[tex]{\boxed{\underline{\tt{ \orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}[/tex] ★GIVEN:- A Polynomial — (x^2 + x –6). ★TO VERIFY:- Whether 2 and –3 are the roots of the Polynomial. ★SOLUTION:- [tex] \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}[/tex] First we will put 2 as the root. [tex] \displaystyle \sf : \longrightarrow{p(x) = x^2 + x – 6 }[/tex] [tex] \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}}}}[/tex] Now we will put –3 as the root. [tex] \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}}}}[/tex] Hence, after putting 2 and –3 The results coming is 0 Therefore, They are the roots of the equation. HENCE VERIFIED Reply
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! Reply
[tex]{\boxed{\underline{\tt{ \orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}[/tex]
★GIVEN:-
★TO VERIFY:-
★SOLUTION:-
[tex] \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}[/tex]
First we will put 2 as the root.
[tex] \displaystyle \sf : \longrightarrow{p(x) = x^2 + x – 6 }[/tex]
[tex] \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}}}}[/tex]
Now we will put –3 as the root.
[tex] \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}}}}[/tex]
Hence, after putting 2 and –3
The results coming is 0
Therefore, They are the roots of the equation.
HENCE VERIFIED
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!