using quadratic equation:x^2-4x-5=0. solve by factoring, solve by using quadratic formula. show work please.?

Answer 1

1.solve by factoring:
x^2 -5x +x-5 =0
x(x-5)+1(x-5)=0
(x+1)(x-5)=0
=> either x-5=0 => x=5
or x+1 =0 => x=-1

2. by using qudrt formula:
for ax^2+bx+c=0
x = [-b+-sqrt[b^2 -4ac]]/2a
here a=1, b=-4, c=-5
therefore x= [4+-sqrt[16+20]]/2
x=[4+-6]/2
=>
x=[4+6]/2=5 and x=[4-6]/2=-1

Answer 2

Quadratic equation is : [-b ± √(b² - 4ac)] / 2a Anything with square roots that can't be factored I'll just be leaving as is. 1. 3x² - 0x - 12 = 0 [ 0 ± √(0² - 4(3)(-12))] / 2(3) [0 ± √(0 + 144)] / 6 ± √(144)] / 6 ±12/ 6 ±2 2. 4x² + 9x + 5 = 0 [-b ± √(b² - 4ac)] / 2a [-9 ± √(9² - 4(4)(5))] / 2(4) [-9 ± √(81 - 80)] / 8 [-9 ± √1]8 [-9 ± 1] / 8 Either -1 or -(5/4) 3. 5x² - 6x - 4 = 0 [6 ± √((-6)² - 4(5)(-4))] / 2(5) [6 ± √(36 + 80] / 10 [6 ± √(116)] / 10 [6 ± (√2 * √2 * √29)] / 10 [6 ± 2√29] / 10 [3 ± √29] / 5 4. -x² + x + 1 [-1 ± √(1² - 4(-1)(1))] / 2(-1) [-1 ± √(1 + 4)] / -2 [-1 ± √5] / -2 5. 2x² + 5x + 0 [-5 ± √(5² - 4(2)(0))] / 2(2) [-5 ± √(25 - 0)] / 4 [-5 ± √25] / 4 [-5 ± 5] / 4 Either 0 or -(5/2) Hopefully my calculations are all correct! 19 - Caucasian (American).

Answer 3

Factoring we have (x-5)(x+1) = 0 and either x-5 = 0 or x + 1 = 0 thus x = 5 or x = -1.

Using the quadratic formula x = 4 (plus or minus)radical (16-4 times 1 times –5)all divided by 2(1)

and x = 4 (plus or minus) (radical 36) all divided by 2 = 4 (plus or minus 6) divided by 2

thus x = 5 or –1.

Answer 4

The formula: ax^2 + bx + c. The equation is aalready in the standard form so all you need to do is factor.

(X-5) (X+1) = 0

X-5=0 X+1=0

X=5 X= -1

When you move the numbers or letters in the equation from one side of the equation to another the sign changes i.e if you had a negative number it becomes positive and vice versa. Hope this helps.

Answer 5

Factoring:
x^2 -4x -5=0
(x-5)(x+1)=0
x-5=0 so, x=5 and x+1=0 so x=-1

Quadratic Formula
x = [-b + or - sqrt(b^2-4ac)]/2a
In this case a=1, b=-4 and c = -5
So x= [4 + or - sqrt(-4^2 -4*1*-5)]
= [4 + or - sqrt(36)]/2
=2 + or - 3
So x= 2-3 = -1 or x = 2+3 = 5

Answer 6

x^2 - 4x - 5 =0

(x +1 ) (x- 5) = 0

if u multiply back u got x^2 - 5x +x - 5 so u got
x^2 - 4x - 5

the answer is x = -1 and x = 5

Answer 7

x^2-4x-5=0
by factoring-
(x+1)(x-5)=0
x+1=0 x-5=0
x=-1 x=5

...you can also just enter in your graphing calculator as is, and find the zeros

Answer 8

(x+1)(x-5)
x= -1 and 5

Answer 9

x2-4x-5=0, (x-5)(x+1)=0, x-5=0 or x+1=0, x=5 or x=-1

Answer 10

x^2-4x-5=0
x^2-5x+x-5=0
x(x-5)+1(x-5)=0
(x-5)(x+1)=0
x=5 or -1

by formula
x=[4+/-rt(16+20)]/2
=10/2 or -2/2
=5 or -1