Question

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

Answer 1

Given:-

• $\alpha$ and $\beta$ are the zeroes of polynomial.

• f(x) = x² - 5x + k

• $\sf \alpha - \beta = 1$

To find:-

• Value of k.

Solution:-

★ x² - 5x + k

The given equation is in the form of standard quadratic equation

that is ax² + bx + c = 0.

Thus, the value of a, b and c are:-

• a = 1
• b = -5
• c = k

✦ We know that:-

Sum of zeroes ( $\sf \alpha + \beta$ ) = $\dfrac{-b}{a}$

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

$:\implies\sf ( \alpha + \beta )$ = 5

Product of zeroes ( $\sf \alpha \beta$ ) = $\dfrac{c}{a}$

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

$:\implies\sf ( \alpha \beta )$ = k

$\rule{200}{2}$

Now, $\sf \alpha - \beta = 1$

✦ Squaring both sides:-

$:\implies\sf ( { \alpha } - { \beta })^2= (1)^2$

$:\implies\sf { \alpha }^2 + { \beta }^2 - 2 \alpha \beta = 1$

✦ Add and subtract $2 \alpha \beta$

$:\implies\sf ( { \alpha }^2 + { \beta }^2 + 2 \alpha \beta ) - 4 \alpha \beta = 1$

$:\implies\sf ( { \alpha }+ { \beta })^2 - 4 \alpha \beta = 1$

Putting value of $( \alpha + \beta )$ and $( \alpha \beta )$ :-

$:\implies\sf (5)^2 - 4 \times k = 1$

$:\implies\sf 25 - 4k = 1$

$:\implies\sf - 4k = 1 - 25$

$:\implies\sf - 4k = -24$

$:\implies\sf k = \cancel{ \dfrac{-24}{-4}}$

$:\implies\sf k = 6$

★ Hence Solved!!

$\rule{200}{2}$

Answer 2

$\huge\bf{ \underline{Question}}$

• If a and b are the zeros of the polynomial $\ x^{2}-5x+k$ such that $a-b = 1$ , then the value of k is ?

$\huge\bf{ \underline{Answer}}$

• k = 6

$\huge\bf{ \underline{Solution}}$

$\bf{ \underline{ \boxed{ \orange{Sum\:of\:zeroes\:=\:\dfrac{-b}{a}}}}}$

$\bf{So\:}$

$\sf{a+b\:=\:\dfrac{-(-5)}{1}}$

$\sf{a+\cancel{b}\:=\:5}$

$\sf{a-\cancel{b}\:=\:1}$

________________________

$\sf{a+a\:=\:5+1}$

$\sf{2a\:=\:6}$

$\sf{a\:=\dfrac{6}{2}}$

$\sf{a\:=\dfrac{\cancel{6}}{\cancel{2}}}$

$\sf{\bold{\boxed{\boxed{\green{\dag{a\:=\:3}}}}}}$

$\mapsto\sf{a-b\:=\:1 }$

$\mapsto\sf{3-b\:=\:1}$

$\mapsto\sf{b\:=\:3-1}$

$\mapsto\sf{b\:=\:2}$

$\sf{\bold{\boxed{\boxed{\green{\dag{b\:=\:2}}}}}}$

$\bf{ \underline{ \boxed{ \orange{Product\:of\:zeroes\:=\:\dfrac{c}{a}}}}}$

$\sf{a\times b\:=\:\dfrac{k}{1}}$

$\sf{3\times2\:=\:k}$

$\huge\sf{\boxed{\boxed{\green{\dag{k\:=\:6}}}}}$