Question

solve: (x+2)(x+3)+(x-3)(x-2)-2×(x+1)=0
$(x + 2)(x + 3) + (x - 3)(x - 2) - 2 x(x + 1) = 0$
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Answer 1

Given:-

• (x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0

To find:-

• Solve the following.

Solution:-

According to the question,

→ (x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0

→ x² + 3x + 2x + 6 + x² - 2x - 3x + 6 - 2x² - 2x = 0

→ x² + 5x + 6 + x² - 5x + 6 - 2x² - 2x = 0

→ x² + x² - 2x² + 5x - 5x - 2x + 6 + 6 = 0

→ -2x + 12 = 0

→ -2x = -12

→ x = (-12)/(-2)

→ x = 6

Hence,

• the value of x is 6.

Answer 2

Answer:

Given to solve,

⇒ $(x+2)(x+3)+(x-3)(x-2)-2(x+1)=0$

⇒ $(x^2 + 3x + 2x + 6) + (x^2 -2x-3x+6) -(2x+2) = 0$

⇒ $x^2 + 5x + 6 + x^2 -5x+6 -2x - 2 = 0$

⇒ $2x^2 - 2x + 10 = 0$

⇒ $x^2 - x + 5 = 0$

We know the formula,

$x = \frac{-b±\sqrt{b^2 - 4ac} }{2a}$

Here a = 1

        b = -1

        c =5

∴ $x = \frac{-(-1)±\sqrt{(-1)^2 - 4(1)(5)} }{2.1}$

  $x = \frac{1±\sqrt{1 - 20} }{2}$

  $x = \frac{1±\sqrt{-19} }{2}$

Therefore it has no real roots.

It has imaginary roots because the value inside the square root has a negative sign.The square root of a negative number is not a real number.

                Hope it helps you^_^