Math, asked by minalmandalkusmi, 3 months ago

the šeros of
verify that 2and -3
are the zeroes of
polynomial x2+x-6​

Answers

Answered by snehitha2
17

Step-by-step explanation:

Given :

quadratic polynomial x² + x – 6

To verify :

2 and –3 are the zeroes of the given polynomial

Solution :

If 'a' is a zero of the polynomial p(x), then p(a) = 0

Let p(x) = x² + x – 6

To check if 2 is a zero of the given polynomial :

Put x = 2,

p(2) = 2² + 2 – 6

= 4 + 2 – 6

= 6 – 6

= 0

p(2) = 0; hence 2 is a zero of the given polynomial.

To check if 3 is a zero of the given polynomial :

Put x = 3,

p(–3) = (–3)² + (–3) – 6

= 9 – 3 – 6

= 6 – 6

= 0

p(–3) = 0; hence –3 is a zero of the given polynomial.

Hence verified!

Answered by diajain01
63

{\boxed{\underline{\tt{ \orange{Required  \:  \: answer \:  \: is \:  \:  as  \:  \: follows:-}}}}}

★GIVEN:-

  • A Polynomial -- (x^2 + x -6).

★TO VERIFY:-

  • Whether 2 and -3 are the roots of the Polynomial.

★SOLUTION:-

 \leadsto \displaystyle \sf{x^2+x-6} \\  \\  \leadsto \displaystyle \sf{x^2  + \pink{3x - 2x} - 6} \\  \\  \leadsto \displaystyle \sf{x(x + 3) - 2(x + 3)} \\  \\  \leadsto \displaystyle \sf{(x - 2)(x + 3)} \\  \\  \therefore \:  \sf{x - 2 = 0 } \\  \\  \sf{ \purple{x = 2}} \:  \\  \\ and \:  \sf{(x + 3) = 0} \\  \\  \sf{ \purple{x  = - 3}} \\  \\   \bf{so \: the \: roots \: are \: 2 \: and \:  - 3}

First we will put 2 as the root.

 \displaystyle \sf  : \longrightarrow{p(x) = x^2 + x  - 6 }

 \displaystyle \sf  : \longrightarrow{p(2) = 2^2 + 2  - 6 } \\  \\  \displaystyle \sf  : \longrightarrow{p(2) = 4 + 2 - 6 } \\  \\  \displaystyle \sf  : \longrightarrow{p(2) = 6 - 6 } \\  \\  \displaystyle \sf  : \longrightarrow{p(2) = x^2 + x  - 6 } \\  \\  \therefore \:{  \boxed{ \underline{ \pink{ p(2) = 0}}}}

Now we will put -3 as the root.

 \displaystyle \sf  : \longrightarrow{ p(-3)= (-3)^2+(-3)-6} \\  \\ \displaystyle \sf  : \longrightarrow{ p(-3)= 9 - 3 - 6} \\  \\\displaystyle \sf  : \longrightarrow{ p(-3)= 9 - 9} \\  \\  \therefore \: \:{  \boxed{ \underline{ \pink{ p( - 3) = 0}}}}

Hence, after putting 2 and -3

The results coming is 0

Therefore, They are the roots of the equation.

HENCE VERIFIED

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