Question

Solve the quadratic equation by completing the square. 2x² - 20x + 48 = 0

Answer 1

Steps

So firstly, we need to isolate the x terms onto one side. To do this, subtract 48 on both sides of the equation:

$2x^2-20x=-48$

Next, divide both sides by 2:

$x^2-10x=-24$

Next, we are gonna make the left side of the equation a perfect square. To find the constant of this soon-to-be perfect square, divide the x coefficient by 2 and then square the quotient. Add the result onto both sides of the equation:

$-10 \div 2=-5\\(-5)^2=25\\\\x^2-10x+25=1$

Now, factor the left side:

$(x-5)^2=1$

Next, square root both sides of the equation:

$x-5=\pm 1$

Next, add 5 to both sides of the equation:

$x=5\pm 1$

Lastly, solve the left side twice -- once with the plus sign and once with the minus sign.

$x=6,4$

Answer

In short, x = 6 and 4

Answer 2

Answer: x = 4   x = 6

Step-by-step explanation:

2x² - 20x + 48 = 0

2x² - 20 x + _____ = -48 + ______      subtracted 48 from both sides

2(x² - 10x + _____ ) = 2(-24 + _____ )   factored out 2 from both sides

x² - 10x + _____  = -24 + _____          divided both sides by 2

x² - 10x + 25  = -24 +25          added 25 to both sides

       ↓        ↑            ↑

     $\dfrac{-10}{2}=(-5)^2$     $\bigg(\dfrac{b}{2}\bigg)^2$ makes a perfect square

(x - 5)² = 1             simplified

$\sqrt{(x-5)^2} =\sqrt{1}$    took square root of both sides

x - 5 = ± 1             simplified

x - 5 = 1       x - 5 = -1     split into two separate equations

   x = 6           x = 4       solved for x