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# help hard algebra

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Consider the system of quadratic equations y =3x^2 - 6x, y = 2x^2 + x - c, where c is a real number.

(a) For what value(s) of c will the system have exactly one solution (x,y)?

(b) For what value(s) of c will the system have more than one real solution?

(c) For what value(s) of c will the system have no real solutions? Solutions to the quadratics are (x,y) pairs. Your answers will be in terms of c, but make sure you address both x and y for each part.

Jun 6, 2022

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Because both equations are equal to y, we can set them equal to each other: \(3x^2 - 6x = 2x^2 + x - c\)

Bringing everything to the right-hand side gives us the quadratic: \(x^2 - 7x + c = 0\)

For the quadratic to have exactly 1 solution, the discriminant (\(b^2 - 4ac\)) must equal 0. This means \(49 - 4c = 0\)

For the quadratic to have more than 1 solution, the discriminant must be greater than 0. This means: \(49 - 4c >0\)

For the quadratic to have no real solutions, the discriminant must be less than 0. This means:\(49 - 4c <0\)

Can you take it from here?

Jun 6, 2022