Question

23.
If a and ß are the zeroes of p(x) = 2x2 + 7x + 5, find the value of a + b + aß.
24. (a) Find the point on the x-axis which is equidistant from (2,-5) and (-2,9).
If ya don’t know don’t answer

Answer 1

Answer:

See the explanation

Step-by-step explanation:

23)

2x²+7x+5=0

By middle term splitting,

2x²+2x+5x+5=0

2x(x+1)+5(x+1)=0

(2x+5)(x+1)=0

x= -5/2 , x= -1

So,now we get the two zeroes which are a & ß.

a= -5/2

ß= -1

So now a+ß+aß = (-5/2)+(-1)+(-5/2)(-1)

→ (-5-2+5)/2

→ -1

Answer = -1

24)

As the point is on x-axis, the y coordinate will be 0,

So let,

A(2,-5)

B(-2,9)

P(X,0) → This is the point on x-axis.

By using the distance formula , we have ,

$ap = \sqrt{(2 - x) ^{2} + ( - 5 - 0) ^{2} } \\ pb = \sqrt{(x - ( - 2)) ^{2} + (0 - 9) ^{2} }$

Also, AP=BP ( As P is equidistant from both points.)

So putting value of AP & PB, and then squaring both sides, we have,

(2-x)²+(-5-0)² = (x-(-2))²+(0-9)²

→ 4+x²-4x+25 = x²+4+4x+81

→ 4+25-4-81 = x²-x²+4x+4x

→ -56 = 8x

→ x = -7

So point is P(-7,0)

Answer 2

Answer:

-1

Step-by-step explanation:

2x²+7x+5=0

By middle term splitting,

2x²+2x+5x+5=0

2x(x+1)+5(x+1)=0

(2x+5)(x+1)=0

x= -5/2 , x= -1

So,now we get the two zeroes which are a & ß.

a= -5/2

ß= -1

So now a+ß+aß = (-5/2)+(-1)+(-5/2)(-1)

→ (-5-2+5)/2

→ -1

Answer = -1