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mreijepatmat
quadratic equation: ax² + bx + c =0

x' = [-b+√(b²-4ac)]/2a   and x" =  [-b-√(b²-4ac)]/2a  

6 = x² – 10x ; x² - 10x -6 =0
(a=1, b= - 10 and c = - 6

x' = [10+√(10²+4(1)(-6)]/2(1)  and x" = [10-√(10²+4(1)(-6)]/2(1)
x' =5+√31  and x' = 5-√31
nuhulawal20

The solutions to the quadratic equation in simplest radical form are; x = 5 + √(31)  Or x = 5 - √(31)

What is a Quadratic Equation?

Quadratic equation is simply an algebraic expression of the second degree in x. Quadratic equation in its standard form is;

ax² + bx + c = 0

Where x is the unknown

To solve for x, we use the quadratic formula

x = (-b±√(b² - 4ac)) / (2a)

Given the data in the question;

6 = x² - 10x

We rearrange

x² - 10x - 6 = 0

Hence;

  • a = 1
  • b = -10
  • c = -6

We substitute this values into the quadratic formula above.

x = (-(-10)±√((-10)² - (4 × 1 × -6))) / (2 × 1)

x = (10±√(100 + 24)) / (2)

x = (10±√(124)) / (2)

x = (10 ± 2√(31)) / (2)

Hence;

x = (10 + 2√(31)) / (2)  Or x = (10 - 2√(31)) / (2)

x = 2(5 + √(31)) / (2)  Or x = 2(5 - √(31)) / (2)

x = 5 + √(31)  Or x = 5 - √(31)

Therefore, the solutions to the quadratic equation in simplest radical form are; x = 5 + √(31)  Or x = 5 - √(31)

Learn more about quadratic equations here: https://brainly.com/question/1863222

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