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[tex]\Large{\underline{\underline{{\mathfrak{{\bigstar}\:Answer}}}}}\\\\[/tex]

[tex]\bf{1.\:\:x^{2}+4x+1=0}[/tex]

[tex]\longrightarrow\:\:\sf{x=\dfrac{-4{\underline{+}}\sqrt{4^{2}-4\times1\times1}}{2\times1}}[/tex]

[tex]\longrightarrow\:\:\sf{x=\dfrac{-4{\underline{+}}\sqrt{16-4}}{2}}[/tex]

[tex]\longrightarrow\:\:\sf{x=\dfrac{-4{\underline{+}}\sqrt{12}}{2}}[/tex]

[tex]\longrightarrow\:\:\sf{x=\dfrac{4{\underline{+}}2\sqrt{3}}{2}}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=-2-\sqrt{3}\:,\:x_{2}= -2+\sqrt{3}}\\\\\\[/tex]

[tex]\bf{2.\:\:x^{2}-9x+14=0}[/tex]

[tex]\longrightarrow\:\:\sf{x^{2}-2x-7x+14=0}[/tex]

[tex]\longrightarrow\:\:\sf{x\times (x-2)-7(x-2)=0}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=2\:,\:x_{2}=7}\\\\\\[/tex]

[tex]\bf{3.\:\:x^{2}+8x+2=22}[/tex]

[tex]\longrightarrow\:\:\sf{x^{2}+8x+2-22=0}[/tex]

[tex]\longrightarrow\:\:\sf{x(x+10)-2(x+10)=0}[/tex]

[tex]\longrightarrow\:\:\sf{(x+10)(x-2)=0}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=-10\:,\:x_{2}=2}\\\\\\[/tex]

[tex]\bf{4.\:\:x^{2}+8x+7=0}[/tex]

[tex]\longrightarrow\:\:\sf{x^{2}+7x+x+7=0}[/tex]

[tex]\longrightarrow\:\:\sf{x(x+7)+x+7=0}[/tex]

[tex]\longrightarrow\:\:\sf{(x+7)(x+1)=0}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=-7\:,\:x_{2}=1}\\\\\\[/tex]

[tex]\bf{5.\:\:x^{2}-10x+25=9}[/tex]

[tex]\longrightarrow\:\:\sf{(x-5)^{2}=9}[/tex]

[tex]\longrightarrow\:\:\sf{x-5={\underline{+}}3}[/tex]

[tex]\longrightarrow\:\:\sf{x-5=-3 \: ; \:x-5=3}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=2\:,\:x_{2}=8}\\\\\\[/tex]

[tex]\bf{6.\:\:x^{2}-10x+16=0}[/tex]

[tex]\longrightarrow\:\:\sf{x^{2}-2x-8x+16=0}[/tex]

[tex]\longrightarrow\:\:\sf{(x-2)(x-8)=0}[/tex]

[tex]\longrightarrow\:\:\sf{x-2=0\:;\:x-8=0}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=2\:,\:x_{2}=8}\\\\\\[/tex]

[tex]\bf{7.\:\:2x^{2}+7x-4=0}[/tex]

[tex]\longrightarrow\:\:\sf{2x(x+4)-(x-4)=0}[/tex]

[tex]\longrightarrow\:\:\sf{(x+4)((2x-1)=0}[/tex]

[tex]\longrightarrow\:\:\sf{x+4=0\:;\:2x-1=0}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=-4\:,\:x_{2}=\dfrac{1}{2}}\\\\\\[/tex]

[tex]\bf{8.\:\:x^{2}-2x+3=0}[/tex]

[tex]\longrightarrow\:\:\sf{x=\dfrac{-(-2){\underline{+}}\sqrt{(-2)^{2}-4\times 1\times 3}}{2\times1}}[/tex]

[tex]\longrightarrow\:\:\sf{\dfrac{2{\underline{+}}\sqrt{4-12}}{2}}[/tex]

[tex]\longrightarrow\:\:\sf{\dfrac{2{\underline{+}}\sqrt{-8}}{2}}[/tex]

[tex]\longrightarrow\:\:\sf{x \notin {\mathbb{R}}}\\\\\\[/tex]

[tex]\bf{9.\:\:3x^{2}+8x+5=0}[/tex]

[tex]\longrightarrow\:\:\sf{x(3x+5)+3x+5=0}[/tex]

[tex]\longrightarrow\:\:\sf{(3x+5)(x+1)=0}[/tex]

[tex]\longrightarrow\:\:\sf{3x+5=0\:;\:x+1=0}[/tex]

[tex]\longrightarrow\:\:\sf{x_{1}=-\dfrac{5}{3}\:,\:x_{2}=-1}\\\\\\[/tex]

[tex]\bf{10.\:\:{\mathbb{\red{1}}}\:\:}[/tex]:')

xero099

The solutions are listed below:

(i) [tex]x = -2\pm \sqrt{3}[/tex], (ii) [tex]x = \frac{9}{2}\pm \frac{5}{2}[/tex], (iii) [tex]x = -4 \pm 6[/tex], (iv) [tex]x = -4\pm 3[/tex], (v) [tex]x = -5\pm 3[/tex], (vi) [tex]x = 5 \pm 3[/tex], (vii) [tex]x = -\frac{7}{4}\pm \frac{9}{4}[/tex], (viii) [tex]x = 1 \pm i\sqrt{2}[/tex], (ix) [tex]x = -\frac{4}{3}\pm \frac{1}{3}[/tex], (x) 1. I do not celebrate Valentine's day at all.

