Find the Bounds of the Zeros p(x)=-1.2x^2+62.5x-491

Math
p(x)=-1.2×2+62.5x-491
Check the leading coefficient of the function. This number is the coefficient of the expression with the largest degree.
Largest Degree: 2
Leading Coefficient: -1.2
Simplify each term.
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Cancel the common factor of -1.2.
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Cancel the common factor.
p(x)=-1.2×2-1.2+62.5x-1.2+-491-1.2
Divide x2 by 1.
p(x)=x2+62.5x-1.2+-491-1.2
p(x)=x2+62.5x-1.2+-491-1.2
Move the negative in front of the fraction.
p(x)=x2-62.5×1.2+-491-1.2
Factor 62.5 out of 62.5x.
p(x)=x2-62.5(x)1.2+-491-1.2
Factor 1.2 out of 1.2.
p(x)=x2-62.5(x)1.2(1)+-491-1.2
Separate fractions.
p(x)=x2-(62.51.2⋅x1)+-491-1.2
Divide 62.5 by 1.2.
p(x)=x2-(52.083‾(x1))+-491-1.2
Divide x by 1.
p(x)=x2-(52.083‾x)+-491-1.2
Multiply 52.083‾ by -1.
p(x)=x2-52.083‾x+-491-1.2
Divide -491 by -1.2.
p(x)=x2-52.083‾x+409.16‾
p(x)=x2-52.083‾x+409.16‾
Create a list of the coefficients of the function except the leading coefficient of 1.
-52.083‾,409.16‾
There will be two bound options, b1 and b2, the smaller of which is the answer. To calculate the first bound option, find the absolute value of the largest coefficient from the list of coefficients. Then add 1.
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Arrange the terms in ascending order.
b1=|-52.083‾|,|409.16‾|
The maximum value is the largest value in the arranged data set.
b1=|409.16‾|
The absolute value is the distance between a number and zero. The distance between 0 and 409.16‾ is 409.16‾.
b1=409.16‾+1
Add 409.16‾ and 1.
b1=410.16‾
b1=410.16‾
To calculate the second bound option, sum the absolute values of the coefficients from the list of coefficients. If the sum is greater than 1, use that number. If not, use 1.
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Simplify each term.
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The absolute value is the distance between a number and zero. The distance between -52.083‾ and 0 is 52.083‾.
b2=52.083‾+|409.16‾|
The absolute value is the distance between a number and zero. The distance between 0 and 409.16‾ is 409.16‾.
b2=52.083‾+409.16‾
b2=52.083‾+409.16‾
Add 52.083‾ and 409.16‾.
b2=461.25
Arrange the terms in ascending order.
b2=1,461.25
The maximum value is the largest value in the arranged data set.
b2=461.25
b2=461.25
Take the smaller bound option between b1=410.16‾ and b2=461.25.
Smaller Bound: 410.16‾
Every real root on p(x)=-1.2×2+62.5x-491 lies between -410.16‾ and 410.16‾.
-410.16‾ and 410.16‾
Find the Bounds of the Zeros p(x)=-1.2x^2+62.5x-491

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