Solve the Quadratic Equation: Balancing and Simplifying to Find x

Question

Solve the following equation:

2x2+6x12=4x2+19x5 -2x^2+6x-12=-4x^2+19x-5

Video Solution

Solution Steps

00:00 Find X
00:03 Arrange the equation so the right side equals 0
00:16 Collect like terms
00:24 Identify equation components
00:33 Use the roots formula
00:42 Substitute appropriate values and solve for X
00:56 Calculate the products and squares
01:15 Calculate the square root of 225
01:19 Find the 2 possible solutions
01:32 This is the solution to the question

Step-by-Step Solution

To solve this quadratic equation, let us first simplify and rearrange the terms:

Start with the original equation:
2x2+6x12=4x2+19x5-2x^2 + 6x - 12 = -4x^2 + 19x - 5

Move all terms to one side to form a standard quadratic equation by adding 4x24x^2, subtracting 19x19x, and adding 55 to both sides:

(2x2+6x12)+4x219x+5=0(-2x^2 + 6x - 12) + 4x^2 - 19x + 5 = 0

This simplifies to:
2x213x7=02x^2 - 13x - 7 = 0

Now, identify the coefficients a=2a = 2, b=13b = -13, and c=7c = -7.

Apply the quadratic formula:
x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Subsitute aa, bb, and cc into the formula:

x=(13)±(13)242(7)22x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2}

Simplify:
x=13±169+564x = \frac{13 \pm \sqrt{169 + 56}}{4}

x=13±2254x = \frac{13 \pm \sqrt{225}}{4}

The square root of 225 is 15, thus:

x=13±154x = \frac{13 \pm 15}{4}

Calculate the two possible solutions:

  • First solution: x1=13+154=284=7x_1 = \frac{13 + 15}{4} = \frac{28}{4} = 7
  • Second solution: x2=13154=24=12x_2 = \frac{13 - 15}{4} = \frac{-2}{4} = -\frac{1}{2}

Therefore, the solutions to the problem are x1=7x_1 = 7 and x2=12x_2 = -\frac{1}{2}.

Thus, the correct answer is option 2: x1=7x_1=7, x2=12x_2=-\frac{1}{2}.

Answer

x1=7 x_1=7 , x2=12 x_2=-\frac{1}{2}