What are the roots of -4 and 3
1 answer:
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Answer:
a = -0.32
b = 16
c = - 168
Step-by-step explanation:
In the figure attached, the quadratic function is shown. We can model it as follows:
y = a*(x - x1)*(x - x2)
where (x1, 0) and (x2, 0) are its roots
From the picture, roots are (15, 0) and (35, 0), replacing them into the equation:
y = a*(x - 15)*(x - 35)
and the vertex is at (25,32), replacing into the equation:
32 = a*(25 - 15)*(25 - 35)
32 = a*(-100)
a = 32/(-100)
a = -0.32
Applying distributive property:
y = -0.32*(x - 15)*(x - 35)
y = -0.32*(x² - 35x - 15x + 525)
y = -0.32x² + 16x - 168
which corresponds to the general form:
y = ax² + bx + c
Hi there!
When we have an equation standard form...
...the formula of the discriminant is
D = b^2 - 4ac
When
D > 0 we have two real solutions
D = 0 we have one real solutions
D < 0 we don't have real solutions
1.) Find the value of the discriminant and the the number of real solutions of
x^2-8x+7=0
Plug in the values from the equation into the formula of the discriminant
D > 0 and therefore we have two real solutions.
2.) Find the value of the discriminant and the number of real solutions of
2x^2+4x+2=0
Again, plug in the values from the equation into the formula of the discriminant.
D = 0 and therefore we have one real solution.
~ Hope this helps you.
When we have an equation standard form...
...the formula of the discriminant is
D = b^2 - 4ac
When
D > 0 we have two real solutions
D = 0 we have one real solutions
D < 0 we don't have real solutions
1.) Find the value of the discriminant and the the number of real solutions of
x^2-8x+7=0
Plug in the values from the equation into the formula of the discriminant
D > 0 and therefore we have two real solutions.
2.) Find the value of the discriminant and the number of real solutions of
2x^2+4x+2=0
Again, plug in the values from the equation into the formula of the discriminant.
D = 0 and therefore we have one real solution.
~ Hope this helps you.