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Can you create a quadratic equation with the solution being -5?
Wiki User
∙ 11y ago
Yes, this can be done this way:
(x-(-5))(x+a)=0, where a is any (positive or negative) number
This simplifies to: (x+5)(x+a) = 0
or written as quadratic: x2+(a+5)x+5a = 0
Solutions: x1 = -5 and x2 = -a,
so if a=5 the equation shortens to (x+5)2=0 ( or x2+10x+25=0 )
and there will be only one solution: x1,2 = -5
Wiki User
∙ 11y ago
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What do you do if the second difference in a quadratic equation is 1?
The second difference of a quadratic equation being one indicates the second derivative at that point is positive. What you do from there depends on what property or transformation you're looking for with respect to the equation.
Solve quadratic equation by using formulae?
with the quadratic equation being ax2+bx+c and the formula being x=[-b±(b2-4ac)1/2]/2a just plug in the values for a b and c the quantity raised to the one half denotes a square root of that quantity.
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