Question

the roots of quadratic equations x²-5x+3(k-1)=0 are such that alpha-bita=1 then value of k is?​

Answer 1

Answer:

k=3

Step-by-step explanation:

(Alpha-beta) ^2=(alpha+beta)^2 -4alpha.beta

(1)^2=(5)^2-4alpha.beta

1=25-4alpha.beta

1-25= -4alpha.beta

-24/-4= alpha. beta

6= alpha.beta

here alpha. beta=3k-3

6=3k-3

6+3=3k

9/3= k

3=k

Answer 2

Step-by-step explanation:

Given:-

• A quadratic equation x² - 5x + 3(k - 1) = 0

• α and β are the roots of the equation.

• α - β = 1

To Find:-

• The value of k

Solution:-

For a quadratic equation ax² + bx + c = 0

Sum of zeroes = $\rm\dfrac{-b}{a}$

Product of zeroes = $\rm\dfrac{c}{a}$

In equation x² - 5x + 3(k - 1)

• a = 1

• b = -5

• c = 3(k - 1)

$\sf sum \: of \: roots \: ( \alpha + \beta ) = \dfrac{ - ( - 5)}{1} = 5$

$\sf \implies \alpha + \beta = 5......(i)$

$\sf \implies \alpha - \beta = 1......(ii)$

Adding equation (i) and (ii)

$\sf \implies \alpha + \beta + \alpha - \beta = 5 + 1$

$\sf \implies 2\alpha = 6$

$\sf \implies \alpha = \dfrac{6}{2} = 3$

$\sf Substitute \: \alpha = 3 \: in \: equation \: (i)$

$\sf \implies 3 + \beta = 5$

$\sf \implies \beta = 5 - 3 = 2$

$\sf \implies \alpha \beta = 2 \times 3 = 6$

Product of zeroes = $\rm\dfrac{c}{a}$

$\sf \implies \dfrac{3(k - 1)}{1} = 6$

$\sf \implies 3k - 3 = 6$

$\sf \implies 3k = 6 + 3$

$\sf \implies 3k = 9$

$\sf \implies k = \dfrac{9}{3} = 3$

$\large\longrightarrow \underline{\boxed{\sf \therefore k = 3}}$