Question

If the roots of the equation 6x2 - 13x + m = 0 are reciprocal of each other, find the
value of m.
​

Answer 1

Answer:

» Let one root of the equation be $\bf{y}$ .

So,

» Second root = $\dfrac{1}{\bf y}$ .

Now,

★ Standard Quadratic Equation is in the form :

$\mapsto\:\boxed{\sf ax^2 + bx + c}$

» Compare it with given equation

We get :

• a = 6
• b = -13
• c = m

Now,

We know

» Product of the zeroes / roots = $\dfrac{\sf c}{\sf a}$

So,

$\colon \longrightarrow \bf \: y \times \dfrac{1}{y} = \dfrac{m}{6}$

$\colon \longrightarrow \bf \: 1 = \dfrac{m}{6}$

$\colon \longrightarrow \bf \: 1 \times 6 = m$

$\purple{ \colon \longrightarrow} \bf \: \underline{ \boxed{\pink{ \bf{m = 6}}}} \quad \blue \bigstar$

$\green\therefore\;\underline{\sf The\:value\:of\:m\:is\:\red{\bf 6}.}$

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Answer 2

Answer:

$\sf» Let \: one \: root \: of \: the \: equation \: be  \: \bf{y}$

So,

$\sf» Second\: root = \dfrac{1}{\bf y}y1$

$\sf \red» Second \: root = \dfrac{1}{\bf y}y \: 1$

Now,

★ Standard Quadratic Equation is in the form :

$\mapsto\:\boxed{\sf ax^2 + bx + c}\begin{gathered} \\ \end{gathered}$

$\sf \frak \orange{» Compare \: it \: with \: given \: equation}$

We get :

a = 6

b = -13

c = m

Now,

We know,

$\sf \purple{» Product \: of \: the \: zeroes \: roots = \dfrac{\sf c}{\sf a}ac}$

So,

$\pink{\begin{gathered} \colon \longrightarrow \bf \: y \times \dfrac{1}{y} = \dfrac{m}{6} \\ \\ \end{gathered}}$

$\sf \begin{gathered} \colon \longrightarrow \bf \: 1 = \dfrac{m}{6} \\ \\ \end{gathered}$

$\green{\begin{gathered} \colon \longrightarrow \bf \: 1 \times 6 = m \\ \\ \end{gathered}}$

$\purple{ \colon \longrightarrow} \bf \: \underline{ \boxed{\pink{ \bf{m = 6}}}} \quad \blue \bigstar$

$\green\therefore\;\underline{\sf The\:value\:of\:m\:is\:\red{\bf 6}.}$

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