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IMO E

Eq I. \(x^2\) - 6x + 55 = 0

On computing D = \(b^2\) - 4ac, we get D = -184; hence no real roots

Eq II. 2\(x^2 \)- 4x - 70 = 0 (Divide eqn by 2)
On computing D = \(b^2\) - 4ac = 144 ==>. Real roots
On solving, two roots are 7 & -5

Eq III. x(x + 7) = 44
On computing D = \(b^2\) - 4ac = 225 ==>. Real roots
On solving, two roots are -11 & 4

No Equation has any roots common
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Bunuel
I. x^2 - 6x + 55 = 0
II. 2x^2 - 4x - 70 = 0
III. x(x + 7) = 44

Given the three quadratic equations above, which pair of equations has at least one common root?

A. I and II
B. II and III
C. I and III
D. I, II and III
E. None of the above

Solution:

Let’s solve the equations.

I. x^2 - 6x + 55 = 0
(x - 11)(x + 5) = 0
x = 11 or x = -5

II. 2x^2 - 4x - 70 = 0
x^2 - 2x - 35 = 0
(x - 7)(x + 5) = 0
x = 7 or x = -5

III. x(x + 7) = 44
x^2 + 7x - 44 = 0
(x + 11)(x - 4) = 0
x = -11 or x = 4

We see that I and II have a common root, namely, -5.

Answer: A
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I made a similar mistake here, but for statement one does not factor into (x - 11)(x + 5) = 0 because that gives you x^2 - 6x - 55 = 0 when the original equation is x^2 - 6x + 55 = 0, you can use the quadratic formula to find the roots but I just guessed there was no real roots and then marked it as answer E None of the above after solving the other statements. Is there a more sure proof way to approach this?
ScottTargetTestPrep


Solution:

Let’s solve the equations.

I. x^2 - 6x + 55 = 0
(x - 11)(x + 5) = 0
x = 11 or x = -5

II. 2x^2 - 4x - 70 = 0
x^2 - 2x - 35 = 0
(x - 7)(x + 5) = 0
x = 7 or x = -5

III. x(x + 7) = 44
x^2 + 7x - 44 = 0
(x + 11)(x - 4) = 0
x = -11 or x = 4

We see that I and II have a common root, namely, -5.

Answer: A
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