How to solve polynomials by completing the square

Completing the square consists in applying algebraic operations to transform part of the second order polynomial into a perfect square trinomial and simplify the resulting expression. Now we proceed to present the corresponding solutions:

(i) [tex]x^{2}+4\cdot x + 1 = 0[/tex]

[tex]x^{2}+4\cdot x +4 = 3[/tex]

[tex](x+2)^{2} = 3[/tex]

[tex]x+2 = \pm \sqrt{3}[/tex]

[tex]x = -2\pm \sqrt{3}[/tex]

(ii) [tex]x^{2}-9\cdot x + 14 = 0[/tex]

[tex]x^{2}-9\cdot x + 14+\frac{25}{4} = \frac{25}{4}[/tex]

[tex]x^{2}-9\cdot x +\frac{81}{4} = \frac{25}{4}[/tex]

[tex]\left(x-\frac{9}{2} \right)^{2} = \frac{25}{4}[/tex]

[tex]x -\frac{9}{2} = \pm \frac{5}{2}[/tex]

[tex]x = \frac{9}{2}\pm \frac{5}{2}[/tex]

(iii) [tex]x^{2}+8\cdot x + 2 = 22[/tex]

[tex]x^{2}+8\cdot x +16 = 36[/tex]

[tex](x+4)^{2} = 36[/tex]

[tex]x+4 = \pm 6[/tex]

[tex]x = -4 \pm 6[/tex]

(iv) [tex]x^{2}+8\cdot x + 7 = 0[/tex]

[tex]x^{2}+8\cdot x + 16 = 9[/tex]

[tex](x+4)^{2} = 9[/tex]

[tex]x+4 = \pm 3[/tex]

[tex]x = -4\pm 3[/tex]

(v) [tex]x^{2}+10\cdot x + 25 = 9[/tex]

[tex](x+5)^{2} = 9[/tex]

[tex]x+5 = \pm 3[/tex]

[tex]x = -5\pm 3[/tex]

(vi) [tex]x^{2}-10\cdot x + 16 = 0[/tex]

[tex]x^{2}-10\cdot x + 25 = 9[/tex]

[tex](x-5)^{2} = 9[/tex]

[tex]x-5 = \pm 3[/tex]

[tex]x = 5 \pm 3[/tex]

(vii) [tex]2\cdot x^{2} + 7\cdot x - 4 = 0[/tex]

[tex]x^{2}+\frac{7}{2}\cdot x -2 = 0[/tex]

[tex]x^{2} + \frac{7}{2}\cdot x -2 +\frac{81}{16} = \frac{81}{16}[/tex]

[tex]x^{2}+\frac{7}{2}\cdot x +\frac{49}{16} = \frac{81}{16}[/tex]

[tex]\left(x+\frac{7}{4} \right)^{2} = \frac{81}{16}[/tex]

[tex]x + \frac{7}{4} = \pm \frac{9}{4}[/tex]

[tex]x = -\frac{7}{4}\pm \frac{9}{4}[/tex]

(viii) [tex]x^{2}-2\cdot x + 3 = 0[/tex]

[tex]x^{2}-2\cdot x + 1 + 2 = 0[/tex]

[tex](x-1)^{2} = -2[/tex]

[tex]x-1 = \pm \sqrt{-2}[/tex]

[tex]x = 1 \pm i\sqrt{2}[/tex]

(ix) [tex]3\cdot x^{2} + 8\cdot x + 5 = 0[/tex]

[tex]x^{2}+\frac{8}{3}\cdot x +\frac{5}{3} = 0[/tex]

[tex]x^{2}+\frac{8}{3}\cdot x +\frac{5}{3}+\frac{1}{9} = \frac{1}{9}[/tex]

[tex]x^{2}+\frac{8}{3}\cdot x + \frac{16}{9} = \frac{1}{9}[/tex]

[tex]\left(x+\frac{4}{3} \right)^{2} = \frac{1}{9}[/tex]

[tex]x + \frac{4}{3} = \pm \frac{1}{3}[/tex]

[tex]x = -\frac{4}{3}\pm \frac{1}{3}[/tex]

(x) 1. I do not celebrate Valentine's day at all.

To learn more on polynomials, we kindly invite to check this verified question: https://brainly.com/question/17822016

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