hi,
I am stuck on this question. y=x^2+4x and y^2= 25x^2+20x+3
how many of the solutions are real? and how many of the real roots are positive and how many are negative?
thank you for any suggestions
I suggest you solve for the roots.
y=x(x+4) At y=0, what are the roots?
Y= (5x+3)(5x+1)
To find the solutions to the equation y = x^2 + 4x, you can set y equal to 0 and solve for x.
Step 1: Set y = 0:
0 = x^2 + 4x
Step 2: Factor the equation:
0 = x(x + 4)
To find the roots, set each factor equal to 0 and solve for x:
x = 0 (root 1)
x + 4 = 0
x = -4 (root 2)
So, the equation y = x^2 + 4x has two real roots, one at x = 0 and another at x = -4.
Now let's move on to the second equation, y^2 = 25x^2 + 20x + 3.
Step 1: Expand the equation:
y^2 = 25x^2 + 20x + 3
Step 2: Set y^2 = 0:
0 = 25x^2 + 20x + 3
Step 3: Factor the equation:
0 = (5x + 3)(5x + 1)
To find the roots, set each factor equal to 0 and solve for x:
5x + 3 = 0
x = -3/5 (root 3)
5x + 1 = 0
x = -1/5 (root 4)
So, the equation y^2 = 25x^2 + 20x + 3 has two real roots, one at x = -3/5 and another at x = -1/5.
To determine whether the real roots are positive or negative, you can check the sign of each root by substituting the values back into the original equations.
For the first equation, y = x^2 + 4x:
- Substitute x = 0:
y = 0^2 + 4(0)
y = 0
- Substitute x = -4:
y = (-4)^2 + 4(-4)
y = 16 - 16
y = 0
Both roots 0 and -4 produce a y-value of 0, so they are neither positive nor negative.
For the second equation, y^2 = 25x^2 + 20x + 3:
- Substitute x = -3/5:
y^2 = 25(-3/5)^2 + 20(-3/5) + 3
y^2 = (9/5)^2 + (12/5)(-3) + 3
y^2 = 81/25 - 36/25 + 3
y^2 = 81/25 - 36/25 + 3/1
y^2 = (81 - 36 + 75)/25
y^2 = 120/25
y^2 = 24/5
- Substitute x = -1/5:
y^2 = 25(-1/5)^2 + 20(-1/5) + 3
y^2 = (1/5)^2 - 4/5 + 3
y^2 = 1/25 - 4/5 + 3/1
y^2 = (1 - 20 + 75)/25
y^2 = 56/25
By evaluating the y-values for both roots, we can see that the y-values for both roots are positive. Therefore, both roots x = -3/5 and x = -1/5 are negative.
To summarize:
- The equation y = x^2 + 4x has two real roots, one at x = 0 and another at x = -4.
- The equation y^2 = 25x^2 + 20x + 3 also has two real roots, one at x = -3/5 and another at x = -1/5.
- Both roots x = -3/5 and x = -1/5 of y^2 = 25x^2 + 20x + 3 are negative.