Question

Find the zeroes of quadratic polynomials 6y2-3-7y and verify the relatinship between the zeros and the coefficient

Answer 1

Question should be:

• Find the zeroes of quadratic polynomials 6y²-3-7y and verify the relationship the zeros and the coefficient.

ANSWER:

• Zeros are 3/2 and -1/3

GIVEN:

• P(x) = 6y²-3-7y

TO FIND:

• Relationship between zeros and the coefficient.

SOLUTION:

=> 6y²-3-7y = 0

=> 6y²-7y-3 = 0

=> 6y²-9y+2y-3 = 0

=> (6y²-9y)+(2y-3) = 0

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

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

Either (2y-3) = 0

=> y = 3/2

Either (3y+1) = 0

=> y = -1/3

Sum of zeros:

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

Product of zeros:

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