Question

Write a quadratic equation in standard form that has 5 3 and 7 3 as its roots. A) 9x2 − 36x + 35 = 0 B) 9x2 + 36x − 35 = 0 C) 9x2 + 36x + 35 = 0 D) 9x2 − 36x − 35 = 0

Answer 1

Answer:

Option A)

$9x^2 - 36x + 35 = 0$

Step-by-step explanation:

We are given the following in the question:

Roots of quadratic equation are:

$\alpha = \dfrac{5}{3}, \beta = \dfrac{7}{3}$

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

$\alpha + \beta = \dfrac{5+7}{3} = 4\\\\\alpha\beta = \dfrac{5}{3}\times \dfrac{7}{3} = \dfrac{5}{9}$

Standard form of quadratic equation:

$x^2-(\alpha + \beta)x+\alpha\beta = 0$

Putting values, we get,

$x^2 - 4x + \dfrac{35}{9} = 0\\\\9x^2 - 36x + 35 = 0$

is the required quadratic equation.

Thus, the correct answer is

Option A)

$9x^2 - 36x + 35 = 0$

Answer 2

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