Question12 Mark for Review $x(x+1)-56=4 x(x-7)$
What is the sum of the solutions to the given equation? $\square$

Studdy Solution

STEP 1

1. The given equation is a quadratic equation.
2. The equation can be simplified and solved using algebraic manipulation.
3. The sum of the solutions of a quadratic equation $ax^2 + bx + c = 0$ is given by the formula $-\frac{b}{a}$ .

STEP 2

1. Expand and simplify both sides of the equation.
2. Move all terms to one side to form a standard quadratic equation.
3. Identify the coefficients of the quadratic equation.
4. Use the formula for the sum of the solutions of a quadratic equation.

STEP 3

Expand the left-hand side of the equation:
$x(x+1) - 56 = x^2 + x - 56$

STEP 4

Expand the right-hand side of the equation:
$4x(x-7) = 4x^2 - 28x$

STEP 5

Move all terms to one side to form a standard quadratic equation. Subtract the right-hand side from the left-hand side:
$x^2 + x - 56 - (4x^2 - 28x) = 0$

STEP 6

Simplify the equation by combining like terms:
$x^2 + x - 56 - 4x^2 + 28x = 0$ $-3x^2 + 29x - 56 = 0$

STEP 7

Identify the coefficients of the quadratic equation $ax^2 + bx + c = 0$ :
$a = -3, \quad b = 29, \quad c = -56$

STEP 8

Use the formula for the sum of the solutions of a quadratic equation, $-\frac{b}{a}$ :
$\text{Sum of solutions} = -\frac{b}{a} = -\frac{29}{-3} = \frac{29}{3}$
The sum of the solutions to the given equation is:
$\boxed{\frac{29}{3}}$

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