Find the equation whose roots are \(-\frac{2}{3}\) and 3
Question 1 Report
Find the equation whose roots are \(-\frac{2}{3}\) and 3
Answer Details
The equation whose roots are -2/3 and 3 is of the form:
a(x + 2/3)(x - 3) = 0
where 'a' is a constant. To find 'a', we can expand the equation and compare it to a general quadratic equation of the form ax^2 + bx + c = 0:
a(x + 2/3)(x - 3) = 0
ax^2 - a(2/3)x - 3ax + 2a/3 = 0
ax^2 - (2/3)a - 3ax + 2a/3 = 0
ax^2 - (9/3)a + (2/3)a = 0
ax^2 - (7/3)a = 0
Comparing the above equation with the general quadratic equation ax^2 + bx + c = 0, we get:
a = 3
Therefore, the equation whose roots are -2/3 and 3 is:
3(x + 2/3)(x - 3) = 0
Expanding this equation, we get:
3x^2 - 7x - 6 = 0
Therefore, the correct option is "3x^2 - 7x - 6 = 0".
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