Question

iff p and q are the zeros of polynomial f(x)=$x^{2} -5x +k$ such that p-q = 0, find the value of k

Answer 1

Answer:

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$(p - q {})^{2} = ({p + q})^{2} - 4pq$

And we know that

$p + q \: = - \frac{b}{a}$

where b is the coefficient of x and a is the coefficient of x^2

$p \times q = \frac{c}{a}$

where c is the constant term .

By substituting the values in above equation

We get

$( {p - q})^{2} = (\frac{5}{1} ) {}^{2} - 4 \times \frac{k}{1}$

And

According to question

$p - q = 0$

Therefore ,

after substituting its value ,

we get

$0 = 25 - 4k$

$k = \frac{25}{4}$

Hence value of k is 25/4

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Answer 2

AnswEr :

$\bf{\large{\green{\underline{\underline{\bf{Given\::}}}}}}$

If p and q are the zeroes of polynomial f(x) = x² - 5x + k such that p-q = 0.

$\bf{\large{\red{\underline{\underline{\bf{To\:find\::}}}}}}$

The value of k.

$\bf{\large{\orange{\underline{\underline{\bf{Explanation\::}}}}}}$

We have a quadratic polynomial x² - 5x + k such that p - q = 0.

As we compared by ax² + bx + c = 0.

• a = 1
• b = -5
• c = k

Now,

$\bf{\Large{\green{\underline{\underline{\blacksquare{\sf{Sum\:of\:the\:zeroes\::}}}}}}}$

$\longrightarrow\tt{p+q=\dfrac{-b}{a} }\\\\\\\\\longrightarrow\tt{p+q=\dfrac{-(-5)}{1} }\\\\\\\\\longrightarrow\tt{p+q=5}\\\\\\\\\longrightarrow\tt{\red{p=5-q............................(1)}}$

We have p - q = 0 so, putting this from equation (1), we get;

$\longrightarrow\tt{5-q-q=0}\\\\\\\\\longrightarrow\tt{5-2q=0}\\\\\\\\\longrightarrow\tt{\cancel{-}2q=\cancel{-}5}\\\\\\\\\longrightarrow\tt{\green{q\:=\:\dfrac{5}{2} }}$

putting the value of q in equation (1), we get;

$\longrightarrow\tt{p\:=\:5-\dfrac{5}{2} }\\\\\\\\\longrightarrow\tt{p\:=\:\dfrac{10-5}{2} }\\\\\\\\\longrightarrow\tt{\green{p\:=\:\dfrac{5}{2} }}$

$\bf{\Large{\green{\underline{\underline{\blacksquare{\sf{Product\:of\:zeroes\::}}}}}}}$

$\longrightarrow\tt{\alpha \times \beta =\dfrac{c}{a}} \\\\\\\\\longrightarrow\tt{\alpha \times \beta =\dfrac{k}{1} }\\\\\\\\\longrightarrow\tt{\alpha \times \beta =k}\\\\\\\\\longrightarrow\tt{\dfrac{5}{2} \times \dfrac{5}{2} =k}\\\\\\\\\longrightarrow\tt{\red{\dfrac{25}{4} =k}}$

∴ The value of k is $\bf{\dfrac{25}{4} }}$.