High School

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------------------------------------------------ Write a quadratic equation in standard form that has \(5\sqrt{3}\) and \(7\sqrt{3}\) as its roots.

A) \(9x^2 - 36x + 35 = 0\)
B) \(9x^2 + 36x - 35 = 0\)
C) \(9x^2 + 36x + 35 = 0\)
D) \(9x^2 - 36x - 35 = 0\)

Answer :

Answer:

A) 9x^2 − 36x + 35 =0

Step-by-step explanation:

9x^2 − 36x = −35

Using the zero-product property, write the solutions as factors of the quadratic equation and multiply:

(x − 5/3)(x − 7/3) = 0

9x2 − 36x + 35 = 0

9x2 − 36x = −35

Answer:

Option A)

[tex]9x^2 - 36x + 35 = 0[/tex]

Step-by-step explanation:

We are given the following in the question:

Roots of quadratic equation are:

[tex]\alpha = \dfrac{5}{3}, \beta = \dfrac{7}{3}[/tex]

The sum of the roots and the product of the roots can be calculated as:

[tex]\alpha + \beta = \dfrac{5+7}{3} = 4\\\\\alpha\beta = \dfrac{5}{3}\times \dfrac{7}{3} = \dfrac{5}{9}[/tex]

Standard form of quadratic equation:

[tex]x^2-(\alpha + \beta)x+\alpha\beta = 0[/tex]

Putting values, we get,

[tex]x^2 - 4x + \dfrac{35}{9} = 0\\\\9x^2 - 36x + 35 = 0[/tex]

is the required quadratic equation.

Thus, the correct answer is

Option A)

[tex]9x^2 - 36x + 35 = 0[/tex]