Question

For the equation 3x^2+ px+3=0, p>0 if one root is square of the other then p is equal to
a)1/3 b) 1 c)3 d)2/3

Answer 1

Answer:

Option (c) 3

Step-by-step explanation:

Given the quadratic equation

$3x^2+px+3=0$

Let one root of the equation be 'a' then the other root will be a²

Sum of the roots = -p/3

⇒ a + a² = -p/3            ..................... (1)

Product of roots = 3/3 = 1

⇒ a × a² = 1

$\implies a ^3=1$

$\implies a ^3-1=0$

Thus the roots of the above will be 1, ω, ω² which are called cube roots of unity

If we take a = 1

then from eq (1)

1+1 = -p/3

or, p = -6

but given that p> 0

If we take a = ω

then from eq (1)

ω + ω² = -p/3

But we know that for cube roots of unity

1 + ω + ω² = 0

or, ω + ω² = -1

Thus we get

-1 = -p/3

or, p = 3

Again if we take

a = ω²

then from eq (1) again

ω² + ω⁴ = -p/3

But ω⁴ = ω × ω³

or, ω⁴ = ω            (∵ ω³ = 1)

Thus we again get

ω + ω² = -p/3

which will again give

p = 3

Therefore, the value of p is 3

Answer 2

Solution: As we know that

there is a relation between roots of Quadratic equation and coefficient of Quadratic equation.

if
$\alpha \: \: \: and \: \: \: \beta$
are the roots of Quadratic equation than

$\alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a}$
here

$a = 3 \\ \\ b = p \\ \\ c = 3$
if one root is
$\alpha$
then other will be
${ \alpha }^{2}$
So

$\alpha + { \alpha }^{2} = \frac{ - p}{3} \\ \\ \alpha \times { \alpha }^{2} = \frac{3}{3} \\ \\ { \alpha }^{3} = 1 ...eq2\\ \\ \alpha = \sqrt[3]{1} \\ \\ \alpha = 1 \\ \\ so \\ \\ 1 + {1}^{2} = \frac{ - p}{3} \\ \\ 2 \times 3 = - p \\ \\ p = - 6$

which can not be the case,since p> 0

so from equation eq2

$({ \alpha }^{3} -1 )=0$
find cube roots of unity,as we know that $1,\omega,{\omega}^2$

are cube roots of unity,and the properties are

$1+\omega+{\omega}^2=0\\\\or\\\\\omega+{\omega}^2=-1\\\\{\omega}^3= 1$
put this value in equation below

$\alpha + { \alpha }^{2} = \frac{ - p}{3} \\ \\ -1= \frac{ - p}{3} \\ \\p=3$

Hence option C is correct.