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705-805 (Hard)|   Algebra|      
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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 05 Jun 2020, 08:55
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If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?

A. -7
B. −5/2
C. 5/2
D. 7
E. 21
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A
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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 05 Jun 2020, 09:06
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If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?

A. -7 --> correct: let's say p & q are the roots of the equation of \(x^2 + 28x – 588 = 0\) => (x-p)(x-q)=0 => x^2-(p+q)x+pq = 0, so p+q=-28, And let's say s & t are the roots of the equation of \(x^2 + bx + 98 = 0\) => so s*t = 98 and s = (p+q)/2=-28/2=-14, so -14t=98=>t = -7, so the other root of the equation \(x^2 + bx + 98 = 0\)is -7
B. −5/2
C. 5/2
D. 7
E. 21
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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 05 Jun 2020, 09:14
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If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?

A. -7
B. −5/2
C. 5/2
D. 7
E. 21
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Asked: If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?

Sum of roots of the equation (x^2 + 28x – 588 = 0 ) = -28
Average of the roots of the equation (x^2 + 28x – 588 = 0 ) = -28/2 = -14

One root of the equation (x^2 + bx + 98 = 0) = - 14
Let the other root of the equation (x^2 + bx + 98 = 0) be x
-14x = 98
x = - 7

IMO A
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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 05 Jun 2020, 09:21
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X^2+28x-588 = 0
D = b^2 - 4ac
D = 784-4*588
D = -1568 = √D = -28√2
Roots = -b±√D/2a
Root a = -28+28√2/2 = -14+14√2
Root b = -28-28√2/2 = -14-14√2
Average of root a and b = one of the root of equation = x^2+bx+98
(-14+14√2-14-14√2)/2
-28/2 = -14

-14 is one if the root of X^2+bx+98
Product of two roots = c/a = 98/1 = -14*x
-7=X
Answer is A

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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 05 Jun 2020, 11:05
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IMO A

Sum of roots for second equation = -\(\frac{b}{a}\) = -28
AM = -\(\frac{28}{2}\) = -14

One root = -14
Let second root = x

Product of roots of first eqn = \(\frac{c}{a}\) = 98
-14 * x = 98
x = -7
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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 05 Jun 2020, 11:23
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Bunuel
If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?

A. -7
B. −5/2
C. 5/2
D. 7
E. 21
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Note for any equation \(x^2 + bx + c = 0\), \(b\) is the negative sum of roots. \(c\) is the product of roots.

The sum of roots of \(x^2 + 28x – 588 = 0\) would be -28 then, so the average of the roots would be -14. Since one of the roots of \(x^2 + bx + 98 = 0\) is -14, we can find the other root by using \(-14*x = 98\). The other root would be x = -7.

Ans: A
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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 09 Jun 2020, 05:20
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Bunuel
If one of the roots of the quadratic equation \(x^2 + bx + 98 = 0\) is the average (arithmetic mean) of the roots of the equation \(x^2 + 28x – 588 = 0,\) what is the other root of the equation \(x^2 + bx + 98 = 0\)?

A. -7
B. −5/2
C. 5/2
D. 7
E. 21
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Solution:

Let f(x) = ax^2 + bx + c be a quadratic function (a ≠ 0). The mean of the two real roots of any quadratic equation (or function) is the x-value of the vertex of that function. Therefore, instead of solving the equation x^2 + 28x - 588 = 0, we can just determine the x-value of the vertex of f(x) = x^2 + 28x - 588. Recall that the formula for the x-value of the vertex of f(x) = ax^2 + bx + c is x = -b/(2a). Therefore, the x-value of the vertex of f(x) = x^2 + 28x - 588 is x = -28/(2*1) = -14.

Now we can say that x = -14 is one of the roots of x^2 + bx + 98 = 0. In other words, x + 14 must be a factor of x^2 + bx + 98. In that case, x + 7 must be the other factor since only 14 * 7 = 98. In other words, x^2 + bx + 98 = 0 must be factored as (x + 14)(x + 7) = 0, regardless of the value of b. Setting x + 7 = 0, x = -7 must be the other root of the equation.

Answer: A
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If one of the roots of the quadratic equation x^2 + bx + 98 = 0 is the 01 Dec 2024, 08:58
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