How do you solve by completing the square $a^{2} + 8 a + 11 = 0$ $a^2+ 8a + 11 = 0$ ?

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Answer 1 · 2018-04-30T18:27:35

$a = 4 + \sqrt{5}, 4 - \sqrt{5}$ $a=4+sqrt5, 4-sqrt5$

$a^{2} + 8 a + 11 = {(a + 4)}^{2} - 16 + 11 = {(a + 4)}^{2} - 5$ $a^2+8a+11=(a+4)^2-16+11=(a+4)^2-5$

Checking:

$(a + 4) (a + 4) = a^{2} + 8 x + 16 - 5 = a^{2} + 8 x + 11$ $(a+4)(a+4)=a^2+8x+16-5=a^2+8x+11$

Solve:

${(a - 4)}^{2} - 5 = 0$ $(a-4)^2-5=0$

${(a - 4)}^{2} = 5$ $(a-4)^2=5$

$a - 4 = \pm \sqrt{5}$ $a-4=pmsqrt5$

$a = 4 \pm \sqrt{5}$ $a=4pmsqrt5$

How do you solve by completing the square a2+8a+11=0a^2+ 8a + 11 = 0?