Question

If one of the roots of the equation 4x2 – 24x + p = 0 is thrice the other root, then
value of p is​

Answer 1

Answer:

Step-by-step explanation:

x^2 +12x - k = 0.

one root is thrice the other

Let one root be =α

Other root=β

3β=α

As this is a quadratic equation thus the sum of roots will be

α+β= -b/a

α+ 3α= -12/1

4α= -12

α= -3

Thus one root is -3

Putting the value of -3 in the polynomial

x^2 +12x -K

(-3)^2 +12(-3) -K=0

9 -36 -K=0

-27 -K=0

K=27

Answer 2

Answer:

The value of p is 27.

Step-by-step explanation:

Given the quadratic equation:

$4x^2 - 24x + p = 0$

Given one of the roots of the equation is thrice the other root.

Let one of the root be, $\alpha =a$.

Then the root which is thrice of the root be, $\beta =3a$.

For a quadratic equation, $ax^2 + bx + c = 0$, if $\alpha$ and β are the roots of the equation, then the sum of the roots of the equation is given by -b/a, i.e.,

$\alpha+ \beta =-\dfrac{b}{a}$

and the product of the roots of the equation is given by c/a, i.e.,

$\alpha \beta =\dfrac{c}{a}$

In the given equation,

$4x^2 - 24x + p = 0$, we have a = 4, b = -24 and c = p

Thus, the sum of the roots is given by

$a+ 3a =-\dfrac{(-24)}{4}$

$\implies 4a =\dfrac{24}{4}\implies a =\dfrac{24}{4\times 4}=\dfrac{3}{2}$

And the products of the roots is given by

$a \times 3a =\dfrac{p}{4}$

$\implies 3a^{2} =\dfrac{p}{4}\implies a^{2} =\dfrac{p}{4\times 3}=\dfrac{p}{12}$

Using the value of $a = 3/2$

$\bigg(\dfrac{3}{2}\bigg)^{2} =\dfrac{p}{12}\implies \dfrac{9}{4} =\dfrac{p}{12}$

$\implies 4\times p =9\times 12$

$\implies p =\dfrac{9\times 12}{4} =9\times 3=27$

Therefore, the value of p is 27.

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