Question

If 2/3 and -3 are the zeroes of the quadratic polynomial ax^2+7x+b then find the values of a and b.

Answer 1

Answer:

Hope this helps you ✌️✌️

Answer 2

Answer:

The value of a = 3 & b = -6.

Step-by-step explanation:

We know for any quadratic polynomial, when zeroes replace x the value of the expression is zero.

So for $a x^2 + 7x + b$

By substituting 2/3 we will have,

$a (\frac{2}{3})^2 + 7 (\frac{2}{3}) + b = 0$

$a (\frac{4}{9}) + \frac{14}{3} + b = 0$

$4a + 42 + 9b = 0$

$4a + 9b = -42$ ... (i)

And substituting -3 we will have,

$a (-3)^2 + 7 (-3) + b = 0$

$9a - 21 + b = 0$

$b = 21 - 9a$

Substituting the value of b in equation (i)

$4a + 9(21 - 9a) = -42$

-77a = -42 - 189

-77a = -231

a = 3 & b = 21 - 9(3) = -6.

Hence a = 3 & b = -6 using the property that zeroes on substitution gives 0 as the value.