Question

I've got 2 questions I would like some help with!

Any help is much appreciated!

Answer 1

Question 1)

One possible equation is:

`\displaystyle 11x^2 - 15x + 10 = 0`

Question 2)

Choice C

The equation is:

`2x^2 + 6x + 3=0`

Explanation:

Question 1)

Recall that the quadratic formula is given by:

`\displaystyle x = (-b\pm√(b^2 -4ac))/(2a)`

We want to find a quadratic with the solutions:

`\displaystyle x = (15\pm√(-215))/(22)`

Each value must be equal to its corresponding expression. That is:

`\displaystyle -b = 15,\, 2a = 22, \text{ and } b^2 -4ac = -215`

We can solve for b and a:

`\displaystyle b = -15 \text{ and } a = 11`

Now, we can solve for c:

`\displaystyle \begin{aligned} b^2 - 4ac &= -215 \\ \\ (-15)^2 - 4(11)c &= -215 \\ \\ (225) - 44c &= -215 \\ \\ -44c &= -440 \\ \\ c &= 10 \end{aligned}`

Hence, a = 11, b = -15, c = 10.

The quadratic formula is applied to quadratics in the form:

`\displaystyle ax^2 + bx + c =0`

Substitute. Hence, one possible equation is:

`\displaystyle 11x^2 - 15x + 10 = 0`

Note: There are infinitely many equations that will have the given solutions. The new equations will simply be the above equation multiplied by a constant.

Question 2)

We are given the equation:

`ax^2 + 6x + c= 0`

And we want to find two integer values for a and c such that the equation has two real solutions.

Recall that the number of solutions of a quadratic is given by its discriminant:

`\displaystyle \Delta = b^2 - 4ac`

The quadratic will have two real solutions for positive discriminants. In other words:

`b^2 - 4ac > 0`

We know that b = 6. Substitute and simplify:

`\displaystyle \begin{aligned}b^2 - 4ac & >0 \\ \\ (6)^2 - 4ac & > 0 \\ \\ 36 - 4ac &> 0 \\ \\ -4ac &> -36 \\ \\ ac &< 9 \end{aligned}`

So, the product of a and c must be less than 9.

From the given answer choices, only Choice C is correct.

Therefore, a = 2 and b = 3.

Then our equation is:

`2x^2 + 6x + 3=0`