Question

If a quadratic equation $x^2 + 3x + 7 = 0$ & $ax^2 + bx + c = 0$ have one root in common, then find $\dfrac{a^3+b^3+c^3}{3abc}$.

Here, $\Big\{a,b \: {\sf{and }}\: c\in {\mathbb{R}}\Big\}$​

Answer 1

Step-by-step explanation:

Given :-

x²+3x+7 = 0 and ax²+bx+c = 0 have one root in common .

To find :-

Find the value of (a³+b³+c³)/3abc ?

Solution :-

Given quadratic equations are

x²+3x+7 = 0 and ax²+bx+c = 0

First we have to find the discriminant for the existing the roots of the equation x²+3x+7 = 0

a = 1

b = 3

c = 7

We know that

The discriminant (D) = b²-4ac

=> (-3)²-4(1)(7)

=> 9-28

=> -19 < 0

Since , the discriminant is less than zero, then the roots are imaginary roots .

They are in the form of conjugate .

But given that a,b,c are Real numbers

Given that both equations have one root in common .

We know that

If two equations have one root in common then the ratio of coefficients of both equations are equal.

=> a/1 = b/3 = c/7

Let a/1 = b/3 = c/7 = k

On taking a/1 = k

=> a = k --------------(1)

On taking b/3 = k

=> b = 3k ------------(2)

On taking c/7 = k

=> c = 7k ------------(3)

Now,

The value of (a³+b³+c³)/3abc

On substituting the values of a , b and c then

=> [(k)³+(3k)³+(7k)³]/[(3)(k)(3k)(7k)]

=> (k³+27k³+343k³)/(63k³)

=> k³(1+27+343)/(63k³)

=> 371k³/63k³

=> 371/63

The value of (a³+b³+c³)/3abc is 371/63

Answer:-

The value of (a³+b³+c³)/3abc for the given problem is 371/63

Used formulae:-

→ The discriminant of ax²+bx+c = 0 is

D = b²-4ac

→ If two equations have one root in common then the ratio of coefficients of both equations are equal.

→ If D < 0 then the equation has imaginary roots .