Question

find the zeros of the quadratic polynomial and verify the relation between zeros and its Coefficient 6x² + 3​

Answer 1

Answer:

⇒ x = -1/3

Hence, the zeros of the quadratic equation are 3/2 and -1/3. Therefore, the relationship between zeros and their coefficients is verified.

Step-by-step explanation:

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Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆⠀GIVEN  POLYNOMIAL :  6x² - 3 = 0

$\qquad \dashrightarrow \sf 6x^2 - 3 \: =\:\:0\: \: \\\\\qquad \dashrightarrow \sf 3 \:\bigg( 2x^2 - 1 \:\bigg) \: =\:\:0\: \: \\\\\qquad \dashrightarrow \sf 2x^2 - 1 \: =\:0\:\times \: 3\: \: \\\\\qquad \dashrightarrow \sf 2x^2 - 1 \: =\:\:0\: \: \\\\\qquad \dashrightarrow \sf 2x^2 \: =\:\:\:1\: \: \\\\\qquad \dashrightarrow \sf \:x^2\:=\:\dfrac{1}{2}\: \:\\\\\qquad \dashrightarrow \sf \:x\:=\:\pm \:\sqrt{\dfrac{1}{2}}\:\:\\\\\qquad \dashrightarrow \sf \:x\:=\: \:\sqrt{\dfrac{1}{2}}\:\:\:or \:\:-\:\:\sqrt{\dfrac{1}{2}} \: \: \\\\\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:x\:= \: \:\sqrt{\dfrac{1}{2}}\:\:\:or \:\:-\:\:\sqrt{\dfrac{1}{2}} \:\:\: }}}}}\:\:\bigstar$

$\qquad \therefore \:\underline { \sf The \:zeroes \:of \:Polynomial \:are \: \bf \sqrt{1 /2} \:\: \sf and \:\bf -\sqrt{1/2} \:\:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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

$\qquad \underline {\boxed {\pmb {\red { \:\maltese \:\: Sum \: of \: zeroes \:}\: \:\purple{ ( \: \alpha \:+\beta \:)}\red{\:\::\:}}}}\\\\\qquad \dashrightarrow \sf \bigg( \:\: \alpha \:\:+ \: \beta \:\:\bigg) \:=\: \dfrac{-(Cofficient \:of\:x)\:}{Cofficient \:of \:x^2 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg( \:\: \:\sqrt{\dfrac{1}{2}}\:\:\bigg) \:+\:\bigg( \:-\sqrt{\dfrac{1}{2}}\:\:\bigg) \:=\: \dfrac{-(0)\:}{6 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg( \:\cancel {\: \:\sqrt{\dfrac{1}{2}}}\:\:\bigg) \:+\:\bigg( \cancel {\:-\sqrt{\dfrac{1}{2}}}\:\:\bigg) \:=\: \dfrac{-(0)\:}{6 \:\:}\:\:$

$\qquad \dashrightarrow \sf 0 \:=\: \dfrac{-(0)\:}{6 \:\:}\:\:$

$\qquad \dashrightarrow \sf 0 \:=\: 0\:\:$

$\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:0 \:\:= \:\:0\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

$\qquad \underline {\boxed {\pmb {\red { \:\maltese \:\: Product \: of \: zeroes \:}\: \:\purple{ ( \: \alpha \:\beta \:)}\red{\:\::\:}}}}\\\\\qquad \dashrightarrow \sf \bigg( \:\: \alpha \:\: \: \beta \:\:\bigg) \:=\: \dfrac{ \:Constant \:Term\:}{Cofficient \:of \:x^2 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg( \:\: \:\sqrt{\dfrac{1}{2}}\:\:\bigg) \:\times\:\bigg( \:-\sqrt{\dfrac{1}{2}}\:\:\bigg) \:=\: \dfrac{-3\:}{6 \:\:}\:\:$

$\qquad \dashrightarrow \sf \:\: \:\sqrt{\dfrac{1}{2}}\:\: \:\times\: \:-\sqrt{\dfrac{1}{2}}\:\: \:=\: \dfrac{-3\:}{6 \:\:}\:\:$

$\qquad \dashrightarrow \sf \:\: \dfrac{-1}{2}\:\: \:=\: \dfrac{-3\:}{6 \:\:}\:\:$

$\qquad \dashrightarrow \sf \:\: \dfrac{-1}{2}\:\: \:=\: \cancel {\dfrac{-3\:}{6 \:\:}}\:\:$

$\qquad \dashrightarrow \sf \:\: \dfrac{-1}{2}\:\: \:=\: \dfrac{-1\:}{2 \:\:}\:\:$

$\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:\dfrac{-1}{2} \:\:= \:\:\dfrac{-1}{2}\: }}}}}\:\:\bigstar$

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