# What is the value(s) of x? x^2 = 10x + 16

*print*Print*list*Cite

### 4 Answers

Well, the first thing is to move the terms from the right side, to the left side, changing their signs. Since both terms are positive to the right side, they will become negative, to the left side.

x^2 = 10x + 16

x^2 - 10x - 16 = 0

Now, we'll solve the quadratic equation, using the quadratic formula.

x1 = [-b+sqrt(b^2 -4ac )]/2a

x2 = [-b-sqrt(b^2 -4ac )]/2a

The coefficients a,b,c are the coefficients of the quadratic: ax^2 + bx + c = 0

We'll identify a,b,c:

a = 1

b = -10

c = -16

delta = b^2 - 4ac

delta = 100 + 64

delta = 164

sqrt delta = 2sqrt41

x1 = (10+2sqrt41)/2

x1 = 5+sqrt41

x2 = 5-sqrt41

**The values of x that verify the expression x^2 = 10x + 16 are {5-sqrt41 ; 5+sqrt41}.**

To find x .

x^2=10x+16

Solution:

x^2=10x+16 .

Subtract 10x from both sides:

x^2-10x = 16 .

Add 5^2 to both sides so that left side becomes A perfect square.

x-5*2x+5^2 = 5^2+16.

(x-5)^2 = 5^2+16 .

(x-5)^2 = 41.

We take square root:

x-5 = sqrt41 or x-5 = -sqrt41.

x = 5+sqrt41 or x = 5-sqrt41.

To find the values that x can take from the equation x^2 = 10x + 16

First bring all the terms to one side.

=> x^2 - 10x -16= 0

Now we know that the roots of a quadratic equation are given by:

[-b-srqt(b^2-4ac]/2a and [-b+srqt(b^2-4ac]/2a

Here a = 1, b=-10 and c=-16

=> [-b-srqt(b^2-4ac]/2a = [10- sqrt( 100+ 64)]/2a

= [ 10 - sqrt 164]/2 = -1.403

and [-b+srqt(b^2-4ac]/2a =[10+ sqrt( 100+ 64)]/2a

= [ 10 + sqrt 164]/2 = 11.403

**The values of x are -1.403 and 11.403**

Given:

x^2 = 10x + 16

To find the value of x wetake all the terms of the equation on left hand side andhen factorize the expresson o left hand side. Thus

x^2 - 10x - 16 = 0

==> x^2 - 10x + 25 - 41 = 0

==> (x - 5)^2 - 41 = 0

==> [x - 5 + 41^(1/2)][x - 5 - 41^(1/2)]= 0

Therefore x has two possible roots or values.

x = 5 - 41^(1/2) and 5 + 41^(1/2)