Answer:
Question 1)
One possible equation is:
[tex]\displaystyle 11x^2 - 15x + 10 = 0[/tex]
Question 2)
Choice C
The equation is:
[tex]2x^2 + 6x + 3=0[/tex]
Step-by-step explanation:
Question 1)
Recall that the quadratic formula is given by:
[tex]\displaystyle x = \frac{-b\pm\sqrt{b^2 -4ac}}{2a}[/tex]
We want to find a quadratic with the solutions:
[tex]\displaystyle x = \frac{15\pm\sqrt{-215}}{22}[/tex]
Each value must be equal to its corresponding expression. That is:
[tex]\displaystyle -b = 15,\, 2a = 22, \text{ and } b^2 -4ac = -215[/tex]
We can solve for b and a:
[tex]\displaystyle b = -15 \text{ and } a = 11[/tex]
Now, we can solve for c:
[tex]\displaystyle \begin{aligned} b^2 - 4ac &= -215 \\ \\ (-15)^2 - 4(11)c &= -215 \\ \\ (225) - 44c &= -215 \\ \\ -44c &= -440 \\ \\ c &= 10 \end{aligned}[/tex]
Hence, a = 11, b = -15, c = 10.
The quadratic formula is applied to quadratics in the form:
[tex]\displaystyle ax^2 + bx + c =0[/tex]
Substitute. Hence, one possible equation is:
[tex]\displaystyle 11x^2 - 15x + 10 = 0[/tex]
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:
[tex]ax^2 + 6x + c= 0[/tex]
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:
[tex]\displaystyle \Delta = b^2 - 4ac[/tex]
The quadratic will have two real solutions for positive discriminants. In other words:
[tex]b^2 - 4ac > 0[/tex]
We know that b = 6. Substitute and simplify:
[tex]\displaystyle \begin{aligned}b^2 - 4ac & >0 \\ \\ (6)^2 - 4ac & > 0 \\ \\ 36 - 4ac &> 0 \\ \\ -4ac &> -36 \\ \\ ac &< 9 \end{aligned}[/tex]
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:
[tex]2x^2 + 6x + 3=0[/tex]
