Can someone helpp pls

Step-by-step explanation:

You have to solve this as a system. We have 3 coordinates given; the first 2 are given as zeros of the quadratic, the 3rd one is given as a regular coordinate. The zeros of a quadratic are where the y values of a coordinate are 0's because the x-intercepts of a quadratic (aka zeros) exist where y = 0. We use each of those points in the standard form of a quadratic to solve for a, b, and c. First, (0, 0):

which gives us that c = 0. We'll use that value of c as we move forward with the problem. Next we'll use (3, 0):

Hold that thought for a minute or 2. Next we'll use the coordinate (1, 6):

Now we have a system that we can solve using either substitution or elimination for a and b. I used elimination:

6  =  a  +  b

0  = 9a + 3b  and multiply the top equation by -3:

-18  =  -3a  -  3b

 0  =   9a  +  3b

so

-18 = 6a and

a = -3. Now back solve for b:

6 = -3 + b so

b = 9 and our quadratic is

or just

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)