes \: of \\given \: polynomial \end{gathered}$

➪$\sf Sum \:of \: the \: zeroes = \frac{-b}{a}$

➪$\sf\begin{gathered} \implies \alpha + \beta = \frac{-(-7)}{1} \\= 7 \: --(1) \end{gathered}$

$\sf⟹ Product \:of \: the \: zeroes = \frac{c}{a}$

$\sf\begin{gathered} \implies \alpha \beta = \frac{k}{1} \\= k \: --(2) \end{gathered}$

$\tt But , \: \alpha - \beta = 5 \: ---(3) \: (given)But,α−β=5−−−(3)(given)$

$\tt Add \: equations \: (1) \:and \: (3)$

$\implies 2\beta = 12⟹2β=12$

$\tt\implies \beta = \frac{12}{2}$

➪$\tt Put \: \beta = 6 \: in \: Equation \:(1) ,we \:6\:in\:Equation(1),$

➪$\tt\implies \alpha + 6 = 7⟹α+6=7$

➪$\tt\implies \alpha = 7 - 6⟹α=7−6$

➪$\tt\implies \alpha = 1⟹α=1$

➪$\tt Put \: \alpha = 1 \: and \: \beta = 6 , we \:get$

$\tt\implies 1 \times 6 = k⟹1×6=k$

➪$\tt\implies k = 6⟹k=6$

Therefore.,

➪$\tt\red { Value \: of \: k } \green {= 6}$

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