Question

If apha and beta are the zeroes of the quadratic polynomial f(x) = 4x2 + x - 2, find the value of alpha/beta+beta/alpha

Answer 1

EXPLANATION.

• GIVEN

a and b are the zeroes of quadratic equation

=> f(x) = 4x² + x - 2

Find the value of a/b + b/a.

According to the question,

equation = f(x) = 4x² + x - 2

sum of zeroes of quadratic equation

=> a + b = -b/a

=> a + b = -1/4

products of zeroes of quadratic equation

=> ab = c/a

=> ab = -2/4 = -1/2

value of = a/b + b/a

=> a² + b² / ab

apply formula of a² + b²

=> a² + b² = ( a + b) ² - 2ab

=> ( a + b)² - 2ab / ab

=> (-1/4)² - 2 (-1/2) / -1/2

=> 1/16 + 1 / -1/2

=> 17 / 16 / -1/2

=> -17 / 8

value of = a/b + b/a = -17/8

Answer 2

Solution :

We have quadratic polynomial f(x) = 4x² + x - 2,zero of the polynomial f(x) = 0

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

• a = 4
• b = 1
• c = -2

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

$\mapsto\tt{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x} {Coefficient\:of\:x^{2}}\bigg\rgroup}\\\\\\\mapsto\tt{\alpha +\beta =\dfrac{-1}{4} }$

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

$\mapsto\tt{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup \dfrac{Constant\:term} {Coefficient\:of\:x^{2}}\bigg\rgroup}\\\\\\\mapsto\tt{\alpha \times \beta =\cancel{\dfrac{-2}{4} }}\\\\\\\mapsto\tt{\alpha \times \beta =-1/2}$

Now;

$\longrightarrow\sf{\dfrac{\alpha }{\beta } + \dfrac{\beta}{\alpha } }\\\\\\\longrightarrow\sf{\dfrac{(\alpha)^{2} + (\beta)^{2} }{\alpha \beta } }\\\\\\\longrightarrow\sf{\dfrac{(\alpha+ \beta)^{2} -2\alpha\beta }{\alpha \beta }}\\\\\\\longrightarrow\sf{\frac{\bigg(-\dfrac{1}{4} \bigg)-2\times \bigg(-\dfrac{1}{2} \bigg)}{-\dfrac{1}{2} }} \\\\\\\longrightarrow\sf{\frac{\dfrac{1}{16}-\bigg(-\cancel{\dfrac{2}{2}} \bigg) }{-1/2} }\\\\\\\longrightarrow\sf{\dfrac{1/16 -(-1) }{-1/2} }$

$\longrightarrow\sf{\dfrac{1/16 + 1}{-1/2} }\\\\\\\longrightarrow\sf{\frac{\dfrac{1+16}{16} }{-1/2} }\\\\\\\longrightarrow\sf{\dfrac{17/16}{-1/2} }\\\\\\\longrightarrow\sf{\dfrac{17}{\cancel{16}} \times \dfrac{\cancel{2}}{-1} }\\\\\\\longrightarrow\bf{-17/8}$