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If q is one root of the equation x2 + 18x + 11c = 0, where

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If q is one root of the equation x2 + 18x + 11c = 0, where

by varun289 » Thu Apr 04, 2013 11:26 am
If q is one root of the equation x2 + 18x + 11c = 0, where -11 is the other root and c is a constant, then q2 - c2 =
A

98
B

72
C

49
D

0

Correct Answer
E

It cannot be determined from the information given.
Explanation:
No. If you have one root (or factor), you don't need all three terms of the quadratic to solve find the other. If -11 is one of the roots, the factor that yields that root is (x + 11), and the other factor must be (x + 7), so q = -7. If the factors of the quadratic are (x + 11) and (x + 7), then the last term of the quadratic is 77, and c = 7. Thus q2 - c2 = 0. The correct answer is D.




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by srcc25anu » Thu Apr 04, 2013 12:09 pm
Eqn: x^2 + 18x + 11c = 0 where -11 is one root and q is other root, c is constant

A quadratic equation is of the form X^2 + (a+b)*X + (a*b) = 0 where a and b are the two roots of the equation.

we know -11 is one root
x^2 + (q - 11)*x + q*(-11) = 0

q - 11 = 18 => q = 29 Thus the second root is 29
29 * (-11) = 11C
thus c = -29

what we need to find = q^2 - c^2
29^2 - (-29)^2 = 0
hence answer is D
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by Anju@Gurome » Thu Apr 04, 2013 9:13 pm
varun289 wrote:If q is one root of the equation x² + 18x + 11c = 0, where -11 is the other root and c is a constant, then q² - c² =
As -11 is a root of the equation x² + 18x + 11c = 0
--> (-11)² + 18*(-11) + 11c = 0
--> 11*11 - 11*18 + 11c = 0
--> 11 - 18 + c = 0
--> c = 7

Also, now product of the roots of the equation = q*(-11) should be equal to 11c = 77
So, q = -7

Hence, q² - c² = (-7)² - 7² = 0

The correct answer is D.
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by Anju@Gurome » Thu Apr 04, 2013 11:46 pm
srcc25anu wrote:A quadratic equation is of the form X^2 + (a+b)*X + (a*b) = 0 where a and b are the two roots of the equation.
That will be x² - (a + b)x + ab = 0

So, q - 11 = -18 ---> q = -7
And, (-7)*(-11) = 11c ---> c = 7

Hope that helps.
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by srcc25anu » Fri Apr 05, 2013 12:21 am
Anju@Gurome wrote:
srcc25anu wrote:A quadratic equation is of the form X^2 + (a+b)*X + (a*b) = 0 where a and b are the two roots of the equation.
That will be x² - (a + b)x + ab = 0

So, q - 11 = -18 ---> q = -7
And, (-7)*(-11) = 11c ---> c = 7

Hope that helps.
I think the base form of a quadratic equation is ax^2 + bx + c = 0
And if you want to write it as a function of its roots X and Y that will be ax^2 + (X+Y)x + (XY) =0
Or we can say that b = X+Y and C = XY
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by Anju@Gurome » Fri Apr 05, 2013 12:31 am
srcc25anu wrote:I think the base form of a quadratic equation is ax^2 + bx + c = 0
And if you want to write it as a function of its roots X and Y that will be ax^2 + (X+Y)x + (XY) =0
Or we can say that b = X+Y and C = XY
No.
It will be x² - (X + Y)x + XY = 0
Or, b/a = -(X + Y) and c/a = XY

Let's see why?
If X and Y are the roots of the equation ax² + bx + c = 0, then (x - X)(x - Y) = 0
Now, (x - X)(x - Y) = x² - (X + Y)x + XY ............ (1)

As the coefficient of x² in the equation is 1, let us divide the original equation by a to make the coefficient of x equal to 1.
So, ax² + bx + c = 0
--> x² + (b/a)x + c/a = 0 ............ (2)

Comparing the coefficients of (1) and (2)
--> (b/a) = -(X + Y) and c/a = XY

To sum up, in any quadratic equation where the coefficient of x² is 1,
  • The sum of the roots = Negative of the coefficient of x
    The product of the roots = The constant term
Hope that helps.
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