Question

9. If 4-i3 is a root of quadratic equation, then the equation is 1) x2 - 8x + 13 = 0 2) x2 – 8x + 19 = 0 3) x2 – 8x - 13 = 0 4) x2 - 8x -19 = 0 answer is what​

Answer 1

Answer:

$Complete \: \: step \: \: by \: \: step \: \: explanation</p><p>We \: \: know \: \: that \: \: if \: \: a \: \: polynomial \: \: equation \: \: (with \: \: real \: \: coefficients) \: \: has \: \: a \: complex \: \: root, \: \: then \: \: the \: \: complex \: \: conjugate \: \: of \: \: thwe \: \: root \: \: is \: \: also \: \: the \: \: solution \: \: of \: \: the \: \: equation. \: \: Since \: \: the \: \: quadratic \: \: equation \: \: has \: \:  4−i3–√4−i3 \: \:  as \: \: one \: \: of \: \: the roots, the other root of the equation is 4−i3–√¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=4+i3–√4−i3¯=4+i3</p><p>Hence, we have</p><p>Sum of roots =4+i3–√+4−i3–√=8=4+i3+4−i3=8</p><p>Product of roots =(4+i3–√)(4−i3–√)=42−(i3–√)2=42+3=19=(4+i3)(4−i3)=42−(i3)2=42+3=19</p><p>We know that a quadratic equation whose sum of roots is “a” and product of roots is “b” is x2−ax+0x2−ax+0 .</p><p>Hence the required quadratic equation is x2−8x+19=0x2−8x+19=0</p><p>Hence option bb is correct.</p><p></p><p>Note: Verification:</p><p>We can verify the correctness of our solution by checking that 4−i3–√4−i3 is a root of the equation.</p><p>We have</p><p> (4−i3–√)2=16−3−83–√i=13−83–√(4−i3)2=16−3−83i=13−83</p><p>Hence, we have</p><p> x2−8x+9=(4−i3–√)2−8(4−i3–√)+19=13−83–√i−32+8i3–√+19=0</p><p></p><p></p><p>$

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Step-by-step explanation:

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