Question

2. यदि बहुपद p(x) = ax² -3(a-1)x-1 का एक शून्यक 1 है, तो a का मान ज्ञात कीजिए।​

Answer 1

Answer:

see it's too easy the answer is

"a=1"

Step-by-step explanation:

now I will explain u how

$since \: one \: zeros \: of \: x \: is \: one (1)\: then \: ... \\ \\ p(x) = a {x}^{2} - 3(a - 1)x - 1 \\ = > a {x}^{2} - 3(a - 1)x - 1 =0 \\ = > a \times {1}^{2} - 3(a - 1) \times 1 - 1 = 0 \: \: \: \: \: \: \: (since \: x = 1 \: \: given) \\ = > a - 3a +3 - 1 = 0 \\ = > - 2a + 2 = 0 \\ = > - 2a = - 2 \\ = > 2a = 2 \\ = > a = \frac{2}{2} =1$

therefore a=1 hope this answer will help u

Answer 2

Answer:

a=1

Step-by-step explanation:

a. 1^2-3(a-1) 1-1=0

a-3a+3-1=0

-2a+2=0

-2a=-2

a=-2/-2

a=1