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Solution 2.2:2d

From Förberedande kurs i matematik 1

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Current revision (14:39, 23 September 2008) (edit) (undo)
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First, we move all the terms over to the left-hand side
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<center> [[Image:2_2_2d-1(2).gif]] </center>
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{{Displayed math||<math>(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}=0\,\textrm{.}</math>}}
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<center> [[Image:2_2_2d-2(2).gif]] </center>
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As the equation stands now, it seems that the best approach for solving the equation is to expand the squares, simplify and see what it leads to.
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When the squares are expanded, each term inside a square is multiplied by itself and all other terms
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{{Displayed math||<math>\begin{align}
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(x^{2}+4x+1)^{2} &= (x^{2}+4x+1)(x^{2}+4x+1)\\[5pt]
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&= x^{2}\cdot x^{2}+x^{2}\cdot 4x+x^{2}\cdot 1+4x\cdot x^{2}+4x\cdot 4x+4x\cdot 1\\[5pt]
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&\qquad\quad{}+1\cdot x^{2}+1\cdot 4x+1\cdot 1\\[5pt]
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&= x^{4}+4x^{3}+x^{2}+4x^{3}+16x^{2}+4x+x^{2}+4x+1\\[5pt]
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&= x^{4}+8x^{3}+18x^{2}+8x+1\,,\\[10pt]
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(2x^{2}+2x+3)^{2} &= (2x^{2}+2x+3)(2x^{2}+2x+3)\\[5pt]
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&= 2x^{2}\cdot 2x^{2}+2x^{2}\cdot 2x+2x^{2}\cdot 3+2x\cdot 2x^{2}+2x\cdot 2x\\[5pt]
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&\qquad\quad{}+2x\cdot 3+3\cdot 2x^{2}+3\cdot 2x+3\cdot 3\\[5pt]
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&= 4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\[5pt]
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&= 4x^{4}+8x^{3}+16x^{2}+12x+9\,\textrm{.}
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\end{align}</math>}}
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After we collect together all terms of the same order, the left hand side becomes
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{{Displayed math||<math>\begin{align}
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&(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}\\[5pt]
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&\qquad{}= (x^{4}+8x^{3}+18x^{2}+8x+1)+3x^{4}-2x^{2}\\[5pt]
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&\qquad\qquad{}-(4x^{4}+8x^{3}+16x^{2}+12x+9)\\[5pt]
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&\qquad{}= (x^{4}+3x^{4}-4x^{4})+(8x^{3}-8x^{3})+(18x^{2}-2x^{2}-16x^{2})\\[5pt]
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&\qquad\qquad{}+(8x-12x)+(1-9)\\[5pt]
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&\qquad{}= -4x-8\,\textrm{.}
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\end{align}</math>}}
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After all simplifications, the equation becomes
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{{Displayed math||<math>-4x-8=0\quad \Leftrightarrow \quad x=-2\,\textrm{.}</math>}}
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Finally, we check that <math>x=-2</math> is the correct answer by substituting
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<math>x=-2</math> into the equation
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{{Displayed math||<math>\begin{align}
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\text{LHS} &= \bigl((-2)^{2}+4\cdot(-2)+1\bigr)^{2}+3\cdot (-2)^{4}-2\cdot (-2)^{2}\\[5pt]
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&= (4-8+1)^{2} + 3\cdot 16 - 2\cdot 4 = (-3)^{2} + 48 - 8 = 9 + 48 - 8 = 49\,,\\[10pt]
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\text{RHS} &= \bigl(2\cdot(-2)^{2}+2\cdot (-2)+3\bigr)^{2} = (2\cdot 4-4+3)^{2} = 7^{2} = 49\,\textrm{.}
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\end{align}</math>}}

Current revision

First, we move all the terms over to the left-hand side

(x2+4x+1)2+3x4−2x2−(2x2+2x+3)2=0.

As the equation stands now, it seems that the best approach for solving the equation is to expand the squares, simplify and see what it leads to.

When the squares are expanded, each term inside a square is multiplied by itself and all other terms

(x2+4x+1)2(2x2+2x+3)2=(x2+4x+1)(x2+4x+1)=x2x2+x24x+x21+4xx2+4x4x+4x1+1x2+14x+11=x4+4x3+x2+4x3+16x2+4x+x2+4x+1=x4+8x3+18x2+8x+1=(2x2+2x+3)(2x2+2x+3)=2x22x2+2x22x+2x23+2x2x2+2x2x+2x3+32x2+32x+33=4x4+4x3+6x2+4x3+4x2+6x+6x2+6x+9=4x4+8x3+16x2+12x+9.

After we collect together all terms of the same order, the left hand side becomes

(x2+4x+1)2+3x4−2x2−(2x2+2x+3)2=(x4+8x3+18x2+8x+1)+3x4−2x2−(4x4+8x3+16x2+12x+9)=(x4+3x4−4x4)+(8x3−8x3)+(18x2−2x2−16x2)+(8x−12x)+(1−9)=−4x−8.

After all simplifications, the equation becomes

−4x−8=0x=−2.

Finally, we check that x=−2 is the correct answer by substituting x=−2 into the equation

LHSRHS=(−2)2+4(−2)+12+3(−2)4−2(−2)2=(4−8+1)2+316−24=(−3)2+48−8=9+48−8=49=2(−2)2+2(−2)+32=(24−4+3)2=72=49.
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