The roots of the quadratic equation x^2-mx+n=0 ,where m and n are integers ,are a and 1/4a.
(i)Find the sum and product of its roots
(ii)show that 4m^2=25n
How to do part 2 ?
You must be studying the property of the roots of a quadratic, where
sum of roots = -b/a
product of roots = c/a
so for x^2 - mx + n
the sum of the roots = m
product of the roots = n
but you are given the roots as a and 1/4a
sum of roots = a + 1/4a = (5/4)a
product of roots = (1/4)a^2
so (5/4)a = m --- >a = 4m/5 or a2 = 16m^2/25
(1/4)a^2 = n --- a^2 = 4n
thus: 16m^2/25 = 4n
16m^2 = 100n
4m^2 = 25n
Voila!!!!
ohh I get it !! thanks a lot :))
To solve part 2, we need to use the relationship between the roots and coefficients of a quadratic equation.
Given that the roots of the equation are "a" and "1/4a," we can write the equation as:
(x - a)(x - 1/4a) = 0
Expanding this equation gives:
x^2 - (a + 1/4a)x + (a * 1/4a) = 0
Simplifying further:
x^2 - [(4a + a)/4a]x + (a^2/4) = 0
Now, we can compare this equation with the given quadratic equation:
x^2 - mx + n = 0
From the comparison, we can equate the coefficients of the variables:
- (a + 1/4a) = -m (equating coefficients of x)
a + 1/4a = m (dividing by -1)
a * 1/4a = n (equating constant terms)
1/4 = n (dividing by a^2)
Now, to show that 4m^2 = 25n, we substitute the values derived from the equations above:
4m^2 = 4(a + 1/4a)^2
= 4(a^2 + 2 * a * 1/4a + (1/4a)^2)
= 4(a^2 + 1/2 + 1/16a^2)
= 4a^2 + 2 + 1/4a^2
25n = 25 * (1/4)
= 25/4
To equate both sides, we need to have the same expression. Here, we can see that:
4m^2 = 4a^2 + 2 + 1/4a^2
25n = 25/4
Therefore,
4m^2 = 25n
So, we have shown that 4m^2 is equal to 25n, as required.