How do you complete the square to solve $4 x^{2} - 7 x - 2 = 0$ $4x^2 - 7x - 2 = 0$ ?

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How do you complete the square to solve $4 x^{2} - 7 x - 2 = 0$ $4x^2 - 7x - 2 = 0$ ?

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Answer 1 · 2015-05-23T14:37:38

In this way:

$4 x^{2} - 7 x - 2 = 0 \Rightarrow 4 (x^{2} - \frac{7}{4} x) - 2 = 0 \Rightarrow$ $4x^2-7x-2=0rArr4(x^2-7/4x)-2=0rArr$

$4 (x^{2} - \frac{7}{4} x + \frac{49}{64} - \frac{49}{64}) - 2 = 0 \Rightarrow$ $4(x^2-7/4x+49/64-49/64)-2=0rArr$

$4 (x^{2} - \frac{7}{4} x + \frac{49}{64}) - \frac{49}{16} - 2 = 0 \Rightarrow$ $4(x^2-7/4x+49/64)-49/16-2=0rArr$

$4 {(x - \frac{7}{8})}^{2} = \frac{49}{16} + 2 \Rightarrow 4 {(x - \frac{7}{8})}^{2} = \frac{49 + 32}{16} \Rightarrow$ $4(x-7/8)^2=49/16+2rArr4(x-7/8)^2=(49+32)/16rArr$

${(x - \frac{7}{8})}^{2} = \frac{81}{16} \cdot \frac{1}{4} \Rightarrow {(x - \frac{7}{8})}^{2} = \frac{81}{64} \Rightarrow$ $(x-7/8)^2=81/16*1/4rArr(x-7/8)^2=81/64rArr$

$x - \frac{7}{8} = \pm \frac{9}{8} \Rightarrow x = \frac{7}{8} \pm \frac{9}{8} \Rightarrow$ $x-7/8=+-9/8rArrx=7/8+-9/8rArr$

$x_{1} = \frac{7}{8} - \frac{9}{8} = - \frac{2}{8} = - \frac{1}{4}$ $x_1=7/8-9/8=-2/8=-1/4$

$x_{2} = \frac{7}{8} + \frac{9}{8} = \frac{16}{8} = 2$ $x_2=7/8+9/8=16/8=2$ .

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How do you complete the square to solve $4 x^{2} - 7 x - 2 = 0$ $4x^2 - 7x - 2 = 0$ ?

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How do you complete the square to solve $4 x^{2} - 7 x - 2 = 0$ $4x^2 - 7x - 2 = 0$ ?