Question

if one root of the quadratic polynomial 2 x square - 3 x + P is 3 find the other route also find the value of p​

Answer 1

Given Quadratic Polynomial is

$\sf 2 {x}^{2} - 3x + p$

As one Root is 2

So,

$\sf2(2 {)}^{2} - 3(2) + p = 0 \\ \\ \implies \sf8 - 6 + p = 0 \\ \\ \implies \sf \bold{\boxed{ \boxed{p = - 2}}}$

So,

Given Quadratic

Polynomial is:-)

$\sf2 {x}^{2} - 3x - 2$

Let Second Root be k.

As we know that

$\sf SUM \: OF \: ROOTS = \frac{3}{2} \\ \\ \sf 2 + k = \frac{3}{2} \\ \\ \sf \boxed{ \boxed{ k = \frac{ - 1}{2} }}$

Which is the second root.

Answer 2

Given : One Root of Quadratic Polynomial is 2x² - 3x + p is 2 .

Exigency To Find : The other root of Quadratic polynomial.

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⠀⠀⠀⠀⠀GIVEN QUADRATIC POLYNOMIAL : 2x² - 3x + p ,

⠀⠀⠀For , Finding other first we should find value of p :

$\qquad:\implies \sf 2x^2 - 3x + p = 0 \:\:$

⠀⠀ Given that ,

⠀⠀⠀⠀⠀⠀▪︎⠀⠀One Zero of the polynomial is 2 .

Therefore,

$\qquad:\implies \sf 2x^2 - 3x + p = 0 \:\:$

$\qquad:\implies \sf 2(2)^2 - 3(2) + p = 0 \:\:$

$\qquad:\implies \sf 2(4) - 3(2) + p = 0 \:\:$

$\qquad:\implies \sf 2(4) - 6 + p = 0 \:\:$

$\qquad:\implies \sf 8 - 6 + p = 0 \:\:$

$\qquad:\implies \sf p = 6 - 8 \:\:$

$\qquad:\implies \sf p = - 2 \:\:$

$\qquad \therefore \pmb{\underline{\purple{\:p\: =\: -2 }} }\:\:\bigstar$

⠀⠀⠀⠀⠀Therefore , By placing the value of p in Polynomial we get :

⠀⠀⠀⠀⠀⠀▪︎⠀⠀2x² - 3x - 2 = 0 [ as , Quadratic Polynomial ]

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❍ Let's Consider the other root of polynomial be a .

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ QUADRATIC POLYNOMIAL : 2x² - 3x - 2 = 0

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

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

$\qquad \dashrightarrow \sf 2 + a \:=\: \dfrac{ \:\:\: -(-3)\:\:}{\:\: 2 \:\;\:}$

$\qquad \dashrightarrow \sf 2 + a \:=\: \dfrac{ \:\:\: 3\:\:}{\:\: 2 \:\;\:}$

$\qquad \dashrightarrow \sf a \:=\: \dfrac{ \:\:\: 3\:\:}{\:\: 2 \:\;\:} - 2$

$\qquad \dashrightarrow \sf a \:=\: \dfrac{ \:\:\: 3 - 4 \:\:}{\:\: 2 \:\;\:}$

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

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

⠀⠀⠀⠀⠀⠀▪︎⠀⠀Here , a denotes Other root of Polynomial which is -1 / 2

$\therefore \:\underline {\sf Hence, \: The \:other \:root\:of\:Polynomial \:is \: \bf -1/2 .}$

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⠀⠀⠀⠀⠀$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\qquad \qquad \boxed {\begin{array}{cc} \bf{\underline {\bigstar\:\: For \: a \:Quadratic \:Polynomial \::}}\\\\ \sf{ Whose \:\:zeroes \:\:are\:\:\alpha \:\&\;\: \beta\:\:} \\\\ 1)\:\: \alpha + \beta \: =\:\dfrac{-b}{a} \quad \bigg\lgroup \bf Sum\:of\;Zeroes \bigg\rgroup \\\\ 2)\:\: \alpha \times \beta \: =\:\dfrac{c}{a} \quad \bigg\lgroup \bf Product \:of\;Zeroes \bigg\rgroup \\\\ \end{array}}$

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