Question

Find the zeroes and verify the relation 6x square-7x-3

Answer 1

ANSWER:

• Zeros are 3/2 and -1/3.

GIVEN:

• P(x) = 6x²-7x-3

TO FIND:

• Zeros and verify the relationship between the zeroes and coefficients.

SOLUTION:

=> 6x²-7x-3 = 0

=> 6x²-9x+2x-3 = 0

=> (6x²-9x)+(2x-3) = 0

=> 3x(2x-3)+1(2x-3) = 0

=> (2x-3)(3x+1) = 0

Either (2x-3) = 0

=> 2x = 3

=> x = 3/2

Either 3x+1 = 0

=> x = -1/3

Formula:

=> Sum of zeroes (α+β) = -(Coefficient of x)/Coefficient of x²

=> Product of zeroes (αβ) = Constant term/ Coefficient of x²

Sum of zeros:

$= \dfrac{3}{2} + \dfrac{ - 1}{3} \\ \\ = \frac{9 - 2}{6} \\ \\ \implies \frac{ - ( - 7)}{6} = \frac{ - coefficient \: of \: x }{coefficient \: of \: x {}^{2} }$

Product of zeros:

$= \dfrac{3}{2} \times \dfrac{ - 1}{3} \\ \\ \implies \frac{ - 3}{6} = \dfrac{constant \: term}{coefficient \: of \: x {}^{2} }$

Verified.