Question

In a Quadratic equation
$\sf 4x^2-13x+k=0$

One root of the equation is 12times more than the another root.Find the value of k

Note:-

Answer minimum 45+words​

Answer 1

Question :-

In a Quadratic equation 4x² - 13x + k = 0 , one root of this equation is 12 times more than the another root. Find the value of k.

Given :-

One root of quadratic equation is 12 times more than the another root.

Solution :-

Let , the one root of the quadratic equation is m and the another is n then according to the Question ,

• $\star \sf \: n = 12m - - - - (i)$

Given quadratic equation is 4x² - 13x + k = 0

• Now comparing the given equation by ax² + bc + c = 0,

$\begin{gathered} \star \sf \: a = 4 \\ \\ \star \sf \: b = - 13 \\ \\ \star \sf \: c = k\end{gathered}$

Now , we know that :-

$\sf \: sum \: of \: roots = - \dfrac{b}{a}$

So,

$\begin{gathered} \star \sf \: m + n = - \frac{( - 13)}{4} \\ \\ \sf \: substituting \: the \: value \: of \: n \: from \: equation \: (i) \\ \\ \mapsto \sf \: m + 12m = \frac{13}{4} \\ \\ \mapsto \sf \: 13m = \frac{13}{4} \\ \\ \mapsto \sf m = \frac{13}{4 \times 13} \\ \\ \mapsto \boxed{ \sf m = \frac{1}{4} }\end{gathered}$

• Hence the first root of the equation of 1/4 so , the second root is -

$\begin{gathered} \sf \star \: n = 12m \\ \\ \mapsto \sf \: n = 12 \times \frac{1}{4} \\ \\ \mapsto \boxed{\sf n = 3}\end{gathered}$

• So the second root of the quadratic equation is 3.

Now, we know that :-

$\sf \: product \: of \: roots = \dfrac{c}{a}$

So,

$\begin{gathered} \star \sf \: m \times n = \frac{k}{4} \\ \\ \mapsto \sf \frac{1}{4} \times 3 = \frac{k}{4} \\ \\ \mapsto \sf k = \frac{1}{4} \times 4 \times 3 \\ \\ \mapsto \boxed{\sf k = 3 }\end{gathered}$

Hence, the value of k is 3.