Question

The roots α and β of the quadratic equation x2 – 5x + 3(k - 1) = 0 are such that α – β = 1.

Find the value of k.

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Answer 1

x2 – 5x + 3(k - 1) = 0

sum of roots = a+b = 5/1 = 5 .........1

a-b = 1 (given) .........2

add eq1 and eq2

2a = 6

a= 3

put a in eq2 for b

3 -b = 1

-b = -2

b= 2

product of roots = a×b = 3(k-1)/1 .........3

but the product of roots = 6 ...........4

form eq1 and eq2

3(k-1) = 6

3k -3 = 6

3k = 6+3

3k =9.

k=3

Answer 2

${x}^{2} - 5x + 3(k - 1) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ a {x}^{2} + bx + c = 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ a = 1 \: \: \: \: \:b = - 5 \: \: \: \: \: c = 3(k - 1) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \pink{let \: \: \: \alpha \: \: \: nd \: \: \: \: \beta \: \: \: be \: \: \: roots} \\ \\ \alpha + \beta = \frac{ - b}{a} = \frac{5}{1} \\ \\ \alpha + \beta = 5 \\ \\ \alpha \beta = \frac{c}{a} = \frac{3(k - 1)}{1} \\ \\ \alpha \beta = 3k - 3 \\ \\ \red{since \: \: \: \: given} \\ \\ \alpha - \beta = 1 \\ \\ \: \: \: nd \: \: \: we \: \: \: got \\ \\ \alpha + \beta = 5 \\ \\ \blue{hence \: \: \: \: adding \: \: \: \: this \: \: \: \: both } \\ \\ \alpha - \beta = 1 \\ \alpha + \beta = 5 \\ ..................... \\ 2 \alpha = 6 \\ \\ \alpha = \frac{6}{2} \\ \\ \alpha = 3 \\ \\ 3 - \beta = 1 \\ \\ \beta = 2 \\ \\ \pink{now} \\ \\ \alpha \beta = 3k - 3 \\ \\ 3 \times 2 = 3k - 3 \\ \\ 6 = 3k - 3 \\ \\ 3k = 9 \\ \\ k = \frac{9}{3} \\ \\ \green{ k = 3}$

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