Question 829838: Use the equation x^2+6x+c=0
A) Find a value of c that results in two real roots.
B) Find a value of c that results in one real root.
C) Find a value of c that results in two complex roots.
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
given:
y = x^2 + 6x + c = 0
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A) Find a value of c that results in two real roots:
c = 0
y = x^2 + 6x + 0 = 0
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the above quadratic equation is in standard form, with a=1, b=6, and c=0
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
1 6 0
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two real roots at:
x = 0
x = -6
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the quadratic vertex is a minimum at: ( x= -3, y= -9 )
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B) Find a value of c that results in one real root:
c = 9
y = x^2 + 6x + 9 = 0
---
the above quadratic equation is in standard form, with a=1, b=6, and c=9
---
to solve the quadratic equation, by using the quadratic formula, copy and paste this:
1 6 9
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
---
the quadratic has one real roots at:
x = -3
x = -3
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C) Find a value of c that results in two complex roots:
c = 10
y = x^2 + 6x + 10 = 0
---
the above quadratic equation is in standard form, with a=1, b=6, and c=10
---
to solve the quadratic equation, by using the quadratic formula, copy and paste this:
1 6 10
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
---
the quadratic has two complex roots at:
x = -3 + (1)i
x = -3 - (1)i
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