Find atleast one solution of x^2 -4x=10?

Question

please explain

Answer 1

x² -4x=10

subtract 10 from both sides

x² - 4x - 10 = 0

factor: no obvious solution, use

quadratic equation:

to solve ax + by + c = 0

x = [-b ±√(b²-4ac)] / 2a

x = [4 ±√(16+4*10)] / 2

x = [4 ±√56] / 2

x = [4 ±2√14] / 2

x = 2 + √14 or 2 - √14

edit, correction

x² -4x=12

subtract 12 from both sides

x² - 4x - 12 = 0

factor

(x - 6)(x + 2) = 0

x = 6, -2

.

Answer 2

x^2-4x=12

Many methods can be used to solve. Here are some:

1) Complete the square

Look at the closest square from x^2-4x.

This is x^2-4x+4=(x-2)^2

So, add 4 to get:

x^2-4x+4=12+4

(x-2)^2=16

Square root

x-2=+/-sqrt16

x-2=+/-4

Add 2

x=2+/-4

so,

x=2+4=6

x=2-4=-2

2) Factor

Set the equation equal to 0 by subtracting 12.

x^2-4x-12=0

Now, this is in the form of:

x^2+bx+c=0

this is factored by  finding two numbers, such that:

r+s=b

rs=c

So,

r+s=-4

rs=-12

r=-6, s=2 is valid.

Therefore rewrite:

x^2+2x-6x-12=0

x(x+2)-6(x+2)=0

(x-6)(x+2)=0

Also, if ab=0, either a or b must be 0.

therefore,

x-6=0, x=6

x+2=0, x=-2

3) Quadratic formula

For:

ax^2+bx+c=0

x=[-b+/-sqrt{b^2-4ac}]/2a

So,

-b=4

b^2-4ac=16+48=64--->sqrt of this is +/-8

and 2a=2

So, rewrite:

x=(4+/-8)/2

x=(4/2)+/-(8/2)

x=2+/-4

x=2+4=6

x=2-4=-2

4) Graph

All you need to do is graph:

x^2-4x=y

And

y=12

http://graph.seriesmathstudy.com/

This can be used to graph.

You will see that at x=-2 and x=6, y=12

Hope this helped. peace

Answer 3

x² - 4x - 12 = 0

(x - 6)(x + 2) = 0

x = 6 , x = - 2

Answer 4

x2-4x-12=0

(x-6)(x+2) =0

so, x can either be 6 or -2

Answer 5

rearrange

x^2-4x-12 = 0

factor out quadratic

(x-6)(x+2) = 0

x= 6 x= -2

hope this helps