Question

a and b are zeroes of polynomial 6x^2+x-2 find a÷b+b÷a​

Answer 1

ATQ, 'a' and 'b' are the zeroes of polynomial 6x² + x - 2

so let's find the zeroes of the polynomial by splitting method.

= 6x² + x - 2

= 6x² + (4x - 3x) - 2

= 6x² + 4x - 3x - 2

= 2x(3x + 2) - 1(3x + 2)

= (3x + 2) ( 2x - 1)

equating both factors by 0.

• 3x + 2 = 0

➡ 3x = -2

➡ x = -2/3

• 2x - 1 = 0

➡ 2x = 1

➡ x = 1/2

a = -2/3 and b = 1/2

hence, value of a/b + b/a

= (-2/3 × 2/1) + (1/2 × -3/2)

= -4/3 + (-3/4)

LCM of 3 and 4 = 3 × 4 = 12

= -16/12 - 9/12

= -25/12

Answer 2

Answer:

$\large{\underline{\boxed{\sf -\dfrac{25}{12}}}}$

Step-by-step explanation:

Given that;

a and b are the zeroes of the polynomial 6x² + x - 2. On comparing the given equation with ax² + bx + c, we get -

• a = 6
• b = 1
• c = - 2

Sum of zeroes = $\bf - \dfrac{b}{a}$

⇒ a + b = $\sf - \dfrac{1}{6}$

Product of zeroes = $\bf \dfrac{c}{a}$

⇒ ab = $\sf - \dfrac{2}{6}$

⇒ ab = $\sf - \dfrac{1}{3}$

Now,

$\sf \frac{a}{b} + \frac{b}{a} \\ \\ \implies \sf \frac{a {}^{2} + b {}^{2} }{ab}$

$\implies \sf \dfrac{(a + b) {}^{2} - 2ab }{ab}$

Substituting the value of above in the equation;

$\implies \sf \frac{ (- \frac{1}{6}) {}^{2} - 2 \times ( - \frac{1}{3}) }{ - \frac{ 1}{3} } \\ \\ \implies \sf \frac{ \frac{1}{36} + \frac{2}{3} }{ - \frac{1}{3} } \\ \\ \implies \sf \left( \frac{1}{36} + \frac{2}{3} \right) \div \left(- \frac{1}{3} \right) \\ \\ \implies \sf \left( \frac{1 + 24}{36} \right) \div \left(- \frac{1}{3} \right) \\ \\ \implies \sf \left( \frac{25}{36} \right) \div \left(- \frac{1}{3} \right) \\ \\ \implies \sf \frac{25}{36} \times ( - 3) \\ \\ \implies \sf - \frac{25}{12}$