Find the zeroes v^2+4root3v-15 & also verify relation b/w zeroes & its coefficient.
If I read you right, we have
v^2 + 4√3 v - 15 = 0
(v-√3)(v+5√3) = 0
Not sure what you mean by "its coefficient".
To find the zeroes of a quadratic equation, v^2 + 4√3v - 15, we can use the quadratic formula:
For a quadratic equation of the form ax^2 + bx + c = 0, the quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
By substituting the values from the given equation into the quadratic formula, we get:
v = (-(4√3) ± √((4√3)^2 - 4(1)(-15))) / (2(1))
= (-4√3 ± √(48 + 60)) / 2
= (-4√3 ± √108) / 2
= (-4√3 ± √(36 * 3)) / 2
= (-4√3 ± 6√3) / 2
= (2√3)(-2 ± 3) / 2
= -√3(2 ± 3)
The two solutions for v are:
v = -√3(2 + 3) = -5√3
v = -√3(2 - 3) = √3
To verify the relationship between the zeroes and its coefficients, we can use Vieta's formulas, which state that for a quadratic equation of the form ax^2 + bx + c = 0, the sum of the roots (zeroes) is -b/a and their product is c/a.
In our equation, the sum of the roots is:
(-5√3 + √3) = -4√3
The product of the roots is:
(-5√3)(√3) = -15
Now, let's compare the sum and the product of the roots with the coefficients of the quadratic equation:
-4√3 = sum of the roots = -b/a
-15 = product of the roots = c/a
We see that the relationship between the sum of the roots and the coefficients is correct, as well as the relationship between the product of the roots and the coefficients.