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Can you help me with this math question?
- Thread starter SucAtMath
- Start date
The first one is easy - just equate the powers of 2 and equate the powers of 3.
So solve: x=-y+2, 3y=x
Multiplying the second one by [MATH]\sqrt{2}[/MATH][MATH]2\sqrt{1-x^2}+6=\sqrt{2}\sqrt{1+\sqrt{1-x^2}}+3\sqrt{1+x}+2\sqrt{1-x}[/MATH]Note [MATH]1-x^2=(1-x)(1+x)[/MATH]I would introduce 2 new variables, [MATH]u=\sqrt{1-x}, v=\sqrt{1+x}[/MATH]
Then the equation is:
[MATH]2uv+6=\sqrt{2}\sqrt{1+uv}+3v+2u[/MATH] (1), noting that [MATH]u^2+v^2=2[/MATH] (2)
Now also, [MATH](u+v)^2=u^2+v^2+2uv=2(1+uv)[/MATH] because of (2) above
so [MATH]\sqrt{1+uv}=\frac{(u+v)}{\sqrt{2}}[/MATH](1) becomes [MATH]2uv+6=u+v+3v+2u[/MATH]
[MATH]2uv-4v-3u+6=0\\ 2v(u-2)-3(u-2)=0\\ (2v-3)(u-2)=0[/MATH][MATH]v=\frac{3}{2}[/MATH] or [MATH]u=2[/MATH]Now go back to v and u in terms of x and find x.
(Note if you want the original equation to involve only real numbers, you require [MATH]-1≤x≤1[/MATH]).
So solve: x=-y+2, 3y=x
Multiplying the second one by [MATH]\sqrt{2}[/MATH][MATH]2\sqrt{1-x^2}+6=\sqrt{2}\sqrt{1+\sqrt{1-x^2}}+3\sqrt{1+x}+2\sqrt{1-x}[/MATH]Note [MATH]1-x^2=(1-x)(1+x)[/MATH]I would introduce 2 new variables, [MATH]u=\sqrt{1-x}, v=\sqrt{1+x}[/MATH]
Then the equation is:
[MATH]2uv+6=\sqrt{2}\sqrt{1+uv}+3v+2u[/MATH] (1), noting that [MATH]u^2+v^2=2[/MATH] (2)
Now also, [MATH](u+v)^2=u^2+v^2+2uv=2(1+uv)[/MATH] because of (2) above
so [MATH]\sqrt{1+uv}=\frac{(u+v)}{\sqrt{2}}[/MATH](1) becomes [MATH]2uv+6=u+v+3v+2u[/MATH]
[MATH]2uv-4v-3u+6=0\\ 2v(u-2)-3(u-2)=0\\ (2v-3)(u-2)=0[/MATH][MATH]v=\frac{3}{2}[/MATH] or [MATH]u=2[/MATH]Now go back to v and u in terms of x and find x.
(Note if you want the original equation to involve only real numbers, you require [MATH]-1≤x≤1[/MATH]).
The first one is easy - just equate the powers of 2 and equate the powers of 3.
So solve: x=-y+2, 3y=x
Multiplying the second one by [imath]\sqrt{2}[/imath]
[imath]2\sqrt{1-x^2}+6=\sqrt{2}\sqrt{1+\sqrt{1-x^2}}+3\sqrt{1+x}+2\sqrt{1-x}[/imath]
Note [imath]1-x^2=(1-x)(1+x)[/imath]
I would introduce 2 new variables, [imath]u=\sqrt{1-x}, v=\sqrt{1+x}[/imath]
Then the equation is:
[imath]2uv+6=\sqrt{2}\sqrt{1+uv}+3v+2u[/imath] (1), noting that [imath]u^2+v^2=2[/imath] (2)
Now also, [imath](u+v)^2=u^2+v^2+2uv=2(1+uv)[/imath] because of (2) above
so [imath]\sqrt{1+uv}=\frac{(u+v)}{\sqrt{2}}[/imath]
(1) becomes [imath]2uv+6=u+v+3v+2u[/imath]
[imath]2uv-4v-3u+6=0\\ 2v(u-2)-3(u-2)=0\\ (2v-3)(u-2)=0[/imath]
[imath]v=\frac{3}{2}[/imath] or [imath]u=2[/imath]
Now go back to v and u in terms of x and find x.
(Note if you want the original equation to involve only real numbers, you require [imath]-1≤x≤1[/imath]).
So solve: x=-y+2, 3y=x
Multiplying the second one by [imath]\sqrt{2}[/imath]
[imath]2\sqrt{1-x^2}+6=\sqrt{2}\sqrt{1+\sqrt{1-x^2}}+3\sqrt{1+x}+2\sqrt{1-x}[/imath]
Note [imath]1-x^2=(1-x)(1+x)[/imath]
I would introduce 2 new variables, [imath]u=\sqrt{1-x}, v=\sqrt{1+x}[/imath]
Then the equation is:
[imath]2uv+6=\sqrt{2}\sqrt{1+uv}+3v+2u[/imath] (1), noting that [imath]u^2+v^2=2[/imath] (2)
Now also, [imath](u+v)^2=u^2+v^2+2uv=2(1+uv)[/imath] because of (2) above
so [imath]\sqrt{1+uv}=\frac{(u+v)}{\sqrt{2}}[/imath]
(1) becomes [imath]2uv+6=u+v+3v+2u[/imath]
[imath]2uv-4v-3u+6=0\\ 2v(u-2)-3(u-2)=0\\ (2v-3)(u-2)=0[/imath]
[imath]v=\frac{3}{2}[/imath] or [imath]u=2[/imath]
Now go back to v and u in terms of x and find x.
(Note if you want the original equation to involve only real numbers, you require [imath]-1≤x≤1[/imath]).