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2. Nora made an incorrect statement when using factoring to solve the equationx2 + 2x – 12 = 3. Put an X next to the incorrect statement. Correct her error.The standard form of the equation is x2 + 2x – 15 = 0.The factored form of the equation is (x – 3)(x+ 5) = 0.Because (x - 3)(x + 5) = 0, you can use the Zero-Product Property to write(x - 3) = 0 or (x + 5) = 0.The solutions of the equation are -3 and 5.The x-coordinate of the vertex of the related function is -1.

User Gabriel Gava
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1 Answer

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13 votes

Given:


x^2+2x-12=3

To solve the questions, evaluate each statement.

(a) The standard form of the equation.

The standard form of a quadratic equation is ax² + bx + c = 0. So, to find the standard form, subtract 3 from both sides of the equation.


\begin{gathered} x^2+2x-12-3=3-3 \\ x^2+2x-15=0 \end{gathered}

So, the statement is correct.

(b) The factored form.

The factored form of a quadratic equation is (x-a)(x-b) = 0, where a and b are the zeros of the equations.

To find the zeros, use the quadratic formula.

For ax² + bx + c = 0, the zeros are:


x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}^{}

So, substituting the values:


\begin{gathered} x=\frac{-2\pm\sqrt[]{2^2-4\cdot1\cdot(-15)}}{2\cdot1}^{} \\ x=\frac{-2\pm\sqrt[]{4+60}}{2}^{}=\frac{-2\pm\sqrt[]{64}}{2} \\ x=(-2\pm8)/(2) \\ x_1=(-2-8)/(2)=-(10)/(2)=-5 \\ x_2=(-2+8)/(2)=(6)/(2)=3 \end{gathered}

The zeros are -5 and 3. So, the factored form is (x-3)(x+5) = 0

So, the statement is correct.

(c) Zero-Product Property.

Since (x-3)(x+5) = 0, then:

(x - 3) = 0

or

(x + 5) = 0

So, the statement is correct.

(d) Solutions of the equation.

The solutions of the equation are the zeros: -5 and 3 (shown in part B).

Thus, this statement is false. The correct would be the solution is -5 and 3.

(e) x-coordinate of the vertex.

The x-coordinate of the vertex is:


\begin{gathered} x_v=(-b)/(2a) \\ x_v=(-2)/(2\cdot1) \\ x_v=(-2)/(2) \\ x_v=-1 \end{gathered}

So, the statement is correct.

Answer: Statement D is wrong. The correct statement is: the solution is -5 and 3.

User Cheiron
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