x^2 + 0.4x + 0.04 = 0
You can now solve it as a quadratic "without a number in front"
(x + 0.2)(x + 0.2) = 0
the only value that works is
x = -0.2
Go back to the original equation and see if it works if you use x = -0.2
25x^2 + 10x + 1 = 0
becomes
25(0.04) + 10(-0.2) + 1 = 0
1 - 2 + 1 = 0
0 = 0
it works!
Therefore x = -0.2 is a valid solution for the original equation (with 25 in front) as well as its "monic" equivalent.
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Como say: 1.
x = [- b ± √ (b² - 4ac) ] / 2a
x = [ 48 ± √ (2304 + 196) ] / 14
x = [ 48 ± √ (2500) ] / 14
x = [ 48 ± 50 ] / 14
x = 98/14 , - 2/14
x = 7 , x = - 1/7
2.
x = [- b ± √ (b² - 4ac) ] / 2a
x = [ 15 ± √ (225- 144) ] / 18
x = [ 15 ± √ 81 ] / 18
x = 4/3 , 1/3
3.
25x² + 10x + 1 = 0
[ 5x + 1 ] [ 5x + 1 ] = 0
x = - 1/5
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la console say: 7v² - 48v - 7 = 0
Polynomial like: ax² + bx + c, where:
a = 7
b = - 48
c = - 7
Δ = b² - 4ac (discriminant)
Δ = (- 48)² - 4.(7 * - 7) = 2304 + 196 = 2500 = 50²
v₁ = [- b - √Δ] / 2a → x₁ = [48 - 50] / (2 * 7) = - 2/14 = - 1/7
v₂ = [- b + √Δ] / 2a → x₂ = [48 + 50] / (2 * 7) = 98/14 = 7
9h² - 15h + 4 = 0
Polynomial like: ax² + bx + c, where:
a = 9
b = - 15
c = 4
Δ = b² - 4ac (discriminant)
Δ = (- 15)² - 4.(9 * 4) = 225 - 144 = 81 = 9²
h₁ = [- b - √Δ] / 2a → x₁ = [15 - 9] / (2 * 9) = 6/18 = 1/3
h₂ = [- b + √Δ] / 2a → x₂ = [15 + 9] / (2 * 9) = 24/18 = 4/3
25x² + 10x + 1 = 0
Polynomial like: ax² + bx + c, where:
a = 25
b = 10
c = 1
Δ = b² - 4ac (discriminant)
Δ = (10)² - 4.(25 * 1) = 100 - 100 = 0
x₁ = x₂ = x = - b / 2a → x = - 10/(2 * 25) = - 10/50 = - 1/5
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Robert say: The quadratic formula is derived from the