Question

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Answer 1

Answer:

1. 3 zeroes are given in the graph.

p( x ) = ax^ 2 - 3( a - 1)x - 1

x = 1

put value of x in p( 1 )

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

a - 3a + 3 - 1= 0

- 2a +2 = 0

- 2a = - 2

a = 1.

Answer 2

Answer:

Step-by-step explanation:

1

Number of zeroes of a polynomial refers to the number of points its

graph intersects the x axis

Here it intersects the x axis at 3 points

Yherefore the number of zeroes of the graph are 3.

2

given

$p(x)=ax^2-3(a-1)x-1\\\\given\\\\1\;is\;a\;zero\;of\;p(x)\\\\=>\;p(1)=0\\\\a-3(a-1)-1=0\\\\-2a+2=0\\\\a=1$

3

given

$\alpha+\beta =6\\\\\alpha \beta =6\\\\WKT\;polynomial\;with\;zeroes\;\alpha\;and\;\beta\;is\\\\x^2-(\alpha+ \beta)+\alpha \beta  =0\\\\x^2-6x+6=0\\\\is\;the\;required\;polynomial.$

4

given

$\alpha+\beta =-\frac{9}{2}\\\\\alpha \beta =-\frac{3}{2}\\\\WKT\;polynomial\;with\;zeroes\;\alpha\;and\;\beta\;is\\\\x^2-(\alpha+ \beta)+\alpha \beta  =0\\\\2x^2+9x-3=0\\\\is\;the\;required\;polynomial.$

5

given

$\sqrt{3}x^2-8x+4\sqrt{3}=0\\\\\sqrt{3}x^2-6x-2x+4\sqrt{3}=0\\\\\sqrt{3}x(x-2\sqrt{3})-2(x-2\sqrt{3})=0\\\\(\sqrt{3}x-2)(x-2\sqrt{3})=0\\\\x=\frac{2}{\sqrt{3}}\\\\x=2\sqrt{3}$

HOPE IT HELPS