MATH help quadratic equations!!!!!?

Question

please help..can u answer these?

find the roots of the quadratic equation by completing the square

X^2+4x-7=0

2nd question---- find the roots of the quadratic equation. put answer in simplest form

X^2-4x-6=0

thanK you very, very, very much

Answer 1

To complete the square, you want the constant value to be one-quarter of the square of the coefficient of x. The idea is that if you have (x+a)^2 you get x^2 + 2ax + a^2. The coefficient of x (2a), squared ((2a*2a) = 4a^2), and divided by four (4a^2/4 = a^2) is the constant term that makes the quadratic a perfect square.

In the first problem, the coefficient of x is 4, so you want the constant term on the left-hand side to be 4^2 * 1/4 = 16/4 = 4. Get that by adding 11 to both sides:

x^2 + 4x - 7 = 0
x^2 + 4x + 4 = 11

Now that the left-hand side is a perfect square, and you can take the square root to solve, remembering that you can have plus-or-minus then:

(x+2)^2 = 11
sqrt(x+2)^2 = +/- sqrt(11)
x+2 = +/- sqrt(11)
x = -2 +/- sqrt(11)

The roots are -2+sqrt(11) and -2-sqrt(11).

For the second one, you don't say you also have to complete the square, but let's do it the same way.

The coefficient of x is -4, so you want the constant term to be (-4)^2 *1/4 = 4. Get that by adding 10 to both sides.

x^2 - 4x - 6 = 0
x^2 - 4x + 4 = 10

Now that the left-hand side is a perfect square, and you can take the square root to solve, remembering that you can have plus-or-minus then:

(x - 2)^2 = 10
sqrt((x-2)^2) = +/- sqrt(10)
x-2 = +/- sqrt(10)
x = 2 +/- sqrt(10)

The roots are 2+sqrt(10) and 2-sqrt(10).

Answer 2

X^2+4x-7=0
( X^2+4x+4 ) -4-7=0
( x + 2 )^2 - 11 = 0
( x + 2 )^2 = 11
x+ 2 = +/- sqrt(11)

x+ 2 = sqrt(11) x = -2 +sqrt(11)
or x+2 = -sqrt(11) x = -2 -sqrt(11)

X^2-4x-6=0
( X^2-4x + 4 ) - 4 -6 = 0
( x - 2 )^2 - 1 0 = 0
( x - 2 )^2 = 10
x-2 = +/- sqrt ( 10 )
x - 2 = sqrt ( 10 ) x = 2 + sqrt ( 10 )
x - 2 = -sqrt( 10 ) x = 2 -sqrt( 10 )
any time

Answer 3

Question 1
(x² - 4x + 4) - 4 - 7 = 0
(x - 2)² = 11
x - 2 = ± √11
x = 2 ± √11

Question 2
x = [ 4 ± √(16 + 24) ] / 2
x = [4 ± √(40)] / 2
x = [4 ± 2√10 ] / 2
x = 2 ± √10

Answer 4

1) Complete the square:

(x+2)^2 -7 + B = x^2 +4x -7

x^2 +4x + 4 - 7 + B = x^2 +4x -7

4 -7 + B = -7

B = -4

so:

(x+2)^2 -7 - 4 = (x+2)^2 - 11

[(x+2)+sqrt(11)]*[(x+2)-sqrt(11)]

Solving for x:

x = -2 + or - sqrt(11)


2)Same approach:

(x -2)^2 - 6 - 4 = (x -2)^2 - 10

= [(x-2) + sqrt(10)]*[(x -2) - sqrt(10)]

x = 2 + or - sqrt10

Answer 5

i just took algebra 1 and from my experience i dont believe you can complete either...in the first one...you dont have any numbers that multiply to make 7 and add to make 4, and same case in the second...its been a few weeks tho i might be a little rusty...
but what do you mean by take the square roots of the equation...you would take them when you solve...you dont necessarliy need a separate step...
heres how you would do it if you could...
y= (x +/- some number) (x +/- some number)
{these numbers have to multiply to make 7 & add to make 4} not possible...
same for second...hope this helped somewhat...for quadratic equations that dont have solutions...i believe you just write no solution...dont write zero...
***Another option to try is simply factoring....****