An SAT Math Question

$K(H(x))$

$= K(x^2+1)$

$= -(x^2+1)^2+4$

$= -(x^4+2x^2+1)+4$

$= -x^4-2x^2+3$

So solve for $x^2$ using the quadratic formula gives:

$x^2 = \frac{-(-2)\pm\sqrt{4+12}}{-2}$

$=\frac{2\pm4}{-2}$

$= -1 \pm 2 = 1, -3$

So $x^2 = 1 or -3$ thus $x = \pm 1 or \pm \sqrt{3}i$

$K(H(x))=0\\ -{ H(x) }^{ 2 }+4=0\\ 4={ H(x) }^{ 2 }\\ H(x)=\pm 2\\ { x }^{ 2 }+1=\pm 2\\$

If ${ x }^{ 2 }+1=2$ , then

${ x }^{ 2 }=1\\ x=\pm 1$

and if ${ x }^{ 2 }+1=-2$ , then

${ x }^{ 2 }=-3\\ x=\pm \sqrt { 3 } i$

The solutions are $x=\pm \sqrt { 3 } i$ or $x=\pm 1$

-(x^2 + 1)^2 + 4 = 0

=> (x^2 + 1)^2 - 2^2 = 0

=> (x^2 + 1 - 2)(x^2 + 1 + 2) = 0

=> (x^2 - 1)(x^2 + 3) = 0

=> x = +/- 1 OR x = +/- j sqrt(3)

$K(H(x))=-(x^{2}+1)^{2}+4$

$K(H(x))=-x^{4}-2x^{2}-1+4$

$K(H(x))=-x^{4}-2x^{2}+3$

$0=-x^{4}-2x^{2}+3$

$0=(-x^{2}-3)(x^{2}-1)$

$-x^{2}-3=0 →x^{2}=-3→x=±\sqrt{3}i$

$x^{2}-1=0→x^{2}=1→x=±1$

$x=±1,± \sqrt{3}i$

Say: $y=H\left( x \right) ={ x }^{ 2 }+1$

$=> K(H\left( x \right) )=K(y)=-{ y }^{ 2 }+4=0$

$=> y=\pm 2$

$=> { x }^{ 2 }+1=\pm 2$

$=> x = \pm 1$ or $x = \pm i\sqrt { 3 }$

H(x) = x^2 + 1; K(x) = - x^2 + 4; K(H(x)) = - (x^2 + 1)^2 + 4 = 0; suppose: x^2 + 1 = t, so: - t^2 + 4 = 0, t^2 = 4; => t1 = + 2; t2 = - 2;
1) x^2 + 1 = 2, x^2 = 1; => x1 = + 1; x2 = - 1;
2) x^2 + 1 = - 2, x^2 = - 3; => x3 = + i sqrt(3); x4 = - i sqrt(3);
Note: next time i will try to use an equation editor.

$K(H(x))\\ \qquad =K(x^{ 2 }+1)\\ \qquad =-(x^{ 2 }+1)^{ 2 }+4\\ \\ \\$

$Now\quad K(H(x))=0 \quad \Rightarrow \quad -(x^{ 2 }+1)^{ 2 }+4=0\\ \\ \\ So\quad (x^{ 2 }+1)^{ 2 }-4=0\quad [Multiplying\quad each\quad side\quad by\quad -1]\\ \qquad (x^{ 2 }+1)^{ 2 }-{ 2 }^{ 2 }=0\\ \qquad (x^{ 2 }+1-{ 2) }*(x^{ 2 }+1+{ 2 })=0\quad [{ a }^{ 2 }-b^{ 2 }\quad =\quad (a+b)(a-b)]\\ \qquad (x^{ 2 }-1{ ) }*(x^{ 2 }+3{ )=0 }\\ \\ \\$

$Hence\quad \\ \qquad Either\quad (x^{ 2 }-1{ ) }=0\quad or\quad (x^{ 2 }+3{ )=0 }\\ \\ if\quad (x^{ 2 }-1{ ) }=0\quad \\ \qquad (x+1)*(x-1)=0\\$

$Hence\quad (x+1)=0\quad or\quad (x-1)=0\\ i.e\quad x=-1\quad or\quad x=+1\quad \quad [solution\quad 1\quad x=\pm 1]$

$\quad \\ \\ if\quad (x^{ 2 }+3{ )=0 }\quad \quad i.e\quad (1)x^{ 2 }+(0)x+(3){ =0 }\\ by\quad quadratic\quad solving\quad formula\quad [x=\frac { -b\pm \sqrt [ 2 ]{ { b }^{ 2 }-4ac } }{ 2a } ]\\ a=1,\quad b=0,\quad c=3\\ \\ \\ \qquad x=\frac { -0\pm \sqrt [ 2 ]{ 0-4*1*3 } }{ 2*1 } \\ \\ \Rightarrow \quad x=\quad \frac { \pm \sqrt { -12 } }{ 2 } \quad =\quad \pm i\sqrt { 3 } \quad [solution\quad 2\quad x=\pm i\sqrt { 3 } ]\\ \\ \\ Hence\quad the\quad solutions\quad are\quad \quad x=\pm 1,\pm i\sqrt { 3 } \\$