Question

Help me please!! what is the solution set on x^2-10=30x?

Answer 1

`What`
`is`
`the`
`solution`
`set`
`of`
`x^(2) -10=30x`
`?`

`Answer`
`is`
`C`
`{15 - √(235) ,15 + √(235)`

What is the solution set of
`x^2 - 10 = 30x?`?

A) {–220, 250}

B) {–250, 220}

✔ C)
`{15 - √(235) ,15 + √(235)`

D)
`{-15 - √(235), -15 + √(235)`

Explanation:

`Solution`
`found`

`x_1=15+√(235)`

`x_2=15-√(235)`

`Decimal form:`

`x_1=30.33`

`x_2=-0.33`

Step-by-step solution

Solving quadratic equations by completing the square

1. Move all terms to the left side of the equation

2. Find the coefficients

To find the coefficients, use the standard form of a quadratic equation:

`ax^2+bx+c=0`

`x^2-30x-10=0`

`a=1`

`b=-30`

`c=-10`

3. Move the constant to the right side of the equation and combine

Add -10 to both sides of the equation:

`x^2-30x-10=0`

`x^2-30x-10+10=0+10`

`x^2-30x=10`

4. Complete the square

To make the left side of the equation into a perfect square trinomial, add a new

constant equal to
`\left((b)/(2)\right)^2` to the equation:

`b=-30`

`((b)/(2))^2=((-30)/(2))^2`

Use the exponents fraction rule
`\left((x)/(y)\right)^2=(x^2)/(y^2)`

`((-30)/(2))^2=(-30^2)/(2^2)`

`(-30^2)/(2^2)=(900)/(4)`

`(900)/(4)=225`

Add 225 to both sides of the equation:

`x^2-30x=10`

`x^2-30x+225=10+225`

Simplify the arithmetic:

`x^2-30x+225=235`

Now we have perfect square trinomial, we can write it as a perfect square form by

adding half of the
`b` coefficient,
`(b)/(2)` :

`b=-30`

`(b)/(2)=(-30)/(2)`

2 additional steps

Find the greatest common factor of the numerator and denominator:

`(b)/(2)=(\left(-15\cdot 2\right))/(\left(1\cdot 2\right))`

Factor out and cancel the greatest common factor:

`(b)/(2)=-15`

`x^2-30x+225=235`

`(x-15)^2=235`

5. Solve for
`x`

Take the square root of both sides of the equation: IMPORTANT: When finding the

square root of a constant, we get two solutions: positive and negative

`(x-15)^2=235`

`√((x-15)^2)=√(235)`

Cancel out the square and square root on the left side of the equation:

`x-15=±√(235)`

Add
`15` to both sides

`x-15+15=15±√(235)`

Simplify the left side:

`x=15±√(235)`

`x_1=15+√(235)`

`x_2=15-√(235)`

`OS`
`The`
`Answer`
`is`
`C`
`{15 - √(235) ,15 + √(235)` .

Have a Nice Day .

Answer 2

x² - 30x - 10 =0

We are going to use formula

`x= \frac{-b+/- \sqrt{b^(2)-4ac} }{2a} \\ \\ a= 1, b=-30, c= - 10 \\ \\x= \frac{30+/- \sqrt{(-30)^(2)-4(1)(-10)} }{2} \\ \\ (30+/- √(900+40) )/(2)= (30+/- √(940) )/(2)= (30+/- √(235*4) )/(2) = 15+/- √(235) Answer\ is\ C.`