Question

find the zeros of the quadratic polynomials given below and verify the relationship between the zeros and the coefficients of the polynomial

Answer 1

Hii there!!
Here's the solution to your problem --------------------------------------------------------------

(i) 5x^2 - 4 - 8x

=> 5x^2 - 8x -4
=> 5x^2 - 8x - 4 = 0
=> 5x^2 - (10 - 2)x - 4 = 0
=> 5x^2 - 10x + 2x - 4 = 0
=> 5x( x - 2 ) + 2 ( x - 2) =0
=> (5x + 2) (x - 2)= 0

Zeroes = (5x + 2)
=> 5x = -2
=> X = -2/5

(x -2)
=> x = 2

Verification :-

α+β = -b / a
-2/5 + 2 = -8/ 5
=> 8/ 5 = 8/5

Again ,
α× β = c/a
=> -2/5× 2 = -4/5
=> -4/5 = -4/5

Hence, verified

Hope it helps ya
Thank you :-))

Answer 2

⠀☆ GIVEN POLYNOMIAL : 6x² - 3 - 7x ,

❍ Let the given POLYNOMIAL be denoted by f(x) .

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 6x^2 - 3 - 7x \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 6x^2 - 7x - 3 \:\:\qquad \bigg\lgroup \sf{ \:in\:Standard\:form\: }\bigg\rgroup \\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 6x^2 - 7x - 3 \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad \bigstar \:\:\: \sf By \: using\:Middle \:term \:splitting \:method \:\:: \\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 6x^2 - 7x - 3 \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 6x^2 - 9x + 2x - 3 \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad \bigstar \:\:\: \sf By \: Finding \:out \:common \:terms \:\:: \\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 6x^2 - 9x + 2x - 3 \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 3x(2x - 3) + 1 ( 2x - 3 ) \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad \bigstar \:\:\: \sf By \: Rewriting \:them \:in\:factored \:terms \:\:: \\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: 3x(2x - 3) + 1 ( 2x - 3 ) \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf f(x) \:=\: (3x + 1) ( 2x - 3 ) \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad \qquad \therefore \sf \:f(x) \:\:=\:0 \: \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf \: (3x + 1) ( 2x - 3 ) = 0 \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf \: 3x + 1 = 0 \:\:or\:\: 2x - 3 = 0 \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf \: x = \dfrac{\:\:-1\:\: }{\:\:3\:\:} \:\:or\:\: x \:\:=\:\: \dfrac{\:\:3\:\:}{\:\:2\:\:} \:\:\\\\\end{gathered}$

$\begin{gathered}\qquad \therefore \pmb{\underline{\purple{\:x \:\:\:=\: \dfrac{\:\:-1\:\: }{\:\:3\:\:} \:\:or\:\: \dfrac{\:\:3\:\:}{\:\:2\:\:} }} }\:\:\bigstar \\\end{gathered}$

$\begin{gathered}\therefore {\underline{ \sf \:Hence,\:The\: Zeroes \: \:of\:Polynomial \:are\:\bf \: \dfrac{\:\:-1\:\: }{\:\:3\:\:} \:\:and\:\: \dfrac{\:\:3\:\:}{\:\:2\:\:} \:\:}}\\\end{gathered}$

━━━━━━━━━━━━━━━━━━━⠀━⠀

$\begin{gathered}\qquad \maltese \:\:\underline {\sf Verifying \: \:the \: relationship \:between \: zeros \:and\: their\:Cofficients \:\::} \\\\\end{gathered}$

$\\sf \: \dfrac{\:\:3\:\: }{\:\:2\:\:} \:\:-\:\: \dfrac{\:\:1\:\:}{\:\:3\:\:}\:=\: \dfrac{-\:(-7\:)}{6 }\\\\\qquad \dashrightarrow \sf \: \dfrac{\:\:7\:\: }{\:\:6\:\:} \:=\: \dfrac{7\:}{6 }\\\\\qquad \therefore \pmb{\underline{\purple{\: \dfrac{\:\:7\:\: }{\:\:6\:\:} \:=\: \dfrac{7\:}{6 }}} }\:\:\bigstar \\\end{gathered}$

$\dfrac{\:\:3\:\:}{\:\:2\:\:}\:\:=\: \dfrac{\:-3\:}{6 }\\\\\qquad \dashrightarrow \sf \dfrac{\:\:-3\:\:}{\:\:6\:\:}\:\:=\: \dfrac{\:-3\:}{6 }\\\\\qquad \dashrightarrow \sf \cancel {\dfrac{\:\:-3\:\:}{\:\:6\:\:}\:}\:=\: \cancel {\dfrac{\:-3\:}{6 }}\\\\\qquad \dashrightarrow \sf \dfrac{\:\:-1\:\:}{\:\:2\:\:}\:\:=\: \dfrac{\:-1\:}{2 }\\\\\qquad \therefore \pmb{\underline{\purple{\: \dfrac{\:\:-1\:\:}{\:\:2\:\:}\:\:=\: \dfrac{\:-1\:}{2 } }} }\:\:\bigstar \\\\\end{gathered}$

        ━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀M O R E ⠀TO⠀⠀K N O W :

$\:\:are\:\:\alpha \:\&\;\: \beta\:\:} \\\\ 1)\:\: \alpha + \beta \: =\:\dfrac{-b}{a} \quad \bigg\lgroup \bf Sum\:of\;Zeroes \bigg\rgroup \\\\ 2)\:\: \alpha \times \beta \: =\:\dfrac{c}{a} \quad \bigg\lgroup \bf Product \:of\;Zeroes \bigg\rgroup \\\\ \bf{\underline{\bigstar\:\:Quadratic \:Polynomial \:Formation \:\::}}&\:\: \\\\ \sf p(x) \: = \:{ \{ x^2 - ( \alpha + \beta ) \: x \:\:+\: \alpha \beta } \} \\\\ \end{array}}\end{gathered}$