Question

(iii) 6x^2 - 3 - 7x. Find the zeroes of the following quadratic polynomials and vrify the relationship between the zeroes and the coefficients​

Answer 1

Answer:

The roots are $\frac{ 18 }{ 12 } \; \ and\; -\frac{ 4 }{ 12 }$.

Step-by-step explanation:

The formula to compute the roots of a quadratic equation ax² + bx + c = 0 is:

$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }$

Here,

a = 6

b = -7

c = -3

Compute the roots as follows:

$x = \frac{ -(-7) \pm \sqrt{(-7)^2 - 4(6)(-3)}}{ 2(6) }\\x = \frac{ 7 \pm \sqrt{49 - -72}}{ 12 }\\x = \frac{ 7 \pm \sqrt{121}}{ 12 }\\x = \frac{ 18 }{ 12 } \; \; \; x = -\frac{ 4 }{ 12 }$

The roots are $\frac{ 18 }{ 12 } \; \ and\; -\frac{ 4 }{ 12 }$.

In case of a quadratic equations with roots α and β:

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

Check:

$\alpha +\beta =-\frac{b}{a}\\\frac{18}{12}-\frac{4}{12}=-\frac{-7}{6}\\\frac{18-4}{12}=\frac{7}{6}\\\frac{7}{6}=\frac{7}{6}$

$\alpha \times\beta =\frac{c}{a}\\\frac{18}{12}\times-\frac{4}{12}=\frac{-3}{6}\\\frac{-3}{6}=\frac{-3}{6}$