Question

If the largest real root of the equation x⁴ - 4x³ + 5x² - 4x + 1 =0 can be expressed as, (a+√b)/c where b is prime.

Then find (a+b+c).​

Answer 1

Given equation,

$\longrightarrow \small {x}^{4} - 4 {x}^{3} + 5 {x}^{2} - 4x + 1 = 0$

Divide all the terms by $x^2$

$\longrightarrow \small {x}^{2} - 4 {x} + 5 - \dfrac{4}{x} + \dfrac{1}{ {x}^{2} } = 0$

$\longrightarrow \small {x}^{2} + \dfrac{1}{ {x}^{2} } - 4 {x} - \dfrac{4}{x} + 5 = 0$

$\longrightarrow \small \left( {x}^{2} + \dfrac{1}{ {x}^{2} } \right) - 4 \left({x} + \dfrac{1}{x} \right)+ 5 = 0$

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Now, use the identity :

• A² + B² = (A + B)² - 2AB

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${ \longrightarrow \small \left( {x} + \dfrac{1}{ {x} } \right)^{2} - \left(2x \times \dfrac{1}{x} \right) - 4 \left({x} + \dfrac{1}{x} \right)+ 5 = 0}$

${ \longrightarrow \small \left( {x} + \dfrac{1}{ {x} } \right)^{2} - \big(2 \big) - 4 \left({x} + \dfrac{1}{x} \right)+ 5 = 0}$

$\longrightarrow \small \left( {x} + \dfrac{1}{ {x} } \right)^{2}- 4 \left({x} + \dfrac{1}{x} \right)+ 3= 0$

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Assume that,

• $\small \left( {x} + \dfrac{1}{ {x} } \right) = t$

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Therefore,

$\longrightarrow \small \left( {x} + \dfrac{1}{ {x} } \right)^{2}- 4 \left({x} + \dfrac{1}{x} \right)+ 3= 0$

$\longrightarrow \small t^{2}- 4 t+ 3= 0$

$\longrightarrow \small t^{2}- 3 t - t+ 3= 0$

$\longrightarrow \small t(t- 3) -1( t - 3)= 0$

$\longrightarrow \small (t- 3) (t-1)= 0$

$\boxed { \longrightarrow \small t \: = 3 \: and \: 1}$

Now since according to our assumption, there are two cases possible, when t = 3 and when t = 1.

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C A S E - 1

$\implies\small t = \left( {x} + \dfrac{1}{ {x} } \right)$

$\implies\small 3= \left( {x} + \dfrac{1}{ {x} } \right)$

$\implies\small 3= \left( \dfrac{ {x}^{2} + 1}{ {x} } \right)$

$\implies\small 3x= {x}^{2} + 1$

$\implies\small 0= {x}^{2} - 3x + 1$

$\implies\small x= \dfrac{ -( - 3 )\pm \sqrt{ {( - 3)}^{2} - 4(1)(1) } }{2(1)}$

$\implies\small x= \dfrac{ 3\pm \sqrt{ 9 -4} }{2}$

$\boxed{ \implies\small x= \dfrac{ 3\pm \sqrt{5} }{2} }$

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C A S E - 2

$\implies\small t = \left( {x} + \dfrac{1}{ {x} } \right)$

$\implies\small 1= \left( {x} + \dfrac{1}{ {x} } \right)$

$\implies\small 1= \left( \dfrac{ {x}^{2} + 1 }{ {x} } \right)$

$\implies\small x= {x}^{2} + 1$

$\implies\small 0= {x}^{2} - x+ 1$

For this equation, both roots are imaginary ( or not real ), but we have to find largest real root of the equation.

Therefore, the largest real root of the equation is,

$\leadsto\small x= \dfrac{ 3 + \sqrt{5} }{2}$

This is already expressed in the form of $\left( \dfrac{a+\sqrt{b}}{c} \right)$ where b is a prime.

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Therefore,

• a = 3
• b = 5
• c = 2

Now the required answer will be,

$= (a + b + c)$

$= (3 + 2 + 5)$

$= (10)$

Therefore, the required answer is 10.

Answer 2

Step-by-step explanation:

Given :-

The largest real root of the equation x⁴ - 4x³ + 5x² - 4x + 1 =0 can be expressed as, (a+√b)/c where b is prime.

To find :-

Find the value of (a+b+c) ?

Solution :-

Given bi-quadratic equation is

x⁴ - 4x³ + 5x² - 4x + 1 =0

On dividing by x² both sides then

=> (x⁴ - 4x³ + 5x² - 4x + 1)/x² = 0/x²

=> (x⁴ - 4x³ + 5x² - 4x + 1)/x2 = 0

=> (x⁴/x²)-(4x³/x²)+(5x²/x²)-(4x/x²)+(1/x²) = 0

=> x²-4x +5-(4/x)+(1/x²) = 0

It can be arranged as

=> [x²+(1/x²)] -4[x+(1/x)] +5 = 0

=> [{x+(1/x)}²-2] -4[x+(1/x)] +5 = 0

=> [x+(1/x)]² -4[x+(1/x)]+5-2 = 0

=> [x+(1/x)]² -4[x+(1/x)]+3 = 0

Put x+(1/x) = a then

=> a²-4a+3 = 0

=> a²-a-3a+3 = 0

=> a(a-1)-3(a-1) = 0

=> (a-1)(a-3) = 0

=> a-1 = 0 or a-3 = 0

=> a = 1 or a = 3

Now

If a = 1 then we have

=> x+(1/x) = 1

=>(x²+1)/x = 1

=> x²+1 = x

=> x²-x + 1 = 0

=> x²-2(x)(1/2) = -1

=> x²-2(x)(1/2)+(1/2)² = -1+(1/2)²

=> [x-(1/2)]² = -1+(1/4)

=> [x-(1/2)]² = (-4+1)/4

=> [x-(1/2)]² = -3/4

=> x-(1/2) = ±√(-3/4)

=> x -(1/2) = ±√-3/ 2

=> x = ±√(-3)/2+(1/2)

=> x = (1±√-3)/2

and

If a = 3 then we have

> x+(1/x) = 3

=>(x²+1)/x = 3

=> x²+1 = 3x

=> x²-3x + 1 = 0

=> x²-2(x)(3/2) = -1

=> x²-2(x)(3/2)+(3/2)² = -1+(3/2)²

=> [x-(3/2)]² = -1+(9/4)

=> [x-(3/2)]² = (-4+9)/4

=> [x-(3/2)]² = 5/4

=> x-(3/2) = ±√(5/4)

=> x -(3/2) = ±(√5)/ 2

=> x = ±√(5)/2+(3/2)

=> x = (3±√5)/2

Therefore, x = (1+√-3)/2 ; (1-√-3)/2 ; (3+√5)/2 ; (3-√5)/2

The roots of the given equation are (1+√-3)/2 ; (1-√-3)/2 ; (3+√5)/2 ; (3-√5)/2

The largest real roots = (3+√5)/2

According to the given problem

The largest real root of the equation is in the form of (a +√b)/c, where b is a prime

=> (3+√5)/2 = (a+√b)/c

On comparing both sides then

a = 3

b = 5

c = 2

Now,

The value of a +b+c

=> 3+5+2

=> 10

Therefore, a+b+c = 10

Answer:-

The value of (a+b+c) for the given problem is 10

Used Methods :-

→ Splitting the middle term

→ Completing the square method

Used formulae:-

→ (a+b)² = a²+2ab+b²

→ (a-b)² = a²-2ab+b²