Question

find the zeros of the polynomial X ^2 + x and verify the relationship between the zeros and its coefficient ​

Answer 1

Given :

• Polynomial : x² + x = 0

To Find :

• Zeroes of the polynomial.
• Verify relationship between zeroes and coefficients.

Solution :

Let's find the zeroes.

➣ $\mathtt{x^2+x=0}$

➣ $\mathtt{x(x+1)=0}$

➣ $\mathtt{x=0\:\:or\:\:x+1=0}$

➣ $\mathtt{x=0\:\:or\:\:x=-1}$

$\large{\boxed{\mathtt{\red{Zeroes\:of\:the\:polynomial\:=\:0\:and\:-1}}}}$

Relationship between zeroes and coefficients :

Compare the given quadratic polynomial with general form :

• ax² + bx + c = 0

Values of variables :

• a = 1
• b = 1
• c = 0

Let,

• α = 0
• β = - 1

Sum of zeroes :

α + β = $\mathtt{-0+(-1)}$

➣ $\mathtt{0-1}$

➣ $\mathtt{-1}$

➣ $\mathtt{-1}$ ____(1)

We know that, sum of zero is found using the form :

α + β = $\mathtt{\dfrac{-b}{a}}$

➣ $\mathtt{\dfrac{-(1)}{1}}$

➣ $\mathtt{-1}$ ____(2)

From (1) and (2) ,we see that LHS of both equation are -1.

Hence, we conclude that relationship between sum of zeroes and coefficients is satisfied.

Product of zeroes :

αβ = $\mathtt{0\:\times\:-1}$

➣ $\mathtt{0}$ ____(3)

Now, product zeroes is found using the form :

αβ = $\mathtt{\dfrac{c}{a}}$

➣ $\mathtt{\dfrac{0}{1}}$

➣ $\mathtt{0}$ ____(4)

From (3) and (4), we see that LHS of both equation are equal.

Hence, relationship between product of zeroes and coefficients is satisfied.

Answer 2

$\huge \fcolorbox{red}{pink}{Solution :) }$

Given ,

The polynomial is x² + x or x² + x + 0

• Coefficient of (x)² = 1
• Coefficient of x = 1
• Constant term = 0

Zeroes or roots of the polynomial is a value of x where the polynomial is equal to 0 , so

$\sf \hookrightarrow {(x)}^{2} + x = 0 \\ \\ \sf \hookrightarrow</p><p>x(x + 1) = 0 \\ \\ \sf \hookrightarrow </p><p>x = 0 \: and \: x = - 1$

Hence , the zeroes of the given polynomial is 0 and - 1

$\huge \fcolorbox{red}{pink}{ Verification :) }$

We know that ,

$\large \mathtt{ \star \: \: \fbox{ Sum \: of \: zeroes = - \frac{Coefficient \: of \: {x}}{Coefficient \: of \: {(x)}^{2} } }}$

LHS = 0 + (-1) = -1

and

RHS = -1/1 = -1

$\therefore \sf \: LHS = RHS$

$\large \mathtt{ \star \: \: \fbox{Product \: of \: zeroes = \frac{Constant \: term}{Coefficient \: of \: {(x)}^{2} } }}$

LHS = 0 × (-1) = 0

and

RHS = 0/1 = 0

$\therefore \sf \: LHS = RHS$

Hence verified

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