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How do you find the zeros for this equation? y=x (to the power of 2) +3x-5
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High School Teacher
To solve for the zeros, set y = 0 and you'll have:
`x^2 + 3x - 5 = 0`
Then use quadratic formula: `x = (-b+-sqrt(b^2-4ac))/(2a)`
The quadratic equation above is in the form `ax^2 + bx + c = 0.`
You can now apply the formula.
a = 1
b = 3
c = -5
So, `x = (-3+-sqrt(3^2-4(1)(-5)))/((2)(1))`
`x = (-3+-sqrt(9+20))/2`
`x = (-3 + sqrt(29))/2`
`x = 1.19 (rounded to two decimal position)`
`x = -4.19(rounded to two decimal position)`
Since it is a quadratic equation (with 2 as the highest exponent), you should have 2 values for x.
The roots are 1.19 and - 4.19.
Posted by mariloucortez on March 26, 2013 at 11:02 PM (Answer #1)
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