Question

11. If roots of an equation 6x2 – 13x + m = 0 are reciprocal of each other, find the value of m. PLS EXPLAIN STEP BY STEP

Answer 1

$\large\underline{\sf{Answer- }}$

$\sf \: Let \: \alpha \: and \: \beta \: be \: the \: roots \: of \: {6x}^{2} - 13x + m = 0.$

On comparing with ax² + bx + c = 0, we get

• a = 6

• b = - 13

• c = m

According to given statement,

• Roots are Reciprocal of each other.

So,

$\bf\implies \: \beta \: = \: \dfrac{1}{ \alpha } - - - (1)$

We know,

$\: \: \: \: \: \: \: \: \boxed{{\tt Product\ of\ the\ roots=\frac{c}{a}}}$

Or

$\: \: \: \: \: \: \: \boxed{{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

Therefore,

$\rm :\longmapsto\: \alpha \times \dfrac{1}{ \alpha } \: = \: \dfrac{m}{6}$

$\rm :\longmapsto\:1 = \dfrac{m}{6}$

$\bf\implies \:m \: = \: 6$

Short cut trick :-

If the roots of the quadratic equation ax² + bx + c = 0 are Reciprocal of each other, then coefficient of x² = constant term.

• That is , a = c.

Additional Information :-

$\: \: \: \: \: \: \: \boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

OR

$\: \: \: \: \: \: \: \boxed{{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$