Question

Can a quadratic polynomial equation with real coefficients have one real solution and one complex solution? If so,
give an example of such an equation. If not, explain why not.
Recall from Algebra II that every quadratic expression can be written as a product of two linear factors, that is,
ax^2 + bx + c = a(x − r1)(x − r2),
where rr1 and rr2 are solutions of the polynomial equation ax^2 + bx + c = 0.

Answer 1

Answer:

No

Step-by-step explanation:

We have to find that a quadratic polynomial equation with real coefficient can have one real solution and one complex solution.

quadratic equation is given by

$ax^2+bx+c=0$

It can be written as  the product of linear factors

$ax^2+bx+c=a(x-r_1)(x-r_2)$

Where $r_1,r_2$ are solutions of the given  polynomial equation.

No , a quadratic polynomial equation can not  have one real solution and one complex solution because complex root are always in paired not a single.

A quadratic equation have two roots only.

If a quadratic equation have complex root then the equation have both complex root .

If  a equation have real root then it have both real.

Therefore, a quadratic equation can not have one real and one compelx solution.

Answer 2

Answer:

Step-by-step explanation:

We want to solve the equation

$ax2+bx+c=0$

where a≠0.

Instead of dividing by a which is the common procedure,

multiply both sides by 4a. Here we obtain the equivalent equation

$4a2x2+4abx+4ac=0$-----------(1)

Note that $4a2x2+4abx$ is almost the square of $2ax+b$.

Then,

$4a2x2+4abx=(2ax+b)2−b2.$

Then the equation can be rewritten as

$(2ax+b)2−b2+4ac=0$  -----------(2)

or we can also written as

$(2ax+b)2=b2−4ac$----------(3)

We find that

$2 a x+b=\pm \sqrt{b^{2}-4 a c}$−−−−−−−(4)

and therefore

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$−−−−−−(5)

 

I have attempted to show that underlying division by a, when followed by a finishing the square methodology, is definitely not an easiest technique. One may comment moreover that in the event that we first gap by a, we wind up a few extra "variable based math" steps to mostly fix the division to give the arrangements their conventional structure.  

Division by an is unquestionably a right start in case it is trailed by a contention that fosters the association between the coefficients and the aggregate and result of the roots. In a perfect world, each sort of confirmation ought to be introduced, since each associates with a significant group of thoughts. What's more, a twice demonstrated hypothesis is twice as obvious.

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