Polynomial $ f(x)=x^{2}-5 x+k $ has zeroes $ \alpha $ and $ \beta $ such that $ \alpha-\beta=1 . $ Find the value of $ 4 k $.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial \( f(x)=x^{2}-5 x+k \) and \( \alpha-\beta=1 . \)
To do:
We have to find the value of \( 4 k \).
Solution:
We know that,
The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are constants and $a≠0$.
Comparing the given equation with the standard form of a quadratic equation,
$a=1$, $b=-5$ and $c=k$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{-(-5)}{1} = 5$.......(i)
$\alpha-\beta=1$...........(ii)
Adding (i) and (ii), we get,
$\alpha+\beta+\alpha-\beta=5+1$
$2\alpha=6$
$\alpha=3$
$\Rightarrow \beta=\alpha-1=3-1=2$
Product of the roots $= αβ = \frac{c}{a} = \frac{k}{1} = k$
$\Rightarrow 3\times2=k$
$k=6$
$\Rightarrow 4k=4(6)=24$
The value of $4k$ is $24$.
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