Question

Check 3 and -2 are-zeroes of p(x) = x²-x-6 and
verify the relationship between the zeroes and
coefficients
​

Answer 1

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆⠀GIVEN  POLYNOMIAL :  x² - x - 6 = 0

⠀⠀⠀⠀⠀Checking whether 3 and -2 are zeroes of polynomial or not .

$\qquad \dashrightarrow \sf x^2 - x - 6 \: =\:\:0\: \: \\\\\qquad \dashrightarrow \sf x^2 - 3x + 2x - 6 \: =\:\:0\: \: \\\\\qquad \dashrightarrow \sf x (x - 3) + 2 ( x - 3 ) \: =\:\:0\: \: \\\\\qquad \dashrightarrow \sf (x - 3) ( x + 2 ) \: =\:\:0\: \: \\\\\qquad \dashrightarrow \sf \:x\:=\:\: 3 \:\:or\:\:x\:\:=\:\: - 2 \:\:\: \: \\\\\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:x\:= \: 3 \:\: or \: -2 \:\:\: }}}}}\:\:\bigstar$

$\qquad \therefore \:\underline { \sf The \:zeroes \:of \:Polynomial \:are \: \bf 3 \:\: \sf and \:\bf -2 \:\:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \bigstar \:\:\underline {\sf \: Verifying \: \:\bf \:Relationship\: \:\sf\:between \:\:\bf\:zeroes\:\sf\:and\:\bf Cofficients \:\sf\:\:of \:\:\:Polynomial \:\:}:$

$\qquad \underline {\boxed {\pmb {\red { \:\maltese \:\: Sum \: of \: zeroes \:}\: \:\purple{ ( \: \alpha \:+\beta \:)}\red{\:\::\:}}}}\\\\\qquad \dashrightarrow \sf \bigg( \:\: \alpha \:\:+ \: \beta \:\:\bigg) \:=\: \dfrac{-(Cofficient \:of\:x)\:}{Cofficient \:of \:x^2 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \bigg( \:\: 3\:\:\bigg) \:+\:\bigg( \:\:-2 \:\:\bigg) \:=\: \dfrac{-(-1)\:}{1 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \bigg( \:\: 3\:\:- \:\:2 \:\:\bigg) \:=\: \dfrac{-(-1)\:}{1 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \bigg( \:\: 3\:\:- \:\:2 \:\:\bigg) \:=\: \dfrac{1\:}{1 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \bigg( \:\: 1 \:\:\bigg) \:=\: \dfrac{1\:}{1 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \:\: 1 \: \:=\: 1\:\:\\\\\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:1 \:\:= \:\:1\: }}}}}\:\:\bigstar \\\\ \qquad\qquad \bf AND ,$

$\qquad \underline {\boxed {\pmb {\red { \:\maltese \:\: Product \: of \: zeroes \:}\: \:\purple{ ( \: \alpha \:\beta \:)}\red{\:\::\:}}}}\\\\\qquad \dashrightarrow \sf \bigg( \:\: \alpha \:\: \: \beta \:\:\bigg) \:=\: \dfrac{ \:Constant \:Term\:}{Cofficient \:of \:x^2 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \bigg( \:\: 3\:\:\bigg) \:\times\:\bigg( \:\:-2 \:\:\bigg) \:=\: \dfrac{-6\:}{1 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \:\: 3\:\: \:\times\:\bigg( \:\:-2 \:\:\bigg) \:=\: \dfrac{-6\:}{1 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \:\: 3\:\: \:\times\:\bigg( \:\:-2 \:\:\bigg) \:=\: -6\:\:\:\\\\\qquad \dashrightarrow \sf \:\:\:\bigg( \:\:-6 \:\:\bigg) \:=\: -6\:\:\:\\\\\qquad \dashrightarrow \sf \:\:\: \:\:-6 \:\: \:=\: -6\:\:\:\\\\\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:-6 \:\:= \:\:-6\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀$\therefore {\underline {\bf{ Hence, \:Verified \:}}}$

Answer 2

Given :

• Polynomial p( x ) = x²- x - 6

• The zeroes of the polynomial are given as 3 and ( -2 )

‎

‎

To :

• Check whether the zeroes of the given polynomial are 3 and ( -2 )

• Verify the relationship between the zeroes and the coefficient.

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‎

Concept :

Firstly, we would check whether 3 and ( -2 ) are the zeroes of the given polynomial by factorising following middle term breaking. Then after we would verify the relationship between the zeroes and the coefficient by using two formulae. The two formulae are as follows :

❒ $\small{\underline{\boxed{\mathrm{Sum~of~the~zeroes~=~\frac{-b}{a}}}}}$

❒ $\small{\underline{\boxed{\mathrm{Product~of~the~zeroes~=~\frac{c}{a}}}}}$

‎ We would equate this two formulae with the sum and product of the zeroes of the given polynomial and verify if LHS = RHS.

‎ Hope am clear let's solve :D~

‎

Solution :

Checking if 3 and ( -2 ) are the zeroes of the given polynomial

$\tt\:Polynomial~:~x^{2}-x-6$

Following middle term breaking method

• Middle term that is 1 can be splitted into 3 and ( -2 ) since subtracting them we would get 1 , whereas multiplying them we would get ( -6 )

$\tt:\implies\:x^{2}-(3-2)x-6$

Multiplying the terms and removing the brackets

$\tt:\implies\:x^{2}-3x+2x-6$

Taking x common from first two terms and 2 common from other two terms

$\tt:\implies\:x(x-3)+2(x-3)$

Taking ( x-3 ) common from whole terms

$\tt:\implies\:(x-3)(x+2)$

Either : $\tt\:x-3=0$ ⠀ Or : $\tt\:x+2=0$

$\tt\leadsto\:x=3$ ⠀ $\tt\leadsto\:x=-2$

Therefore,

The two zeroes of the polynomial are $\large{\mathfrak\purple{3}}$ and $\large{\mathfrak\purple{(-2)}}$.

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Lastly let's verify the relationship between the zeroes and the coefficient.

➷ Sum of the zeroes = 3 + ( -2 ) = 3 - 2 = $\small\fbox\red{1}$

➷ Product of the zeroes = 3 × ( -2 ) = $\small\fbox\red{(-6)}$

‎

Given polynomial :

‎$\tt\:x^{2}-x-6$

Here :

• Coefficient of a, b and c are 1 , ( -1 ) and ( -6 )

Now using the formulae and equating it with the sum and product of the zeroes :

$\tt\:Sum~of~the~zeroes~=~1$

$\tt\twoheadrightarrow\:\frac{-b}{a}~=~1$

Substituting the value of b and a

$\tt\twoheadrightarrow\:\frac{-1}{1}~=~1$

$\tt\twoheadrightarrow\:1=~1$

$\tt\twoheadrightarrow\:LHS=~RHS$

Hence Verified !~

‎

Again :

$\tt\:Product~of~the~zeroes~=~(-6)$

$\tt\twoheadrightarrow\:\frac{c}{a}~=~(-6)$

Substituting the value of c and a

$\tt\twoheadrightarrow\:\frac{-6}{1}~=~(-6)$

$\tt\twoheadrightarrow\:(-6)=~(-6)$

$\tt\twoheadrightarrow\:LHS=~RHS$

Hence Verified !~

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