Question

find the zeros of the quatratic polynomial and verify the relationship between the zeros and the cofficients

Answer 1

$4x {}^{2} - 4x - 3 \\ 4 {x}^{2} - 6x + 2x - 3 \\ 2x(2x - 3) + 1(2x - 3) \\ (2x - 3)(2x + 1)$
putting zero in front of equation
$(2x - 3)(2x + 1) = 0 \\ if \: 2x - 3 = 0 \\ 2x = 3 \\ x = \frac{3}{2 } \\ if \: 2x + 1 = 0 \\ 2x = - 1 \\ x = \frac{ - 1}{2}$

Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN QUADRATIC POLYNOMIAL : 4x² - 4x - 3 = 0

$\qquad \dashrightarrow \sf 4x^2 - 4x - 3 \: =\:\:0\: \:$

$\qquad \dashrightarrow \sf 4x^2 - 6x + 2x - 3 \: =\:\:0\: \:$

$\qquad \dashrightarrow \sf 2x( 2x - 3) + 1 ( 2x - 3 )\: =\:\:0\: \:$

$\qquad \dashrightarrow \sf ( 2x - 3) ( 2x + 1 )\: =\:\:0\: \:$

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

$\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:x\:= \: \dfrac{3}{2} \:\: or \: \dfrac{-1}{2} \:\:\: }}}}}\:\:\bigstar$

$\qquad \therefore \:\underline { \sf The \:zeroes \:of \:Quadratic \:are \: \bf 3/2 \:\: \sf and \:\bf -1/2 \:\:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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

$\qquad \underline {\boxed {\pmb {\red { \:\maltese \:\: Sum \: of \: zeroes \:}\: \:\purple{ ( \: \alpha \:+\beta \:)}\red{\:\::\:}}}}$

$\qquad \dashrightarrow \sf \bigg( \:\: \alpha \:\:+ \: \beta \:\:\bigg) \:=\: \dfrac{-(Cofficient \:of\:x)\:}{Cofficient \:of \:x^2 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg( \:\: \dfrac{3}{2}\:\:\bigg) \:+\:\bigg( \:\:\dfrac{-1}{2} \:\:\bigg) \:=\: \dfrac{-(-4)\:}{4 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg(\:\: \dfrac{3}{2}\:\:-\:\dfrac{1}{2} \:\:\bigg) \:=\: \dfrac{-(-4)\:}{4 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg(\:\: \dfrac{3\:-\:1}{2}\:\: \:\:\bigg) \:=\: \dfrac{-(-4)\:}{4 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg(\:\: \dfrac{2}{2}\:\: \:\:\bigg) \:=\: \dfrac{-(-4)\:}{4 \:\:}\:\:$

$\qquad \dashrightarrow \sf \bigg(\:\: \dfrac{2}{2}\:\: \:\:\bigg) \:=\: \dfrac{4\:}{4 \:\:}\:\:$

$\qquad \dashrightarrow\sf \:\: \cancel {\dfrac{2}{2}}\:\: \:\: \:=\: \cancel {\dfrac{4\:}{4 \:\:}}\:\:$

$\qquad \dashrightarrow \sf \:\:1 \:\: \:=\: 1\:\:$

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀AND ,

$\qquad \underline {\boxed {\pmb {\red {\:\maltese \:\: Product \: of \: zeroes \:}\: \:\purple{ ( \: \alpha \:\beta \:)}\red{\:\::\:}}}}\\\\\qquad \dashrightarrow \sf \bigg( \:\: \alpha \:\: \: \beta \:\:\bigg) \:=\: \dfrac{\:Constant \:term\:)\:}{Cofficient \:of \:x^2 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \bigg( \:\: \dfrac{3}{2}\:\:\bigg) \:\times\:\bigg( \:\:\dfrac{-1}{2} \:\:\bigg) \:=\: \dfrac{-3\:}{4 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \:\: \dfrac{3}{2}\:\: \:\times\: \:\:\dfrac{-1}{2} \:\:\:=\: \dfrac{-3\:}{4 \:\:}\:\:\\\\\qquad \dashrightarrow \sf \:\: \dfrac{-3\:}{4 \:\:}\:\:\:\:\:=\: \dfrac{-3\:}{4 \:\:}\:\:\\\\\qquad \therefore \underline {\boxed{\pmb{\purple{\frak{\:\dfrac{-3\:}{4 \:\:}\:\: \:\:= \:\:\dfrac{-3\:}{4 \:\:}\:\:\: }}}}}\:\:\bigstar$

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