QuestionWork out the value of $a$ in the equation below. $x^{2}-14 x+49=(x-a)^{2}$

Studdy Solution

STEP 1

1. The equation $x^2 - 14x + 49 = (x-a)^2$ is a quadratic equation.
2. The left side is already in standard quadratic form.
3. The right side is a perfect square trinomial.
4. Solving the equation involves expanding the right side and equating coefficients.

STEP 2

1. Expand the right side of the equation.
2. Equate the coefficients of the quadratic equation.
3. Solve for the value of $a$ .

STEP 3

Expand the right side of the equation. The expression $(x-a)^2$ can be expanded using the formula $(x-a)^2 = x^2 - 2ax + a^2$ .
$(x-a)^2 = x^2 - 2ax + a^2$

STEP 4

Equate the expanded form of the right side to the left side of the equation. This gives us:
$x^2 - 14x + 49 = x^2 - 2ax + a^2$

STEP 5

Since the quadratic terms $x^2$ are the same on both sides, we focus on the linear and constant terms. Equate the coefficients of the linear terms and the constant terms:
1. Linear terms: $-14 = -2a$
2. Constant terms: $49 = a^2$

STEP 6

Solve the equation for the linear terms to find $a$ :
$-14 = -2a$ $2a = 14$ $a = 7$

STEP 7

Verify the constant term equation with the found value of $a$ :
$49 = a^2$ $49 = 7^2$
The value of $a$ satisfies both conditions.
The value of $a$ is:
$\boxed{7}$

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