Question

if alpha beta are zeros of polynomial x square - 7 x + K such that Alpha minus beta is equal to 5 find the value of k

answer is~ (6)​

Answer 1

Step-by-step explanation:

$\alpha \: and \: \beta are \: the \: \\ zeroes \: of \: the \: polynomial$

${x}^{2} - 7x + k$

we know that,

$\alpha + \beta = \frac{coeffiecnt \: of \: x}{coefficient \: of \: {x}^{2} }$

$\alpha + \beta = \frac{ - b}{a}$

$\alpha + \beta = \frac{ 7}{1}$

$\alpha + \beta = 7........(1)$

$\alpha - \beta = 5......(2)$

so adding eq.1 and eq.2

we get,

$2 \alpha = 12 \\ \alpha = 6$

putting that value of alpha in eq.1

$6 + \beta = 7 \\ \beta = 7 -6 \\ \beta = 1$

we know that,

$\alpha \beta = \frac{constat}{coeffiect \: of \: x}$

$\alpha \beta = \frac{c}{a}$

$\alpha \beta = \frac{k}{1}$

$1 \times ( 6) = k \\ 6 = k$

HOPE IT HELPS.

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Answer 2

Solution :-

x² - 7x + k

Comparing x² - 7x + k with ax² + bx + c

We get the values as

• a = 1

• b = - 7

• c = k

We know that

Sum of zeroes i.e α + β = - b/a

⇒ α + β = - (-7)/1

⇒ α + β = 7 --eq(1)

Given

α - β = 5 ---eq(2)

Adding eq(1) and eq(2)

⇒ α + β + (α - β) = 7 + 5

⇒ α + β + α - β = 12

⇒ 2α = 12

⇒ α = 12/2

⇒ α = 6

Substitute α = 6 in eq(1)

⇒ α + β = 7

⇒ 6 + β = 7

⇒ β = 7 - 6

⇒ β = 1

We know that

Product of zeroes i.e αβ = c/a

⇒ 6(1) = k/1

⇒ 6 = k

⇒ k = 6

Therefore the value of k is 6.