QuestionFactor the quadratic equation $x^{2}-14x+49$ .

Studdy Solution

STEP 1

Assumptions1. The given equation is a quadratic equation in the form $ax^{} + bx + c$ where $a =1$ , $b = -14$ , and $c =49$ . . We are asked to factorize this equation.

STEP 2

In a quadratic equation of the form $ax^{2} + bx + c$ , the factorization can be done if the equation can be written in the form $(dx+e)^{2}$ , where $d$ and $e$ are constants. This is possible if $b =2de$ and $c = e^{2}$ .

STEP 3

From the given equation $x^{2}-14x+49$ , we can see that $b = -14$ and $c =49$ .

STEP 4

We can write $b =2de$ as $-14 =2de$ .

STEP 5

We can write $c = e^{2}$ as $49 = e^{2}$ .

STEP 6

olving $49 = e^{2}$ for $e$ , we get $e = \pm$ .

STEP 7

We choose $e =7$ because if we choose $e = -7$ , then $d$ would be negative, which would not match the original equation.

STEP 8

Substituting $e =7$ into $-14 =2de$ , we can solve for $d$ .
$-14 =2d \cdot7$

STEP 9

olving the above equation for $d$ , we get $d = -$ .

STEP 10

Now that we have found $d$ and $e$ , we can write the given quadratic equation in the form $(dx+e)^{2}$ .

STEP 11

Substituting $d = -$ and $e =7$ into $(dx+e)^{}$ , we get the factorization of the given quadratic equation.
$(dx+e)^{} = (-x+7)^{} = (x-7)^{}$ So, the factorization of the quadratic equation $x^{}-14x+49$ is $(x-7)^{}$ .

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