Question

If product of two roots of the equation 4x4 - 242 - 31x2 - 6x-8 = 0 is 1
then sum of magnitude of each root is equal to​

Answer 1

Correct Question :

If product of two roots of the equation 4x⁴ - 24x³ + 31x² + 6x - 8 = 0 is 1 .

Then the sum of the magnitudes of each root is equal to -

Solution :

The expression is :

4x⁴ - 24x³ + 31x² + 6x - 8 = 0 .

Let us factorise this -

4x⁴ - 24x³ + 31x² + 6x - 8 = 0

> 4x⁴ - 22x³ - 2x³ + 20x² + 11x² + 16x - 6x - 8 = 0

> 4x⁴ - 22x³ + 20x² + 16x - 2x³ + 11x² - 10x - 8 = 0

> 2x ( 2x³ - 11x² + 10x + 8) - 1( 2x³ - 11x² + 10x + 8) = 0

> ( 2x - 1)( 2x³ - 11x² + 10x + 8)

> ( 2x - 1)( 2x³ - 7x² - 4x² + 14x - 4x + 8)

> ( 2x - 1)( 2x³ - 7x² - 4x - 4x² + 14x + 8)

> (2x - 1) [ x( 2x² - 7x - 4) - 2( 2x² - 7x - 4) ]

> ( 2x - 1) [ ( x - 2)( 2x² - 7x - 4 ) ]

> ( 2x - 1)( x - 2) [ 2x² - 7x - 4 ]

> ( 2x - 1)( x - 2) [ 2x² - 8x + x - 4 ]

> ( 2x - 1)( x - 2) [ 2x( x - 4) + 1( x - 4) ]

> ( 2x - 1)( x - 2)( 2x + 1)( x - 4)

The roots are -

> x = ½ , x = 2 , x = -1/2 and x = 4

The product of the first two roots , 2 and ½ is 1.

Sum of all roots is 6.

This is the required answer.

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Answer 2

${\large{\sf{\pmb{\underline{Correct \; question...}}}}}$

★ If the product of two roots of the equation 4x⁴ - 24x³ - 31x² - 6x - 8 = 0 is 1 then sum of magnitude of each root is equal to 6.

${\large{\sf{\pmb{\underline{Using \; concepts...}}}}}$

★ We have to factorise

${\large{\sf{\pmb{\underline{Full \; Solution...}}}}}$

Here according to the question if value of x be two then it will give a zero to us. That's why (x-2) is a root here.

»»» 4x⁴ - 24x³ - 31x² - 6x - 8 = 0.

»»» 4x⁴ - 24x³ - 31x² - 6x - 8 = (x-2)(4x³-16x²-x+4)

»»» (x-2)(2x-1)(2x²-7x-4)

»»» (x-2)(2x-1)(2x²-8x+x-4)

»»» (x-2)(2x-1)[2x(x-4)+1(x-4)]

»»» (x-2)(2x-1)(2x+1)(x-4)

»»» x = 1/2, x =2, x = -1/2 and x = 4

• Henceforth, sum of all magnitude of each root is equal to 6.

${\large{\sf{\pmb{\underline{Extra \; information...}}}}}$

Question 1) What is factorisation?

Answer - Let's see what is factorisation by taking an example.!

When we going to factorise an algebraic expression then we have to write it's factorised products. These factors may be number, algebraic variable or algebraic expression. For example,

→ x²-5x-6=0

→ x²-6x+x-6 = 0

→ x(x-6) + 1(x-6) = 0

→ (x+1)(x-6) = 0

→ x = -1 or x = 6

→ x = -1 or +6

There are many ways to factorise. But the most important way is middle term splitting method.

✴ Algebraic identities -

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{2} \: = \: = A^{2} \: + \: 2AB \: + B^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{2} \: = \: = A^{2} \: - \: 2AB \: + B^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto A^{2} \: - B^{2} \: = \: (A+B) \: (A-B)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{2} \: = (A-B)^{2} \: +4AB}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{2} \: = (A+B)^{2} \: -4AB}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{3} \: = A^{3} + \: 3AB \: (A+B) \:+ B^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{3} \: = A^{3} - \: 3AB \: (A-B) \: + B^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto A^{3} \: + B^{3} = \: (A+B) (A^{2} - AB + B^{2}}}}$

✴ Factorised identities -

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\; =\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)^{2} \: = \: a^{2} + 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a-b)^{2} \: = a^{2} - 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)(a-b) \: = \: a^{2} - b^{2}}\end{gathered}$