Question

find the zeros of the polynomial 6x²-x-2 and verify the relationship between the zeros and the Coefficients.​

Answer 1

$\huge\purple{\boxed{\underline{ANSWER}}}$

GIVEN THAT:

$➲$ Polynomial = 6x²-x-2

TO FIND:

$➲$ The zeros of given polynomial and also to verify relationship between the zeros and the Coefficients.

FORMULA:

$➲$ If any quadratic polynomial is in standard form ax²+ bx + c.

where a and b are the Coefficients of x² and x respectively and c is the constant term

if α and β are the zeros of this polynomial then

$&#10230 \: \: \alpha + \beta = \frac{ - b}{a} \\ \\ &#10230 \: \: \alpha \beta = \frac{c}{a} \: \: \: \: \: \: \: \:$

SOLUTIONS:

$➲$ polynomial = 6x²-x-2

$➲$ After comparing its standard form ax²+ bx + c.

$➲$ we get a = 6, b= -1, c = -2

$&#10230 \: \: 6 {x}^{2} - x - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: 6 {x}^{2} - 4x + 3x - 2 \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: 2x(3x - 2) + 1(3x - 2) \\ \\ &#10230 \: \: (3x - 2)(2x + 1) \: \: \: \: \: \: \: \: \: \: \: \\ \\ if \: (3x - 2)(2x + 1) = 0 \\ \\ &#10230 \: \: 3x - 2 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: x = \frac{2}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: 2x + 1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: x = \frac{ - 1}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$➲$ now we get two zeroes of this polynomial

$&#10230 \: \: \alpha = \frac{2}{3} \: \: \: \: \: \\ \\ &#10230 \: \: \beta = \frac{ - 1}{2}$

VERIFICATION:

$➲$ Sum of the zeros

$&#10230 \: \: \alpha + \beta = \frac{ - b}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: \frac{2}{3} + (\frac{ - 1}{2}) = \frac{ - ( - 1)}{6} \\ \\ &#10230 \: \: \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \frac{4 - 3}{6} = \frac{1}{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: \frac{1}{6} = \frac{1}{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$➲$ Products of the zeros

$&#10230 \: \: \alpha \beta = \frac{c}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: \frac{\cancel2}{3} \times \frac{ - 1}{\cancel2} = \cancel\frac{ - 2}{6} \\ \\ &#10230 \: \: \frac{ - 1}{3} = \frac{ - 1}{3} \: \: \: \: \: \: \: \:$

$➲$ HENCE VERIFIED.