Question

Solve Quadratic Equations (show all work)

Answer 1

(9) The solution is x = 1 and x = 5

(10) The solution is x = 12 and x = -8

Explanation:

(9) The given expression is $(x+3)(x-3)=6 x-14$

We need to determine the solution of the equation.

The term $(x+3)(x-3)$ can be expanded using the formula, $(a+b)(a-b)=a^{2}-b^{2}$

Thus, we have,

$x^{2}-9=6 x-14$

$x^{2}+5=6 x$

$x^{2}-6 x+5=0$

Solving the quadratic equation, we get,

$(x-5)(x-1)=0$

The solution of the quadratic equation are x = 5 and x = 1

(10) The given expression is $3 x^{2}-12 x-288=0$

We need to determine the solution of the equation.

Let us solve the equation using the quadratic formula,

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

where $a=3, b=-12, c=-288$

Thus, we have,

$x=\frac{-(-12) \pm \sqrt{(-12)^{2}-4 \cdot 3(-288)}}{2 \cdot 3}$

Simplifying the terms, we have,

$x=\frac{12 \pm \sqrt{144+3456}}{6}$

$x=\frac{12 \pm \sqrt{3600}}{6}$

$x=\frac{12 \pm60}{6}$

Thus, the roots are

$x=\frac{12 +60}{6}$  and  $x=\frac{12 -60}{6}$

$x=\frac{72}{6}$  and  $x=\frac{-48}{6}$

$x=12$  and  $x=-8$

Thus, the solution is x = 12 and x = -8