How many extraneous solutions does the equation below have?

Mathematics

2 answers:

FrozenT [24]3 years ago

8 0

Answer

Explanation

(2m)/(2m+3)-(2m)/(2m-3)=1

Simplifying the left hand first

(2m)/(2m+3)-(2m)/(2m-3) = {2m(2m-3) - 2m(2m+3)}/(4m²-9)

(4m²-6m-4m²-6m)/(4m²-9)

= (-12m) / (4m²-9)

Now this equet to 1

(-12m) / (4m²-9) = 1

-12m = 4m²-9

4m²+ 12m -9 =0 ⇒⇒⇒ This is a quadratic equation that has 2 real solutions.

4m²+ 12m -9 =0

m² + 3m + (3/2)²= 9/4 + 9/4

(m + 3/2)² = 18/4

m = √18/2 - 3/2 or m = -√18/2 - 3/2

= 0.621 = -3.621

So we can say that the equation has NO extraneous solutions.

Answer = 0

Nadya [2.5K]3 years ago

4 0

Solution:

The given equation with solution is

$\frac{2 m}{2 m+3}-\frac{2 m}{2 m-3}=1\\\\ 2m(\frac{1}{2m+3}-\frac{1}{2 m-3})=1\\\\ 2 m\times\frac{2m-3-2m-3}{(2m+3)(2m-3)}=1\\\\ -12 m=4m^2-9\\\\ 4m^2+12 m-9=0\\\\ m=\frac{-12 \pm\sqrt{(12)^2-4\times 4 \times (-9)}}{2\times 4}\\\\ m=\frac{-12\pm\sqrt{288}}{8}\\\\ m=\frac{12}{8}\times(-1\pm\sqrt{2})\\\\ m=\frac{3}{2}\times(-1\pm\sqrt{2})$

The formula used to find the roots is

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

If it is a quadratic equation of type , a x²+bx+c=0

→→As, both the roots , which is , $m=\frac{3}{2}\times(-1\pm\sqrt{2})$ satisfy the equation given.That is when you substitute the value of m in equation given , we get LHS=RHS.