Question

Use the Quadratic formula to find all real zeros of the 2nd degree polynomial

URGENT!!! What did I do wrong??

Answer 1

x ∈ {-5, -1}

Explanation:

Here's the solution using the quadratic formula:

`x=(-b\pm√(b^2-4ac))/(2a)=(-24\pm√(24^2-4\cdot 4\cdot 20))/(2\cdot 4)\\\\=(-24\pm√(576-320))/(8)=(-24\pm√(256))/(8)\\\\=(-24\pm 16)/(8)=-3\pm 2=\{-5,-1\}`

The real zeros are -5 and -1.

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There are many ways to check your answer. One of them is to look at the given quadratic, which has no changes of sign in its coefficients. (They are all positive.) That means there can be no positive real roots, so already you know that x=0.5 won't work.

Also, the constant in the quadratic is the product of the roots, For your roots, their product is -7/4, so even multiplying by 4 (the leading coefficient in the given quadratic), you don't get anything like 20.