q(x)=√3x^2 + 10x + 7√3

Question

Find the zeros of quadratic polynomials and verify the relationship between the zeros and their coefficients:

q(x)=√3x² + 10x + 7√3

Answer 1

Given, 

q(x) = √3x² + 10x + 7√3 

We put q(x) = 0 

⇒ √3x² + 10x + 7√3 = 0 

⇒  √3x² + 3x + 7x + 7√3x = 0 

⇒ √3x(x + √3) + 7 (x + √3) = 0 

⇒ (x + √3)(√3x + 7) = 0 

This gives us 2 zeros, for 

x = -√3 and x = -7/√3 

Hence, the zeros of the quadratic equation are -√3 and -7/√3. 

Now, for verification 

Sum of zeros = – coefficient of x / coefficient of x² 

-√3 + (-7/√3) = – (10) /√3 

(-3-7)/ √3 = -10/√3 

-10/ √3 = -10/√3 

Product of roots = constant / coefficient of x² 

(-√3) x (-7/√3) = (7√3)/√3 

7 = 7

Therefore, the relationship between zeros and their coefficients is verified.

Answer 2

\(q(x) = \sqrt 3x^2 + 10x + 7\sqrt 3 \)

\(= \sqrt 3x^2 + 7x + 3x + 7\sqrt 3\)

\(= \sqrt 3x(x + \sqrt 3) + 7(x + \sqrt 3)\)

\(= (\sqrt 3x + 7) (x + \sqrt 3)\)

Zeroes of the polynomials are \(-\sqrt 3, \frac{-7}{\sqrt 3}\).

Sum of zeroes = \(\frac{-10}{\sqrt 3}\)

⇒ \(-\sqrt 3 - \frac7{\sqrt 3} = \frac{-10}{\sqrt 3}\)

⇒ \( \frac{-10}{\sqrt 3}= \frac{-10}{\sqrt 3}\)

Product of zeroes = \(\frac{7\sqrt 3}3\)

⇒ \(\frac{\sqrt 3x - 7}{\sqrt {30}} = 7\)

⇒ \(7 = 7\)

Hence, relationship verified.