What will happen if you try to use completing the square to solve an equation of the form x^2 + bx + c = 0 in which c > (b/2) ^2?

A. There will be infinite solutions. After you have completed the square, both sides of the equation are negative.

B. There will be two solutions. After you have completed the square, the right side of the equation will be negative.

C. There will be one solution. After you have completed the square, neither side of the equation is negative.

D. There will be no real solutions. After you have completed the square, the right side of the equation will be negative.

Question

Angelica Salas

Mathematics

22 September, 20:16

What will happen if you try to use completing the square to solve an equation of the form x^2 + bx + c = 0 in which c > (b/2) ^2?

A. There will be infinite solutions. After you have completed the square, both sides of the equation are negative.

B. There will be two solutions. After you have completed the square, the right side of the equation will be negative.

C. There will be one solution. After you have completed the square, neither side of the equation is negative.

D. There will be no real solutions. After you have completed the square, the right side of the equation will be negative.

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Answers (1)

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Brody Jordan · Answer 1

D. There will be no real solutions. After you have completed the square, the right side of the equation will be negative.

Step-by-step explanation:

x² + bx + c = 0

x² + bx = - c

[x²+2 (x) (b/2) + (b/2) ² - (b/2) ²] = - c

[x + (b/2) ]² - (b/2) ² = - c

[x + (b/2) ]² = (b/2) ² - c

Since c > (b/2) ²

(b/2) ² - c < 0

So no real roots because a perfect square can not be negative in real numbers