Question

describe and correct the error a student made in solving a quadratic equation. 0 = 2x²+7x + 5 0 = 2x²+2x+5x + 5 0 = 2x(x + 1)+5(x + 1) 0 = 2x,0 = x + 1,0≠5 0 = x - 1 = x a. in the first step, the student should have found factors of 5 that add to give 7 in order to factor the quadratic expression on the right. b. the contradiction 0≠5 in the fourth - line means there are no real solutions. c. the student should have evaluated the function 2x²+7x + 5 with x = 0 to find the zeros of the equation. d. the student should have evaluated the function 2x²+7x + 5 with x = 0 to find the zeros of the equation. what is the correct solution? x = (use a comma to separate answers as needed.)

Explanation:

Step1: Recall factoring quadratic formula

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 2\), \(b=7\), \(c = 5\)), we need to find two numbers \(m\) and \(n\) such that \(m\times n=ac=2\times5 = 10\) and \(m + n=b = 7\). The numbers are \(2\) and \(5\).

Step2: Rewrite the middle - term

Rewrite \(2x^{2}+7x + 5\) as \(2x^{2}+2x+5x + 5\).

Step3: Group the terms

\((2x^{2}+2x)+(5x + 5)=2x(x + 1)+5(x + 1)=(2x + 5)(x+1)\)

Step4: Set the factored form equal to zero

\((2x + 5)(x + 1)=0\). Then, using the zero - product property, if \(ab = 0\), then \(a=0\) or \(b = 0\). So \(2x+5=0\) gives \(x=-\frac{5}{2}\) and \(x + 1=0\) gives \(x=-1\).

Answer:

\(x=-1,-\frac{5}{2}\)