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sergiy2304 [10]
3 years ago
5

Someone help me out please

Mathematics
1 answer:
uysha [10]3 years ago
3 0

1 step (B): raise both sides of the equation to the power of 2.

(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\   (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4.

2 step (A): simplify to obtain the final radical term on one side of the equation.

-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2.

3 step (F): raise both sides of the equation to the power of 2 again.

(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2.

4 step (E): simplify to get a quadratic equation.

4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0.

5 step (D): use the quadratic formula to find the values of x.

D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}.

6 step (C): apply the zero product rule.

x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}.

Additional 7 step: check these solutions, substituting into the initial equation.

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