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Find the product of all positive integer values of $c$ such that the quadratic equation $3x^2+7x+c=15x-10$ has two real roots.

Aug 28, 2023
edited by jlunn762  Aug 28, 2023

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There seem to be a lot of questions about families of quadratics and their respective number of roots!

I will write the given quadratic equation in standard form.

$$3x^2 + 7x + c = 15x - 10 \\ 3x^2 - 8x + (c + 10) = 0$$

In order for this family of quadratics to have 2 real roots, the discriminant of the quadratic must be greater than zero.

$$a = 3; b = -8; c = c + 10 \\ D = b^2 - 4ac \\ D = (-8)^2 - 4 * 3 * (c + 10) \\ D = 64 - 12(c + 10) \\ D = -12c -56$$

Now that I have simplified the discriminant as much as possible, I will determine what values of c make the discriminant greater than 0.

$$D > 0 \\ -12c - 56 > 0 \\ -12c > 56 \\ c < -\frac{14}{3}$$

There are no positive integers c such that the given quadratic has two real roots! Therefore, I do not think it makes sense to take the product!

Aug 30, 2023