It is given that 2-5i/2 and 2+5i/2 are the root of the quadratic equation ax ^2-bx+29=0?
回答 (2)
2017-01-23 2:44 am
two forays:
find a and b obediently by sum and product of roots (-b/a, c/a)
then use quadratic formula to find the answer
-OR-
alter the formula ax ^2-bx+29 on LHS into (x-2+5i/2)(x-2-5i/2) , then use quadratic formula, same as above
find a and b obediently by sum and product of roots (-b/a, c/a)
then use quadratic formula to find the answer
-OR-
alter the formula ax ^2-bx+29 on LHS into (x-2+5i/2)(x-2-5i/2) , then use quadratic formula, same as above
2017-01-17 8:02 am
It is given that (2－5i)/2 and (2+5i)/2 are the root of the quadratic equation
ax ^2－bx+29=0，Solve the equation ax ^2-bx+29=4x.express your answer
in the form of m+ni.
Sol
(2－5i)+(2+5i)=b/a
4=b/a
b=4a
(2－5i)(2+5i)=29/a
29=29/a
a=1
b=4
ax^2－bx+29=4x
x^2－4x+29=4x
x^2－8x+29=0
x=[8+/-√(64－4*1*29)]/2=4+/-i√13
ax ^2－bx+29=0，Solve the equation ax ^2-bx+29=4x.express your answer
in the form of m+ni.
Sol
(2－5i)+(2+5i)=b/a
4=b/a
b=4a
(2－5i)(2+5i)=29/a
29=29/a
a=1
b=4
ax^2－bx+29=4x
x^2－4x+29=4x
x^2－8x+29=0
x=[8+/-√(64－4*1*29)]/2=4+/-i√13
收錄日期: 2021-04-30 22:00:20
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