Iwill name you the brainliest! one of the roots of the equation 5x2−12x c=0 is three times as big as the other root. find the value of c.

c=5.4

The two roots are 3/5 and 9/5

Step-by-step explanation:

assume 5x2−12x c=0 is supposed to be 5x^2 - 12x + c = 0

p = (12 + sqrt(144-20c))/10

q = (12 - sqrt(144-20c))/10

p-3q=0,

1.2 + 0.1sqrt(144-20c) +

-3.6 + 0.3sqrt(144-20c) = 0

-2.4 + 0.4sqrt(144-20c) +2.4 = 2.4

sqrt(144-20c) = 2.4/0.4 = 6

144-20c=36

144-36=20c

c = 108/20 = 5.4

5x^2-12x+5.4=0

x = 3/5 or x = 9/5

c = 27/5

Step-by-step explanation:

Given data:

Standard form of the quadratic equation

a x² + b x + c = 5 x² - 12 x + c = 0

and relation between the roots x₂ = 3 x₁

From which we conclude the following:

a = 5 , b = -12 and c = ?

To solve this problem we will use the Vieta's formulas, they read:

x₁ + x₂ = - b/a  and x₁ · x₂ = c/a

First we will replace x₂ = 3 x₁ in the first formula and get:

x₁ + 3 x₁ = - (-12)/5  =>  4 x₁ = 12/5 => x₁ = (12/5) : 4 = 3/5

x₁ = 3/5

Now we will replace x₁ = 3/5 in the relation x₂ = 3 x₁ and get:

x₂ = 3 · 3/5 = 9/5

x₂ = 9/5

Now we will replace value for x₁ = 3/5 and x₂ = 9/5 in the second formula

x₁ · x₂ = c/a  and get:

(3/5) (9/5) = c/5 => 27/25 = c/5 => c = (27/25) : 5 = 27/5

c = 27/5

Now the quadratic equation reads:

5 x² - 12 x + 27/5 = 0  / · 5

We will multiply whole equation with number 5 and get:

25 x² - 60 x + 27 = 0

Prove:

Every quadratic equation can be factorized as follows:

a (x - x₁) (x - x₂) = a x² + b x + c

we calculated x₁ = 3/5 and x₂ = 9/5

5 (x - (3/5)) (x - (9/5)) = (5 x - 3) (x - (9/5)) = 5 x² - 9 x - 3 x + 27/5 =

= 5 x² - 12 x + 27/5

God with you!!!

i think c. but don't trust it.

step-by-step explanation:

so, converting firstly the mixed fractions to improper fractions and then "adding".