How can i solve these equations using the quadratic formula?

Question

1.) x^2 -25=0

2.)x^2 + 8x=0

3.)4x^2 - 5x=0

Quadratic equations!
Thanks in advance..

Answer 1

x² - 25 = 0
is equivalent to
x² + 0x - 25 = 0
so, with the usual quadratic formula, set the 'b' value to 0

[-0 ± √0² - 4(1)(25)]/2
[± √100]/2
±10/2
x = ±5
--------------------
x² + 8x = 0
is equivalent to
x² + 8x + 0 = 0
so set the 'c' value to 0

[-8 ± √8² - 4(1)(0)]/2
[-8 ± √8²]/2
[-8 ± 8]/2
-16/2 or 0/2
x = -8 or 0
--------------------
4x² - 5x = 0
is equivalent to
4x² - 5x + 0 = 0
so, like the last question, set 'c' to 0

[5 ± √(-5)² - 4(4)(0)]/2(4)
[5 ± √25]/8
[5 ± 5]/8
10/8 or 0/8
x = 5/4 or 0

Answer 2

Question 1
x = [ ± √100 ] / 2
x = 5 , x = - 5

Question 2
x = [ - 8 ± √ (64) ] / 2
x = [- 8 ± 8 ] / 2
x = 0 , x = - 8

Question 3
x = [ 5 ± √25 ] / 8
x = [ 5 ± 5 ] / 8
x = 10 / 8 , x = 0
x = 5 / 4 , x = 0

Answer 3

1.) x^2 -25=0

a = 1, b = 0, c = -25
then you have the Delta, then you have the solutions.

2.)x^2 + 8x=0
a = 1, b = 8, c = 0

3.)4x^2 - 5x=0
a = 4, b = -5, c = 0.

Answer 4

1) a=1, b=0, c=-25

x(1)= [-0+root{0^2-4*1*(-25)}] / 2*a
x(1)=10/2=5

x(2)=[-0 -root{{0^2-4*1*(-25)}]/2*a
x(2)=-10/2=-5

2) a=1, b=8, c=0
x(1)=[-8+root{64-4*0*1}]/2*1
x(1)=0

x(2)=(-8-8)/2= -8

3) D=25-0=25
x(1)=(5+5)/8=5/4
x(2)=(5-5)/8=0

Answer 5

Using the quadratic formula:

1.) x = 0 plus or minus square root (0+4x25) all divided by 2
x=(0 plus or minus 10) divided by 2
x = plus or minus 5

2.) x = -8 plus or minus square root (-8 squared) all divided by 2
x = (-8 plus or minus 8) divided by 2
x = 0 or -8

3.) x = 5 plus or minus square root (5 squared) all divided by 2x4
x = (5 plus or minus 5) divided by 8
x = 10/8 or 0
x = 5/4 or 0

Answer 6

1.
x^2 -25=0
x^2=25
x=5 or x=-5

2.
x^2+8x=0
x(x+8)=0
x=0 or x+8=0
x= -8

4x^2 - 5x=0
x(4x-5)=0
x=0 or 4x-5=0
4x=5
x=5/4

Answer 7

1)here a=1,b=0,c=-25
formula for roots = -b+(or)- sq root of b^2-4ac/2a
= 0 +(or)-sq root of 0-(4*1*-25)/2*1
=sq root of 100/2
=10/2 or -10/2
=5 or -5
2)here a=1,b=0,c=8
formula for roots = -b+(or)- sq root of b^2-4ac/2a
=0-sq root of 0-32/2
=so the roots are sq root of -32 /2 or sq root 32 /2
3)1)here a=4,b=-5,c=0
formula for roots = -b+(or)- sq root of b^2-4ac/2a
=5+(or)- sq root 25-0
=5+(or)- 5
=5+5 or 5-5
=10 or 0

Answer 8

the roots of AX^2 + BX + C = 0 are given by:

(-B +/- SQRT(B^2 - 4AC))/2A

1.) (0 +/- SQRT(0 - 4*25 ))/2 = 5 and -5
2.) (-8 +/- SQRT(64 - 0))/2 = 0 and -8
3.) (5 +/- SQRT(25 - 0))/8 = 1.25 and 0

Answer 9

The quadratic formula is (-b+- the square root of (b^2-4ac))divided by 2a.
just substitute the values...

Answer 10

Quadratic equastion is:

ax^2+bx+c=0
D (discriminant)=b^2-4ac
x1= -(b+sqrt(D))/2a
x2= -(b-sqrt(D))/2a