If - 5 is root of the quadratic equation 2x^2+Px-15 = 0
and the quadratic equation p(x^2+x)+k=0 has equal
roots, find the value of
k .

9514 1404 393

  k = 7/4

Step-by-step explanation:

In quadratic ax^2+bx+c, the product of the roots is c/a, and their sum is -b/a. We know that one root of 2x^2+px-15 is -5, so the other one is ...

  (-15/2)/(-5) = 3/2

Then the value of b (p) is ...

  -p/2 = (3/2)+(-5) = -7/2

  p = 7

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Moving to the next part of the question, the roots will have equal values if the equation is a multiple of a perfect square. The constant in the squared binomial is half the x-coefficient.

  7(x^2 +x) +k = 7(x +1/2)^2 = 7x^2+7x +7/4

Comparing terms, we see that k = 7/4.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

step-by-step explanation: