One root of the equation x^2+6x+q=0 is twice the other root. Find q and the roots.
Use the properties of roots
for x^2 + bx + c = 0, the sum of the roots is -b and the product of the roots is c
So let the roots be m and 2m
then m + 2m = -6
m = -2
the roots are -2 and -4
m(2m) = q
q = -2(-4) = 8
your equation is x^2 + 6x+ 8 = 0
your roots are -2 and -4
check: if the roots are -2 and -4, we would have
(x+2)(x+4) = 0
x^2 + 6x + 8 = 0 , as needed
Well, let's give those roots some names. Let's call one root "Bob" and the other root "Larry." Now, since one root is twice the other, we can say that Bob = 2Larry.
Now, we know that the sum of the roots of a quadratic equation is -b/a, where a is the coefficient of the x² term and b is the coefficient of the x term. In this case, a = 1 and b = 6.
So, Bob + Larry = -6/1. Substituting in Bob = 2Larry, we get:
2Larry + Larry = -6
3Larry = -6
Larry = -6/3
Larry = -2
And since Bob = 2Larry, we have:
Bob = 2(-2)
Bob = -4
So the roots are Larry = -2 and Bob = -4.
Now, we need to find q. We know that the sum of the roots is -b/a, so:
Larry + Bob = -6/1
-2 + -4 = -6/1
-6 = -6
But we also know that the sum of the roots is -b/a. In this case, a = 1 and b = 6, so:
Larry + Bob = -6/1
-2 + -4 = -6/1
-6 = -6
So, q = -6.
Therefore, the roots are -2 and -4, and q is -6. Voila!
To solve this problem, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the roots can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Since we know that one root is twice the other root, let's assign the smaller root as x and the larger root as 2x.
Now, let's substitute these values into the quadratic equation x^2 + 6x + q = 0:
(x)(2x) + 6(x) + q = 0
2x^2 + 6x + q = 0
Applying the quadratic formula to this equation:
x = (-6 ± √(6^2 - 4(2)(q))) / (2(2))
Simplifying further:
x = (-6 ± √(36 - 8q)) / 4
To simplify the equation further, let's write it in terms of the discriminant:
x = (-6 ± √(8(9 - q))) / 4
x = (-6 ± 2√(9 - q)) / 4
x = (-3 ± √(9 - q)) / 2
Now we have the values of x in terms of q.
Since one root is smaller than the other, we can assign the "+/-" sign to the smaller root, so:
x = (-3 + √(9 - q)) / 2 --> The smaller root
2x = (-3 - √(9 - q)) --> The larger root
Since we know that one root is twice the other root, we can equate 2x to the smaller root:
2x = (-3 + √(9 - q)) / 2
Cross-multiplying:
4x = -3 + √(9 - q)
Simplifying further:
4x + 3 = √(9 - q)
Squaring both sides to eliminate the square root:
(4x + 3)^2 = 9 - q
Expanding the left side:
16x^2 + 24x + 9 = 9 - q
Subtracting 9 from both sides:
16x^2 + 24x = -q
Since both roots satisfy this equation, we can equate it to zero:
16x^2 + 24x + q = 0
Comparing this equation with the given equation (x^2 + 6x + q = 0), we can see that the coefficients of x^2 terms are the same, the coefficients of x terms are the same, and the constants are the same.
Therefore, we can conclude that q = 0.
To solve this problem, we can use the quadratic formula. The quadratic formula is given by:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, the equation x^2 + 6x + q = 0 is in the standard quadratic form ax^2 + bx + c = 0, where a = 1, b = 6, and c = q.
Let's substitute these values into the quadratic formula and solve for x:
x = (-(6) ± sqrt((6)^2 - 4(1)(q))) / (2(1))
Simplifying further:
x = (-6 ± sqrt(36 - 4q)) / 2
We are given that one root is twice the other root. Let's represent the smaller root as r, and the larger root as 2r.
Since the quadratic formula gives us both roots, we can equate them to get two equations:
r = (-6 + sqrt(36 - 4q)) / 2 [Eq. 1]
2r = (-6 - sqrt(36 - 4q)) / 2 [Eq. 2]
Now, let's solve these equations simultaneously to find the value of q and the roots:
From Eq. 1, we can multiply both sides by 2:
2r = -6 + sqrt(36 - 4q) [Eq. 3]
From Eq. 2, we can multiply both sides by -1 to eliminate the negative sign:
-2r = 6 + sqrt(36 - 4q) [Eq. 4]
Adding Eq. 3 and Eq. 4, we get:
2r - 2r = (-6 + sqrt(36 - 4q)) + (6 + sqrt(36 - 4q))
0 = sqrt(36 - 4q) + sqrt(36 - 4q)
0 = 2sqrt(36 - 4q)
Squaring both sides:
0 = 4(36 - 4q)
0 = 144 - 16q
16q = 144
q = 144 / 16
q = 9
Now that we know the value of q is 9, we can substitute it back into either Eq. 1 or Eq. 2 to find the roots:
Let's substitute it into Eq. 1:
r = (-6 + sqrt(36 - 4(9))) / 2
r = (-6 + sqrt(36 - 36)) / 2
r = (-6 + sqrt(0)) / 2
r = (-6 + 0) / 2
r = -6 / 2
r = -3
So, one root is -3.
Now, let's substitute q = 9 into Eq. 2:
2r = (-6 - sqrt(36 - 4(9))) / 2
2r = (-6 - sqrt(36 - 36)) / 2
2r = (-6 - sqrt(0)) / 2
2r = (-6 - 0) / 2
2r = -6 / 2
2r = -3
r = -3 / 2
So, the other root is -3/2.
Therefore, the value of q is 9 and the roots of the equation x^2 + 6x + q = 0 are -3 and -3/2.