Question

3) Solve the following quadratic equations.
i) x2 + 3ix + 10 = 0​

Answer 1

Let's solve for i.

$x^2+3ix+10=0$

Step 1: Add$→ -x^2$to both sides.

$→3ix+x^2+10+−x^2=0+−x^2$

$→3ix+10=−x2$

Step 2: Add -10 to both sides.

$→3ix+10+−10=−x^2+−10$

$→3ix=−x^2−10$

Step 3: Divide both sides by 3x.3ix 3x

$→=−x^2−103xi=−x^2−103x$

Answer:

$→i=−x^2−103x$

Answer 2

Roots for the quadratic equation ax
2
+bx+c , the roots are
x=
2a
−b±
b
2
−4ac
​

​


For the given equation x
2
+3ix+10 ,

b
2
−4ac=(3i)
2
−4(1)(10)

=9i
2
−40

=−9−40=−49

So, the roots are
x=
2
−3i±
−49
​

​


x=
2
−3i±7i
​


x=−5i,2i


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