Question

need to know how to solve quadratic equations by completing the squares

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

ok

Answer 3

\[x^2+8x+7=0\]

Answer 4

This is the original equation. ax2 + bx + c = 0
Move the loose number to the other side. ax2 + bx = –c
Divide through by whatever is multiplied on the squared term.
Take half of the x-term, and square it.

Add the squared term to both sides.
x^2 + (b/a)x + (b^2/4a^2) = –(c/a) + (b^2/4a^2)
Simplify on the right-hand side; in this case, simplify by converting to a common denominator. x^2 + (b/a)x + (b^2/4a^2) = –(4ac/4a^2) + (b^2/4a^2)
Convert the left-hand side to square form (and do a bit more simplifying on the right). (x + b/2a)^2 = (b^2 – 4ac)/4a^2
Square-root both sides, remembering to put the "±" on the right. x + b/2a = ± sqrt(b^2 – 4ac)/2a
Solve for "x =", and simplify as necessary. x = [ –b ± sqrt(b^2 – 4ac) ] / 2a

Answer 5

an example may help

Answer 6

thats how the quadratic formula is derived completing the square

Answer 7

a quadratic equation is a polynomial equation of the second degree. The general form is

ax^2+bx+c=0,\,

where x represents a variable, and a, b, and c, constants, with a ≠ 0. (If a = 0, the equation becomes a linear equation.)
The quadratic formula can be derived by the method of completing the square,
x^2+2xh+h^2 = (x+h)^2.\,\!

Dividing the quadratic equation

ax^2+bx+c=0 \,\!

by a (which is allowed because a is non-zero), gives:

x^2 + \frac{b}{a} x + \frac{c}{a}=0,\,\!

or

x^2 + \frac{b}{a} x= -\frac{c}{a}

The quadratic equation is now in a form to which the method of completing the square can be applied. To "complete the square" is to add a constant to both sides of the equation such that the left hand side becomes a complete square:

x^2+\frac{b}{a}x+\left( \frac{1}{2}\frac{b}{a} \right)^2 =-\frac{c}{a}+\left( \frac{1}{2}\frac{b}{a} \right)^2,\!

which produces

\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}.\,\!

The right side can be written as a single fraction, with common denominator 4a2. This gives

\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.

Taking the square root of both sides yields

x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\ }}{2a}.

Isolating x, gives

x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac\ }}{2a}=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}.

Answer 8

hey dirty i found a mistake in your soduka skills

Answer 9

rly where

Answer 10

lets go back

Answer 11

\[x^2-6x-2=0\]
\[x^2-6x=2\]
\[(x-3)^2=2+3^2=2+9=11\]
\[x-3=\pm\sqrt{11}\]
\[x=3\pm\sqrt{11}\]

Answer 12

satelite how bow this one

Answer 13

\[x^2+8x+7=0\]

Answer 14

\[x^2+8x=-7\]
\[(x+4)^2=-7+16\]
\[(x+4)^2=-1\]

Answer 15

oops

Answer 16

\[(x+4)^2=-7+16=9\]

Answer 17

sorry.

Answer 18

means
\[x+4=3\] or \[x+4=-3\]
so
\[x=3-4=-1\]
\[x=-4-3=-7\]

Answer 19

of course you could have done this by factoring.
\[x^2+8x+7=0]
\[(x+1)(x+7)=0\]
\[x+1=0\]
\[x=-1\]
or
\[x+7=0\]
\[x=-7\]

Answer 20

thanks that is what i got just wanted to make sure i did it right

Answer 21

good!

Answer 22

\[x^2+14x+38=0\]

Answer 23

i got\[x=7+\sqrt{11} x=7-\sqrt{11}\]