Question

If a and B are the roots of the equation 4x2 - 5x +2 = 0, Find the equation whose roots are α+1/α and β+1/β​

Answer 1

Answer:

16x² - 80x + 107 = 0

8x² - 9x + 8 = 0

4x² -5x + 4 = 0

4x² -15x + 23 = 0

Step-by-step explanation:

4x²-5x+2=0

α + β = 5/4

αβ = 2/4 = 1/2

Case 1 -

Roots  α + 3β   , 3α + β

(x -(α + 3β) ( x - (3α + β) = 0

=> x² - x (4α + 4β) + ( 3α² + 3β² + 10αβ) = 0

=> x² -4x(α+β) + 3(α+β)² + 4αβ = 0

=> x² -4x(5/4) + 3×(5/4)² + 4×(1/2) = 0

=> x² - 5x + 75/16 + 2 = 0

=> 16x² - 80x + 75 + 32 = 0

=> 16x² - 80x + 107 = 0

Case 2 -

Roots  α/β   ,  β/α

(x - α/β)(x -β/α) = 0

=> x² -x(α/β + β/α)  + 1 = 0

=> x² - x(α² + β²)/αβ + 1 = 0

=> x² - x( (α + β)²-2αβ)/αβ + 1 = 0

=> x² - x( (5/4)² -2×(1/2))/(1/2) + 1 = 0

=> x² - 2x(25/16 -1) + 1 = 0

=> x² -2x(9/16) + 1 = 0

=> x² - 9x/8 + 1 = 0

=> 8x² - 9x + 8 = 0

Case 3

roots are   α²/β   , β²/α

(x-α²/β)(x - β²/α) = 0

=> x² -x(α²/β + β²/α) + αβ = 0

=> x² -x(α³ + β³)/αβ + αβ = 0

=> x² - x((α+β)³ -3αβ(α+β))/αβ + αβ = 0

=> x² - x((5/4)³ -3×(1/2)×(5/4))/(1/2)  + 1/2 = 0

=> x² -2x ( 125/64 - 15/8) + 1/2 = 0

=> x² -2x (125 - 120)/8  + 1/2 = 0

=> x² - 5x/4 + 1/2 = 0

=> 4x² -5x + 4 = 0

Case 4

Roots  are  α + 1/α  & β + 1/β

(x - (α + 1/α)( x -(β + 1/β) = 0

=> x² - x(α + β + 1/α + 1/β) + αβ + 1/αβ  + α/β + β/α = 0

=>x² - x (5/4 + (β+α)/αβ) + 5/4 + 1/(1/2) + (β+α)/αβ = 0

=> x²- x(5/4 + (5/4)/(1/2) ) + 5/4 + 2 + (5/4)/(1/2) = 0

=> x² - x ( 5/4 + 5/2) + 13/4 + 5/2 = 0

=> x² - 15x/4 + 23/4 = 0

=> 4x² -15x + 23 = 0

Answer 2

Solution :

We have quadratic equation p(x) = 4x² - 5x + 2 = 0;

As we know that given polynomial compared with ax² + bx + c;

• a = 4
• b = -5
• c = 2

So;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}}\bigg\rgroup}\\\\\\\mapsto\sf{\alpha +\beta =\dfrac{-(-5)}{4} }\\\\\\\mapsto\bf{\alpha +\beta =\dfrac{5}{4} }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}}\bigg\rgroup}\\\\\\\mapsto\bf{\alpha \times \beta =\dfrac{2}{4} }$

∴ Other two zeroes are given so, we get new equation :

Let the α = α + 1/α  & β = β+1/β

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\longrightarrow\tt{\alpha +\beta }\\\\\longrightarrow\tt{\bigg(\alpha +\dfrac{1}{\alpha }\bigg) +\bigg(\beta +\dfrac{1}{\beta } \bigg)}\\\\\\\longrightarrow\tt{\alpha +\beta +\dfrac{1}{\alpha } +\dfrac{1}{\beta } }\\\\\\\longrightarrow\tt{\alpha +\beta +\dfrac{\alpha +\beta }{\alpha\beta } }\\\\\\\longrightarrow\tt{\dfrac{5}{4} +\dfrac{5/4}{2/4} }\\\\\\\longrightarrow\tt{\dfrac{5}{4} +\dfrac{5}{\cancel{4}} \times \dfrac{\cancel{4}}{2} }\\\\\\\longrightarrow\tt{\dfrac{5}{4} +\dfrac{5}{2} }$

$\longrightarrow\tt{\dfrac{5+10}{4} }\\\\\\\longrightarrow\bf{\dfrac{15}{4} }$

$\underline{\mathcal{PRODUCT\:OF\:ZEROES\::}}$

$\longrightarrow\tt{\alpha \beta }\\\\\longrightarrow\tt{\bigg(\alpha +\dfrac{1}{\alpha } \bigg)\times \bigg(\beta +\dfrac{1}{\beta } \bigg)}\\\\\\\longrightarrow\tt{\alpha \bigg(\beta +\dfrac{1}{\beta } \bigg)+\dfrac{1}{\alpha } \bigg(\beta +\dfrac{1}{\beta } \bigg)}\\\\\\\longrightarrow\tt{\alpha \beta +\dfrac{\alpha }{\beta } +\dfrac{\beta }{\alpha } +\dfrac{1}{\alpha \beta } }\\\\\\\longrightarrow\tt{\alpha \beta +\dfrac{\alpha^{2} + \beta^{2} }{\alpha\beta } +\dfrac{1}{\alpha \beta } }$

$\longrightarrow\tt{\alpha \beta +\dfrac{(\alpha+\beta )^{2} - 2\alpha\beta }{\alpha\beta } +\dfrac{1}{\alpha\beta } }\\\\\\\longrightarrow\tt{\dfrac{2}{4} +\dfrac{(5/4)^{2} - 2(2/4)}{2/4} + \dfrac{1}{2/4} }\\\\\\\longrightarrow\tt{\dfrac{2}{4} +\dfrac{25/16 - 4/4}{2/4} +\dfrac{1}{2/4} }\\\\\\\longrightarrow\tt{\dfrac{2}{4} +\dfrac{25/16 - 1}{2/4} + \dfrac{1}{2/4} }\\\\\\\longrightarrow\tt{\dfrac{2}{4} +\dfrac{\dfrac{25-16}{16} }{2/4} +\dfrac{1}{2/4} }$

$\longrightarrow\tt{\dfrac{2}{4} +\dfrac{9/16}{2/4} +\dfrac{1}{8} }\\\\\\\longrightarrow\tt{\dfrac{2}{4} +\dfrac{9}{\cancel{16}} \times \dfrac{\cancel{4}}{2} +\dfrac{1}{8} }\\\\\\\longrightarrow\tt{\dfrac{2}{4} +\dfrac{9}{8} +\dfrac{1}{8} }\\\\\\\longrightarrow\tt{\dfrac{4+9+1}{8} }\\\\\\\longrightarrow\bf{\dfrac{14}{8} }$

Thus;

$\boxed{\bf{The\:required\:quadratic\:equation \::}}}$

$\mapsto\sf{x^{2} -(sum\:of\:zeroes)x+(product\:of\:zeroes)=0}\\\\\mapsto\sf{x^{2} -\dfrac{15}{4} x + \cancel{\dfrac{14}{8 }}=0}\\\\\mapsto\sf{x^{2} - 15/4 x + 7/4 = 0}\\\\\mapsto\bf{ 4x^{2} - 15x + 7 = 0}$