Question

Ayuda porfa es muy urgente

Answer 1

Respuesta:

a) $x^{2} -10x+25$ $=0$ :

Factorizamos :

$x^{2} -10x+25=(x-5)(x-5)=(x-5)^{2} \\\\x^{2} -10x+25=0\\(x-5)^{2}=0\\x-5=\sqrt{0}\\ x-5=0\\x=5$

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b) $x^{2} +2x+1$ $=0$ :

Factorizamos :

$x^{2} +2x+1 =( x+1)(x+1)=(x+1)^{2} \\\\x^{2} +2x+1=0\\(x+1)^{2}=0\\x+1=\sqrt{0}\\x+1=0\\x=-1$

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c) $x^{2} -x+\frac{1}{4}$ $=0$ :

Factorizamos :

$x^{2} -x+\frac{1}{4} =(x-\frac{1}{2} )(x-\frac{1}{2} )=(x-\frac{1}{2} )^{2}$

$x^{2}- x+\frac{1}{4} =0\\(x-\frac{1}{2} )^{2} =0\\x-\frac{1}{2} =\sqrt{0}\\\\ x=\frac{1}{2}$

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d) $9x^{2} -6x+1$ $=0$ :

Factorizamos :

$9x^{2} -6x+1=(3x-1)(3x-1)=(3x-1)^{2} \\\\9x^{2} -6x+1=0\\(3x-1)^{2} =0\\3x-1=\sqrt{0} \\3x-1=0\\3x=1\\x=\frac{1}{3}$

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e) $4x^{2} +12x+9$ $=0$ :

Factorizamos :

$4x^{2} +12x+9=(2x+3)(2x+3)=(2x+3)^{2} \\\\4x^{2} +12x+9=0\\(2x+3)^{2}=0\\2x+3=\sqrt{0}\\ 2x+3=0\\2x=-3\\x=\frac{-3}{2}$

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f) $16x^{2} -4x+\frac{1}{4}=0$ :

Factorizamos :

$16x^{2} -4x+\frac{1}{4}=(4x-\frac{1}{2} )(4x-\frac{1}{2})=(4x-\frac{1}{2})^{2} \\\\16x^{2} -4x+\frac{1}{4}=0\\(4x-\frac{1}{2})^{2}=0\\4x-\frac{1}{2}=\sqrt{0}\\ 4x-1/2=0\\4x=1/2\\x=\frac{1/2}{4}\\ x=\frac{1}{8}$