Question #cd174

2 Answers
Dec 14, 2017

x=-4+sqrt41

x=-4-sqrt41

Explanation:

.

x^2+8x-25=0

To solve for x, you use the quadratic formula which is:

for a quadratic equation of general form:

ax^2+bx+c=0

x=(-b+-sqrt(b^2-4ac))/(2a)

Here, a=1, b=8, and c=-25

Therefore,

x=(-8+-sqrt(8^2-4(1)(-25)))/(2(1))=(-8+-sqrt(64+100))/2

x=(-8+-sqrt164)/2=(-8+-sqrt(4(41)))/2=(-8+-2sqrt41)/2

x=(cancelcolor(red)2(-4+-sqrt41))/cancelcolor(red)2=-4+-sqrt41

Answers to other problems:

A)., y=3(x-1)^2

From: (a-b)^2=a^2-2ab+b^2, our a=x, and b=1), we have:

y=3(x^2-2x+1)

y=3x^2-6x+3

B)., y=2(x-3)(x+4)

y=2(x^2+4x-3x-12)=2(x^2+x-12)

y=2x^2+2x-24

A)., y=6(x-7)^2-2

This equation is in the vertex form:

y=a(x-h)^2+k where the coordinates of the vertex are (h,k)

Out h=7, and k=-2, therefore, vertex is (7,-2)

B). y=-1/2(x+10)²-12

Vertex is (-10,-12)

Dec 14, 2017

x = +-sqrt41 -4

Explanation:

The trinomial does not factorise.

In ax^2 +bx +c, if a =1 and b is even, completing the square is a quick method.

x^2 +8x-25=0" "larr move the constant to the right

x^2 +8x " "=25" "larr add on (b/2)^2 to both sides

x^2 +8x +16 =25+16" "larr the left side is a perfect square

(x+4)^2 = 41

x +4 = +-sqrt41" "larr consider positive and negative roots

x = +-sqrt41 -4