Question

Write a quadratic equation with 4/3 and 4 as its root. write the equation in the form ax^2+bx+c=0

Answer 1

If the roots are r1 and r2 then the factored form is 0=a(x-r1)(x-r2)
whwere a is a constant
I don't like fractions so a=3 for now

0=3(x-4/3)(x-4)
0=3(x^2-4/3x-4x+16/3)
0=3x^2-4x-12x+16
0=3x^2-16x+16

Answer 2

When we have two roots that are known, b and c, then we say that we can form a polynomial in the form: a(x - b)(x - c) = 0, where a is a real number.
This means that there are two roots, b and c, and the complexity of the leading coefficient is situated as a, an arbitrary real number.

Thus, we can say that a polynomial with roots 4/3 and 4 is actually:
$a_1(x - \frac{4}{3})(x - 4) = 0$
$a_1(3x - 4)(x - 4) = 0$
$a_1(3x^{2} - 12x - 4x + 16) = 0$
$a_1(3x^{2} - 16x + 16) = 0$

To find it in the form we want, let's distribute the a₁ into each term:
$3a_1x^{2} - 16a_1x + 16a_1 = 0\text{, where }3a_1 = a\text{, } -16a_1 = b\text{, }16a_1 = c$