The roots of the equation x^2+(a+3)x-2b=0 are -2 and -3. Find the values of a and b.
If x = -2, the equation is satisfied
(-2)^2 + (a+3)(-2) = 2 b
4 - 2 a - 6 = 2 b
2 a + 2 b = -2
a + b = -1
so b = -a -1
If x = -3 the equation is satisfied
(-3)^2 +(a+3)(-3) = 2 b
9 - 3 a - 9 = 2 b
3 a + 2 b = 0
3 a + 2 (-a -1) = 0
3 a - 2 a = 2
a = 2
b = -3
another solution:
recall that a quadratic equation can also be rewritten as,
x^2 - (sum of roots)x + (product of roots) = 0
where
the numerical coeff of x = -(sum of roots), and
constant = (product of roots)
therefore,
sum of roots = -2 + (-3) = -5
equating this to the numerical coeff of x,
-(a + 3) = -5
a + 3 = 5
a = 2
product of roots = (-2)(-3) = 6
equating this to constant,
-2b = 6
b = -3
hope this helps~ :)
To find the values of a and b, we can use the information given about the roots of the equation.
The roots of the quadratic equation can be found using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the roots are -2 and -3. We can plug these values into the quadratic formula to form two separate equations:
-2 = (-a - 3 - √((a+3)^2 - 4(-2b))) / 2
-3 = (-a - 3 + √((a+3)^2 - 4(-2b))) / 2
Now we can simplify and solve these equations.
For the first equation:
-2 = (-a - 3 - √((a+3)^2 - 4(-2b))) / 2
Multiply both sides by 2 to eliminate the denominator:
-4 = -a - 3 - √((a+3)^2 - 4(-2b))
Rearrange terms:
-a - 1 = √((a+3)^2 - 4(-2b))
Square both sides:
(-a - 1)^2 = ((a+3)^2 - 4(-2b))^2
Expand and simplify both sides:
a^2 + 2a + 1 = a^2 + 6a + 9 + 16b
We can cancel out the a^2 term:
2a + 1 = 6a + 9 + 16b
Rearrange terms:
-4a - 8 = 16b
Divide both sides by -4:
a + 2 = -4b
Now let's solve the second equation in a similar manner.
For the second equation:
-3 = (-a - 3 + √((a+3)^2 - 4(-2b))) / 2
Multiply both sides by 2:
-6 = -a - 3 + √((a+3)^2 - 4(-2b))
Rearrange terms:
-3a - 3 = √((a+3)^2 - 4(-2b))
Square both sides:
(-3a - 3)^2 = ((a+3)^2 - 4(-2b))^2
Expand and simplify both sides:
9a^2 + 18a + 9 = a^2 + 6a + 9 + 16b
Cancel out the a^2 term:
8a + 16 = 16b
Divide both sides by 8:
a + 2 = 2b
Now we have two equations:
a + 2 = -4b
a + 2 = 2b
We can solve this system of equations to find the values of a and b.
Subtract the second equation from the first equation:
-6b = -4b - 2b
-6b = -6b
This means that b can have any value. Let's substitute this into one of the equations.
Using a + 2 = -4b, we can substitute b with any value, let's say b = 1:
a + 2 = -4(1)
a + 2 = -4
a = -6
Therefore, the values of a and b are a = -6 and b = 1, respectively.