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

Question

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.

last

srcc25anu · Answer

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

Anju@Gurome · Answer

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.

Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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Anju@Gurome · Answer

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.

srcc25anu · Answer

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

Anju@Gurome · Answer

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.

Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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