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

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:

[tex]30\div 2 = 15\\15^2=225\\\\x^2-30x+225=235[/tex]

Now, factor the left side:

[tex](x-15)^2=235[/tex]

Next, square root both sides:

[tex]x-15=\pm\ \sqrt{235}[/tex]

Now, add 15 to both sides of the equation:

[tex]x=15\pm \sqrt{235}[/tex]

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:

[tex]x=30.33,-0.33[/tex]

Answer:

In short:

Exact Solution: [tex]x=15\pm\sqrt{235}[/tex]
Approximate Solution (Rounded to the hundredths): [tex]x=30.33,-0.33[/tex]

gmany gmany · Answer 2 · 2018-12-22T22:07:23+01:00

[tex]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=\dfrac{-b-\sqrt{b^2-4ac}}{2a},\ x_2=\dfrac{-b+\sqrt{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\\\\\sqrt{b^2-4ac}=\sqrt{940}=\sqrt{4\cdot235}=\sqrt4\cdot\sqrt{235}=2\sqrt{235}\\\\x_1=\dfrac{-(-30)-2\sqrt{235}}{2(1)}=\dfrac{30-2\sqrt{235}}{2}=15-\sqrt{235}\\\\x_2=\dfrac{-(-30)+2\sqrt{235}}{2(1)}=\dfrac{30+2\sqrt{235}}{2}=15+\sqrt{235}[/tex]

[tex]Answer:\ \boxed{x=15-\sqrt{235}\ and\ x=15+\sqrt{235}}[/tex]