Question

Solve the quadratic equations in questions 1 – 5 by factoring.Need my answers checked1. x2 – 49 = 02. 3x3 – 12x = 03. 12x2 + 14x + 12 = 184. –x3 + 22x2 – 121x = 05. x2 – 4x = 51.x^2-49=0x^2-7^2=0(x-7)(x+7)=0Solution x=7, x=-7 2.3x^3-12x=03x(x^2-4)=03x(x^2-2^2)=03x(x-2)(x+2)=0 3.12x^2+14x+12=1812x^2+14+12-18=012x^2+14x-6=02(6x^2+7x-3)=02(6x^2-2x+9x-3)=02(2x(3x-1)+3(3x-1)=02(2x+3)(3x-1)=0Solution X=-3/2, X=1/3 4.-x^3+22x^2-121x=0-x(x^2-22x+121)=0-x(x-11)^2=0-x(x-11)(x+11)=0Solutions x=11,x=-11 5.x^2-4x=5x^2-4x-5=0x^2+z-5x-5=0(x^2+x)-(5x+5)=0x(x+1)-5(x+1)=0(x-5)(x+1)=0Solutions x=5,x=-1

Answer 1

Given:

The quadratic equation is:

$\begin{gathered} (1)x^2-49=0 \\ \\ (2)3x^3-12x=0 \\ \\ (3)12x^2+14x+12=18 \\ \\ (4)-x^3+22x^2-121x=0 \\ \\ (5)x^2-4x=5 \\ \end{gathered}$

Find-:

Solve the quadratic equation is:

Explanation-:

The factoring of the equation is:

(1)

$\begin{gathered} x^2-49=0 \\ \\ (x+7)(x-7)=0 \\ \\ x+7=0\text{ and }x-7=0 \\ \\ x=-7\text{ and }x=7 \end{gathered}$

(2)

The equation is:

$\begin{gathered} 3x^3-12x=0 \\ \\ 3x(x^2-4)=0 \\ \\ 3x=0\text{ and }x^2-4=0 \\ \\ x=0\text{ and }(x+2)(x-2)=0 \\ \\ x=-2\text{ and }x=2\text{ and }x=0 \end{gathered}$

(3)

The equation is:

$\begin{gathered} 12x^2+14x+12=18 \\ \\ 12x^2+14x+12-18=0 \\ \\ 12x^2+14x-6=0 \\ \\ 6x^2+7x-3=0 \\ \\ 6x^2+9x-2x-3=0 \\ \\ 3x(2x+3)-1(2x+3)=0 \\ \\ (2x+3)(3x-1)=0 \end{gathered}$

So the value of "x" is:

$\begin{gathered} 2x+3=0\text{ and }3x-1=0 \\ \\ x=-\frac{3}{2}\text{ and }x=\frac{1}{3} \end{gathered}$

(4)

$\begin{gathered} -x^3+22x^2-121x=0 \\ \\ x(-x^2+22x-121)=0 \\ \\ x=0\text{ and }-x^2+22x-121=0 \\ \\ -x^2+22x-121=0 \\ \\ -x^2+11x+11x-121=0 \\ \\ -x(x-11)+11(x-11)=0 \\ \\ (x-11)(-x+11)=0 \\ \\ x=11\text{ and }x=11 \end{gathered}$

So, the value of "x" is:

$x=0\text{ and }x=11$

(5)

$\begin{gathered} x^2-4x=5 \\ \\ x^2-4x-5=0 \\ \\ x^2-5x+x-5=0 \\ \\ x(x-5)+1(x-5)=0 \\ \\ (x-5)(x+1)=0 \\ \\ x=5\text{ and }x=-1 \end{gathered}$

The value of x is:

$x=5\text{ and }x=-1$