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 must be in 45+words ​

Answer 1

Question : -

In a Quadratic equation 4x² - 13x + k = 0 .

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

ANSWER

Given : -

Quadratic equation is 4x² - 13x + k = 0

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

Required to find : -

• Value of k ?

Solution : -

Let,

Second root be "z"

First root is 12 times the 2nd root = "12z"

Now,

Quadratic equation is 4x² - 13x + k = 0

Standard form of an quadratic equation is ax² + bx + c = 0

By comparing the given equation with standard equation !

• 4x² - 13x + k = 0
• ax² + bx + c = 0

Here,

• a = 4
• b = -13
• c = k

Now,

We have some relations between the roots of the equation & coefficient of the quadratic equation.

So, using this concept we can end up getting the required answer ...

sum of the roots = -b/a {or}

sum of the roots = (-coefficient of x)/(coefficient of x²)

Substituting the values , we have

12z + z = -(-13)/(4)

13z = (13)/(4)

z = (13)/(4) ÷ 13

z = (13)/(4) x (1)/(13)

z = (13)/(4 x 13)

z = (1)/(4)

• Value of z = (1)/(4)

Now, substituting this value in roots to find the accurate value of the roots ..

First root = 12z = 12 x (1)/(4) = 3

Second root = z = (1)/(4)

However,

Product of the roots = (c)/(a) {or}

Product of the roots = (constant term)/(coefficient of x²)

Substituting the value we have;

3 x (1)/(4) = (k)(4)

(3)/(4) = (k)/(4)

4 get's cancelled since it is common on both sides

(3)/(1) = (k)/(1)

This implies;

• k = 3

Therefore,

• Value of k is 3 ✓

Answer 2

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,

$\star \sf \: a = 4 \\ \\ \star \sf \: b = - 13 \\ \\ \star \sf \: c = k$

Now , we know that :-

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

So,

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

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

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

So the second root of the quadratic equation is 3.

Now, we know that :-

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

So,

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

Hence, the value of k is 3.

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Hope it will help you :)