Write a quadratic equation whose roots are 2 times less than the roots of the equation 4x ^ {2} -12x + 3 = 0.

1. Find the roots of the equation:
4x² – 12x + 3 = 0.
Discriminant:
D = b² – 4ac = (-12) ² – 4 * 4 * 3 = 144 – 48 = 96.
x = (- b +/- √D) / 2а.
x1 = (- (-12) + √96) / 2 * 4 = (12 + 4√6) / 8 = (3 + √6) / 2.
x2 = (- (-12) – √96) / 2 * 4 = (12 – 4√6) / 8 = (3 – √6) / 2.
2. Since, by condition, the roots of the new equation are 2 times less, then:
x′1 = x1 / 2 = ((3 + √6) / 2) / 2 = (3 + √6) / 4.
x′2 = x2 / 2 = ((3 – √6) / 2) / 2 = (3 – √6) / 4.
3. By Vieta’s theorem:
x′1 + x′2 = – b (coefficient at x in the new equation);
x′1 * x′2 = c (free term in the new equation).
Let’s find the coefficient – b:
– b = (3 + √6) / 4 + (3 – √6) / 4 = (3 + √6 + 3 – √6) / 4 = 6/4 = 3/2.
Find a free term with:
c = ((3 + √6) / 4) * ((3 – √6) / 4) = ((3 + √6) (3 – √6)) / 16 = (9 – 6) / 16 = 3 /sixteen.
The new equation will look like:
x² – 3x / 2 + 3/16 = 0.
Let’s get rid of fractions by multiplying the whole equation by 16:
16 (x² – 3x / 2 + 3/16) = 16 * 0;
16x² – 16 * 3x / 2 + 16 * 3/16 = 0;
16x² – 8 * 3x + 3 = 0;
16x² – 24x + 3 = 0.
Answer: 16x² – 24x + 3 = 0.