Question

If alpha and beta are the zeros of quadratic polynomial p(x)=x^2-5x+k such that alpha -beta =1

Answer 1

QuesTion:

If $\sf{\alpha}$ and $\sf{\beta}$ are the zeroes of quadratic polynomial $\sf{p(x)=x^{2}-5x+k}$ such that $\sf{\alpha-\beta=1}$. Find the value of k.

Answer:

The value of k is 6.

Given:

• The given polynomial is $\sf{p(x)=x^{2}-5x+k}$

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

• $\sf{\alpha-\beta=1}$

To find:

The value of k.

SoLution:

The given quadratic polynomial is $\sf{p(x)=x^{2}-5x+k}$

Here a=1, b=-5 and c=k

Sum of zeroes=$\sf{\frac{-b}{a}}$

$\sf{\therefore{\alpha+\beta=5...(1)}}$

Product of zeroes=$\sf{\frac{c}{a}}$

$\sf{\therefore{\alpha\beta=k...(2)}}$

According to identify:-

$\bold{(a+b)^{2}=(a-b)^{2}+4ab}$

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

...from (1) and (2)

$\sf{5^{2}=1^{2}+4k}$

${\therefore}$ 25=1+4k

${\therefore}$ 4k=25-1

${\therefore{k=\frac{24}{4}}}$

${\therefore}$ k=6

The value of k is 6.

Answer 2

Given ,

The polynomial is (x)² - 5x + k

The difference of zeroes of polynomial is 1

i.e α - β = 1

We know that ,

The sum of zeroes of polynomial is

$\underline{\large \sf \fbox{ \alpha + \beta = \frac{ - b}{a} }}$

Thus ,

$\sf \Rightarrow \alpha + \beta = \frac{ - ( - 5)}{1} \\ \\ \sf \Rightarrow \alpha + \beta = 5$

And the product of zeroes of polynomial is

$\underline{ \large \sf \fbox{ \alpha \beta = \frac{c}{a} }}$

Thus ,

$\sf \Rightarrow \alpha \beta = \frac{k}{1} \\ \\ \sf \Rightarrow \alpha \beta = k$

$\sf Using \: identity \: \sf \fbox{ {(a + b)}^{2} = {(a - b)}^{2} + 4ab } \: , we \: get$

$\sf \Rightarrow {(5)}^{2} = {(1)}^{2} + 4k \\ \\\sf \Rightarrow 25 = 1 + 4k \\ \\ \sf \Rightarrow 4k= 24 \\ \\\sf \Rightarrow k = 6$

$\therefore \underline{ \bold{ \sf{The \: value \: of \: k \: is \: 6}}}$