Question

plz give me correct answer​

Answer 1

$\LARGE\underline{\underline{\sf{\purple{Required\:answer:-}}}}$

Given :-

• The equation x² - 5x + 3(k-1) = 0 has roots α and β such that α - β = 1.

To Find :-

• The value of k

Knowledge required :-

• If a quadratic equation ax² + bx + c has roots α and β . Then the relation between the coefficents and the roots is given by ,

$\\ \implies \boxed{ \sf{ \alpha - \beta = \frac{( \: constant \: term)}{( {x}^{2} \: coefficient) } }}$

Solution :-

We have for the equation x² - 5x + 3(k-1) as α - β = 1.

Now by applying the condition ,

$\\ \implies \sf \: \alpha - \beta = \dfrac{3(k - 1)}{1}$

Substituting the α - β in the equation we get ;

$\\ \implies \sf \: 1 = \frac{3(k - 1)}{1}$

$\\ \implies \sf \: 1 = 3(k - 1)$

$\\ \implies \sf \: \frac{1}{3} = k - 1$

$\\ \implies \sf \: \frac{1}{3} + 1 = k$

$\\ \implies \sf \: \frac{4}{3} = k$

$\\ \implies\underline{\boxed {\bf{k = \frac{4}{3} }}}$

•Hence , The value of k is $\sf{\dfrac{4}{3}}$

Answer 2

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Given:

• Quadratic eqn - $x^2 - 5x + 3(k-1) = 0$
• Roots - $\alpha \& \beta$
• $\alpha - \beta = 1$

To Find:

• Value of k

Solution:

$\implies x^2 - 5x + 3(k-1) = 0$

• a = coefficient of x^2 = 1
• b = coefficient of x = -5
• c = constant = 3(k-1)

The roots of the equation, $x^2 - 5x + 3(k-1) = 0$, are $\alpha \& \beta$.

So,

The sum of zeroes = $\dfrac{-b}{a}$

$\alpha + \beta = \dfrac{-(-5)}{1}$

$\alpha + \beta = 5 -----------(1)$

But,

$\alpha - \beta = 1 -----------(2) (given)$

Adding (1) & (2),

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

$2\alpha = 6$

$\alpha = 3 ------------(3)$

Putting (3) in (1),

$3 + \beta = 5 \\ \beta = 2$

Now,

The product of zeroes = $\dfrac{c}{a}$

$\alpha*\beta = \dfrac{3(k-1)}{1}$

3*2 = 3(k-1)

2 = k - 1

k = 3

The value of k is 3.

Formulae Used:

• The sum of zeroes = $\dfrac{-b}{a}$
• The product of zeroes = $\dfrac{c}{a}$

HOPE IT HELPS!!!