Question

What is the solution set of x2 – 10 = 30x?

Answer 1

Steps:

So for this, I will be completing the square to solve for x. Firstly, add 10 and subtract 30x on both sides of the equation:

`x^2-30x=10`

Next, we want to make the left side of the equation a perfect square. To find the constant of this soon-to-be perfect square, divide the x coefficient by 2 and square the quotient. Once you get that result, add it to both sides of the equation:

`30/ 2 = 15\\15^2=225\\\\x^2-30x+225=235`

Now, factor the left side:

`(x-15)^2=235`

Next, square root both sides:

`x-15=\pm\ √(235)`

Now, add 15 to both sides of the equation:

`x=15\pm √(235)`

This is the exact solution. To find the approximate solution, solve the left side twice -- once with the plus sign, once with the minus sign:

`x=30.33,-0.33`

Answer:

In short:

• Exact Solution:
  `x=15\pm√(235)`
• Approximate Solution (Rounded to the hundredths):
  `x=30.33,-0.33`

Answer 2

`x^2-10=30x\qquad\text{subtract 30x from both sides}\\\\x^2-30x-10=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x_1=(-b-√(b^2-4ac))/(2a),\ x_2=(-b+√(b^2-4ac))/(2a)\\\\\text{We have}\ a=1,\ b=-30,\ c=-10\\\\\text{Substitute.}\\\\b^2-4ac\to(-30)^2-4(1)(-10)=900+40=940\\\\√(b^2-4ac)=√(940)=√(4\cdot235)=\sqrt4\cdot√(235)=2√(235)\\\\x_1=(-(-30)-2√(235))/(2(1))=(30-2√(235))/(2)=15-√(235)\\\\x_2=(-(-30)+2√(235))/(2(1))=(30+2√(235))/(2)=15+√(235)`

`Answer:\ \boxed{x=15-√(235)\ and\ x=15+√(235)}`