Question

If a and B are the zeroes of the polynomial f(x) = x2 – 5x + k such that a - b = 1, then value of k is:
O 8​

Answer 1

$\rule{400}4$

☆ ANSWER:-

▪︎ Given:-

• $\sf f(x) = {x}^{2}-5x+k$

• $\sf\alpha\: And\: \beta\:are\:zeroes$

• $\sf\alpha-\beta = 1$

▪︎ To find:-

• The value of k

▪︎ Concept Used:-

• In a quadratic polynomial sum of zeroes $\sf\large = \frac{-b}{a}$

$\rule{400}2$

▪︎ So Here,

$\tt\mapsto \alpha + \beta = \frac{ - b}{a} \\ \\ \tt\mapsto \alpha + \beta = \frac{ - ( - 5)}{1} \\ \\ \tt\mapsto \alpha + \beta = 5...(1)$

▪︎ Also it is given that:-

$\tt\mapsto \alpha - \beta = 1...(2)$

$\rule{400}2$

▪︎ Now,

• Let us add eq(2) and eq(1)..

$\tt\mapsto( \alpha + \beta ) + ( \alpha - \beta ) = (5) + (1) \\ \\\tt\mapsto \alpha + \cancel \beta + \alpha - \cancel \beta = 5 + 1 \\ \\ \tt\mapsto \alpha + \alpha = 6 \\ \\ \tt\mapsto2 \alpha = 6 \\ \\ \tt\mapsto \alpha = \cancel \frac{6}{2} \\ \\ \tt\large\red{\fbox{\mapsto \alpha = 3}}$

$\large{\mathcal{\blue{\underline{\underline{\pink{\therefore{One\:Zero\:Is\:3}}}}}}}$

$\rule{400}2$

▪︎ So,

• If we put the value of x = 3 then the result of the given f(x) should be equal to 0.

$\sf \mapsto {x}^{2} - 5x + k = 0 \\ \\ \sf \mapsto {3}^{2} - 5(3) + k = 0 \\ \\\sf \mapsto 9 - 15 + k = 0 \\ \\ \sf \mapsto - 6 + k = 0 \\ \\ \sf \mapsto \: k = 0 + 6 \\ \\ \sf\huge\pink{\fbox {\mapsto \: k = 6}}$

$\rule{400}2$

${\small{\green{\boxed{\bold{\mathcal{\therefore{The\:req.\:value\:of\:k\:is\:6.}}}}}}}$

$\rule{400}4$