Question

Joyce is trying to solve the equation y = x2 − 8x + 7 using the quadratic formula. She has made an error in one of the steps below. Find the step where Joyce went wrong. (1 point)

Step 1: x equals negative 8 plus or minus the square root of the quantity eight squared minus four times one times seven, end quantity, all over two times one.

Step 2: x equals negative 8 plus or minus the square root of sixty-four minus twenty-eight all over two times one.

Step 3: x equals negative 8 plus or minus the square root of thirty-six all over two times one.

Step 4: x equals negative 8 plus or minus six all over two.


Step 1

Step 2

Step 3

Step 4

Answer 1

To solve for the roots of a quadratic equation, you use the equation

$x= \frac{-b+/- \sqrt{ b^{2}-4ac } }{2a}$

The parameters a, b and c represent the coefficients of the standard form of the quadratic equation: y = ax^2 + bx + c. Thus, for the equation y = x2 − 8x + 7,

a = 1
b = -8
c = 7

The mistake Joyce made was in Step 1. Because Step 1 is wrong, the rest of the equation is wrong consequently. The mistake was made by Joyce when she assigned b = 8. She forgot to include the negative sign of -8x. 

Answer 2

Her mistake is in the first step, she wrote the equation incorrectly.

Solving quadratic equations:

A general quadratic equation:

ax^2 + bx + c = 0

has the solutions:

x = (-b ± √(b^2 - 4ac))/(2a)

In this case, the equation is:

y = x^2 - 8x + 7

And what Joyce writes in each step is:

Step 1:

x = (-8 ± √(8^2 - 4*1*7))/(2*1)

Then we can already see the mistake.

Notice that in the given equation, we have:

b = -8

And the general solution has:

x = (-b ± √(b^2 - 4ac))/(2a)

-b would be equal to 8, then what Joyce must have written here is:

x = (8 ± √((-8)^2 - 4*1*7))/(2*1)

If you want to learn more about quadratic equations, you can read:

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