Solve by Factoring 2x^2-7x+4=x^2-8

$2 x^{2} - 7 x + 4 = x^{2} - 8$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$2 x^{2} - 7 x + 4 - x^{2} = - 8$

Add $8$ to both sides of the equation.

$2 x^{2} - 7 x + 4 - x^{2} + 8 = 0$

Step 2

Simplify $2 x^{2} - 7 x + 4 - x^{2} + 8$ .

Subtract $x^{2}$ from $2 x^{2}$ .

$x^{2} - 7 x + 4 + 8 = 0$

Add $4$ and $8$ .

$x^{2} - 7 x + 12 = 0$

Step 3

Factor $x^{2} - 7 x + 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$(x - 4) (x - 3) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x - 3 = 0$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 7

The final solution is all the values that make $(x - 4) (x - 3) = 0$ true.

$x = 4, 3$

Solve by Factoring 2x^2-7x+4=x^2-8

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