Question

If one of the zero of the polynomial (a2 + 1)x2 + 13x + 6a is reciprocal of other, find the value of a.

Answer 1

EXPLANATION.

One zeroes of the polynomial.

⇒ (a² + 1)x² + 13x + 6a is reciprocal to other.

As we know that,

Let one zeroes = α.

Other zeroes is reciprocal to other = 1/α

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ α x 1/α = (6a)/(a² + 1).

⇒ 1 = (6a)/(a² + 1).

⇒ a² + 1 = 6a.

⇒ a² - 6a + 1.

As we know that,

⇒ D = b² - 4ac.

⇒ D = (-6)² - 4(1)(1).

⇒ D = 36 - 4.

⇒ D = 32.

⇒ α = - b + √D/2a.

⇒ x = - (-6) + √32/2.

⇒ x = 6 + √32/2.

⇒ x = 6 + 4√2/2.

⇒ x = 2(3 + 2√2)/2.

⇒ x = 3 + 2√2.

⇒ β = - b - √D/2a.

⇒ y = -(-6) - √32/2.

⇒ y = 6 - √32/2.

⇒ y = 6 - 4√2/2.

⇒ y = 2(3 - 2√2)/2.

⇒ y = 3 - 2√2.

Values of a = 3 + 2√2 & 3 - 2√2.

                                                                                                                           

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

Given : One of the zero of the polynomial (a² + 1)x² + 13x + 6a is reciprocal of other .

Exigency To Find : The value of a .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Consider one of the zero of the polynomial be $\alpha$ .

⠀⠀⠀⠀⠀Given that ,

⠀⠀▪︎⠀ One of the zero of the polynomial is reciprocal of other .

Therefore,

⠀⠀▪︎⠀ Second zero of polynomial is $\dfrac{1}{\alpha}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : (a² + 1)x² + 13x + 6a

⠀⠀Here ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Cofficient of x² is ( a² + 1 )

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Cofficient of x is 13

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Constant term is 6a

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\\qquad\: \:\maltese\:\: \bf Product \:of \:zeroes\:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ Product_{( Zeroes \:of\:Polynomial \:)}\:: \:\dfrac{ Constant\:Term }{Cofficient\:of\:x^2\:} }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf Product_{( Zeroes \:of\:Polynomial \:)}\:: \:\dfrac{ Constant\:Term }{Cofficient\:of\:x^2\:}$

$\qquad:\implies \sf \alpha \times \dfrac{1}{\alpha} \:: \:\dfrac{ Constant\:Term }{Cofficient\:of\:x^2\:}$

$\qquad:\implies \sf \alpha \times \dfrac{1}{\alpha} \:: \:\dfrac{ 6a }{a^2 + 1\:}$

$\qquad:\implies \sf \cancel{\alpha} \times \dfrac{1}{\cancel {\alpha}} \:: \:\dfrac{ 6a }{a^2 + 1\:}$

$\qquad:\implies \sf 1 \:: \:\dfrac{ 6a }{a^2 + 1\:}$

$\qquad:\implies \sf 1 \:: \:\dfrac{ 6a }{a^2 + 1\:}$

$\qquad:\implies \sf a^2 - 6a + 1$

$\qquad:\implies \pmb{\underline{\purple{\: a^2 - 6a + 1 \:\:\: }} }\bigstar$

$\qquad:\implies \bf a^2 - 6a + 1 \: \qquad \longrightarrow \:\: New \: Formed \:Polynomial \:$

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Finding zeroes of new formed polynomial :

⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ POLYNOMIAL : a² - 6a + 1

⠀⠀⠀⠀⠀⠀▪︎ Now , By Comparing it with ax² + bx + c :⠀

⠀⠀⠀⠀⠀⠀⠀We get ,

⠀⠀⠀⠀▪︎ ⠀a = 1

⠀⠀⠀⠀▪︎⠀b = -6

⠀⠀⠀⠀▪︎ ⠀c = 1

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\\qquad\: \:\maltese\:\: \bf Zeros \:of \:Polynomial \:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ Zeros _{(Polynomial)} = \: \dfrac { -(b) \pm \sqrt { b^2 - 4ac }}{2a} }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { -(b)\pm \sqrt { b^2 - 4ac }}{2a}$

⠀

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { -(-6) + \sqrt { (-6)^2 - 4(1)(1) }}{2(1)}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 36 - 4(1)(1) }}{2(1)}$

⠀

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 36 - 4 }}{2(1)}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 32 }}{2(1)}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 32 }}{2}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + 4\sqrt { 2 }}{2}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 2 (3 + 2)\sqrt { 2 }}{2}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { \cancel {2} (3 + 2)\sqrt { 2 }}{\cancel {2}}$

$\qquad:\implies \sf Zeros _{(Polynomial)} = \: 3 \pm 2\sqrt { 2 }$

$\qquad:\implies \pmb{\underline{\purple{\: Zeros _{(Polynomial)} = \: 3 \pm 2\sqrt { 2 } \:\: }} }\bigstar$

Therefore,

⠀

$\qquad\qquad \leadsto \bf a \: = \: 3 + 2\sqrt { 2 } \:\: or\:\: 3- 2\sqrt {2}$

$\qquad\leadsto \pmb{\underline{\purple{\: \:a \: = \: 3 + 2\sqrt { 2 } \:\: or\:\: 3- 2\sqrt {2}\: }} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:Value\:of\:a \:is\:\bf{ 3 + 2\sqrt { 2 } \:\: or\:\: 3- 2\sqrt {2},}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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