if one of the root of the equation ax^2+bx+c=0 is four times the other show that 4b^2-25ac=0...
help please show working plz
(x-q)(x-4q) = a x^2 + b x + c
x^2 - 5qx + 4 q^2 =a x^2 + bx + c
if a = 1
b = -5 q
c = 4 q^2
then
4 b^2 = 100 q^2
and
25 a c = 100 q^2
and
100 q^2 -100 q^2 = 0 all right
To solve this problem, we need to use the information given and the quadratic formula. Let's begin:
Given that one of the roots of the equation ax^2 + bx + c = 0 is four times the other, we can express the roots as x and 4x, respectively.
The sum of the roots of a quadratic equation is given by the formula:
Sum of roots (S) = -b/a
We know that the sum of the roots is (x + 4x), which is equal to -b/a.
Therefore, we can write the equation as:
x + 4x = -b/a
Combining the x terms, we have:
5x = -b/a
x = -b/5a
Now, let's use the quadratic formula to find the roots of the equation:
The quadratic formula is:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Since we have one root equal to 4 times the other, we can substitute x = 4(4x) into the quadratic formula:
4(4x) = (-b ± sqrt(b^2 - 4ac)) / (2a)
16x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Simplifying further, we have:
16x = -b/2a ± sqrt(b^2 - 4ac) / (2a)
Multiplying both sides by 2a, we get:
32ax = -b ± sqrt(b^2 - 4ac)
Now, we can substitute x = -b/5a into the equation:
32a (-b/5a) = -b ± sqrt(b^2 - 4ac)
Simplifying, we have:
-32b/5 = -b ± sqrt(b^2 - 4ac)
Multiplying both sides by 5:
-32b = -5b ± 5sqrt(b^2 - 4ac)
10b ± 5sqrt(b^2 - 4ac) = 32b
Rearranging the terms, we have:
10b = 32b ± 5sqrt(b^2 - 4ac)
Subtracting 32b from both sides, we obtain:
-22b = ± 5sqrt(b^2 - 4ac)
Squaring both sides, we get:
484b^2 = 25(b^2 - 4ac)
Expanding, we have:
484b^2 = 25b^2 - 100ac
Combining like terms, we get:
484b^2 - 25b^2 + 100ac = 0
Simplifying further, we have:
459b^2 + 100ac = 0
Therefore, we have shown that 4b^2 - 25ac = 0.
So, the expression 4b^2 - 25ac equals zero.
I hope this step-by-step explanation helps! Let me know if you need further assistance.
To solve this problem, we need to use the concept of the quadratic formula and the fact that one of the roots of the quadratic equation is four times the other.
Let's assume that one root of the equation ax^2 + bx + c = 0 is represented by r. According to the given condition, the other root would be 4r.
The sum of the roots of a quadratic equation is given by the formula:
Sum of roots = -b/a
In this case, the sum of the roots would be:
r + 4r = -b/a
Simplifying, we get:
5r = -b/a
Now, let's find the product of the roots. The product of the roots of a quadratic equation is given by the formula:
Product of roots = c/a
In this case, the product of the roots would be:
r * 4r = c/a
Simplifying, we get:
4r^2 = c/a
Now, we can use these equations to find the values of b and c in terms of a and r.
We'll start by expressing r in terms of b and a using the equation 5r = -b/a. Solving for r, we get:
r = -b/(5a)
Substituting this value of r into the equation 4r^2 = c/a, we can solve for c:
4(-b/(5a))^2 = c/a
(16b^2)/(25a^2) = c/a
Now that we have expressions for b and c, let's substitute them into the equation 4b^2 - 25ac and simplify it.
4b^2 - 25ac = 4b^2 - 25a((16b^2)/(25a^2))
Simplifying further, we get:
4b^2 - (16b^2)
= -12b^2
Therefore, 4b^2 - 25ac = -12b^2, which means the given statement 4b^2 - 25ac = 0 is not true in general.
It's important to note that the original question assumes that the coefficient of x^2 (a) is non-zero. Please double-check the question to ensure that it is correct, or if there may have been a misunderstanding or typo.