Answer:
a>12.5.
Step-by-step explanation:
you need to look at the discriminant in the quadratic formula:
[tex]b^{2} -4ac[/tex]
coming from the quadratic polynomial
[tex]ax^{2} +bx+c=0[/tex]
In our case, b=10 and c=2, hence, the discriminant is
[tex]10^2-4*a*2 \\100-8a[/tex]
We will obtain non-real values if the discriminant is less than zero.
Therefore, this condition is written as
[tex]100-8a<0[/tex]
and we must solve for a. It yields
[tex]100<8a\\a>\frac{100}{8} \\a>12.5[/tex]
Hence, the answer is a>12.5.
The values of a that will result in this equation having non-real solutions must e greater than 12.5
Given the quadratic equation ax^2 + 10x + 2 = 0
For the equation to have non-real solutions, then b^2 - 4ac < 0
From the given equation;
a = a
b = 10
c = 2
Substitute the given parameters into the formula;
10^2 - 4a(2) < 0
100 - 8a < 0
-8a < -100
8a > 100
a > 12.5
Hence the values of a that will result in this equation having non-real solutions must e greater than 12.5
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