Solutions and detailed explanations to quadratic equations problems in sample 3 similar to the problems in the GMAT test.

Solve the equation x^{2} + 5x =  6 for x.
Solution
Write the given equation with right hand side equal to 0 and factor
x^{2} + 5x + 6 = 0
(x + 2)(x + 3)
Solve for x
x + 2 = 0 or x + 3 = 0
Solutions: x =  2 and x =  3

Find the values of the constants k and m so that the equation 2 x^{2} + kx  m = 0 has solutions at x = 1 and x = 2.
Solution
Since x = 1 and x = 2 are solutions to the equation, they must satisfy the equation
2(1)^{2} + k(1)  m = 0 and 2(2)^{2} + k(2)  m = 0
Simplify and rewrite as a system of two equations in two unknowns k and m.
k  m =  2 and 2k  m = 8
Subtract the left and right hand terms of the two eqautions to obtain an equivalent equation.
(k  m)  (2k  m) = (2)  (8)
Simplify.
3k = 6
Solve for k.
k = 2 and
Use equation k  m =  2 to find m.
m = k + 2 = 4
The values of k and m are
k = 2 and m = 4

Solve the equation (x  1)(x + 3) = (1  x) for x.
Solution
Rewrite equation with the right hand side equal to 0.
(x  1)(x + 3)  (1  x) = 0
Rewrite as
(x  1)(x + 3) + (x  1) = 0
Factor and solve
(x  1) [ (x + 3) + 1 ] = 0
(x  1)(x + 4) = 0
x  1 = 0 or x + 4 = 0
Solutions: x = 1 and x = 4

Solve the equation 2  (x  2)^{2} =  18 for x.
Solution
Subtract 2 from both sides of the equation.
 (x  2)^{2} =  20
Multiply both sides of the equation by 1.
(x  2)^{2} = 20
Solve by extracting the square root.
x  2 = ~+mn~ √20 = ~+mn~ 2 √5
solutions: x = 2 + 2 √5 and x = 2  2 √5

Find a positive value of the constant k so that the equation x^{2} + k x + 4 = 0 has only one solution.
Solution
For the given equation to have a solution, the left hand side must be a square. Hence
x^{2} + k x + 4 = (x + 2)^{2}
Expand the right hand side
x^{2} + k x + 4 = x^{2} + 4x + 4
Comparing the left hand side and the right hand sides, the value of k is 4
k = 4

If x = 2 is a solution to the equation (m + 2)x^{2} = 16, where m is a constant, then what is the second solution of this equation?
Solution
x = 2 is a solution and therefore satisfies the equation. Hence
(m + 2)(2)^{2} = 16
Solve for m
4(m + 2) = 16
m + 2 = 4
m = 2
Substitute m by 2 in the given equation and solve it.
(2 + 2)x^{2} = 16
Substitute m by 2 in the given equation and solve it.
4 x^{2} = 16
x^{2} = 4
x = ~+mn~ 2
The second solution is
x =  2

If x =  2 is a solution to the equation (1/2) a x^{2} + 3x  4 = 0, where a is a constatnt, then what is the second solution of this equation?
Solution
x =  2 is a solution and therefore satisfies the given equation. Hence
(1/2) a (2) ^{2} + 3x  4 = 0
Simplify and solve for a
(1/2) a (2) ^{2} + 3(2)  4 = 0
2 a  6  4 = 0
a = 10 / 2 = 5
We now substitute a by 5 in the given equation and then factor knowing that x + 2 is a factor since x =  2 is a solution.
(5/2) x^{2} + 3x  4 = 0
(x + 2)( (5/2)x  2 ) = 0
The second solution is found by setting the second factor equal to 0 and solve
(5/2)x  2 = 0
x = 4 / 5

Find the values of the constants b and c so the equation x^{2} + bx + c = 0 has solutions at x = 2 and x = 1.
Solution
Since the equation has solutions at x = 2 and x =  1, it can be written in factored form as follows
(x  2)(x + 1) = 0
Expand the left hand side
x^{2}  x  2 = 0
If we now compare the given equation x^{2} + bx + c = 0 with the obtained equation x^{2}  x  2 = 0, we can identify constants b and c by their values as follows
b = 1 and c = 2.

X and Y are two real numbers such that X + Y = 11 / 5, X · Y = 2 / 5 and X is greater than 1. Find X and Y.
Solution
The second equation X · Y = 2 / 5 may written as follows
Y = 2 / (5 X)
Substitute Y by 2 / (5 X) in the equation X + Y = 11 / 5
X + 2 / (5 X) = 11 / 5
Multiply all terms of the above equation and simplify
5 X^{2} + 2 = 11 x
5 X^{2}  11 X + 2 = 0
Factor left hand side and solve
(5X  1)(X  2) =
5X  1 = 0 , X = 1 / 5
X  2 = 0 , X = 2
We now calculate the value of Y using the equation X + Y = 11 / 5
For X = 1 / 5, Y = 11 / 5  X = 11 / 5  1 / 5 = 10 / 5 = 2
For X = 2, Y = 11 / 5  X = 11 / 5  2 = 11 / 5  10 / 5 = 1 / 5
The two numbers are X = 2 and Y = 1 / 5

The sum of a number x, such that x is greater than 1, and its reciprocal is equal to 10 / 3. Find x.
Solution
The reciprocal of x is 1 / x. The sum of x and 1 / x is equal to 10 / 3 is translated mathematically as follows.
x + 1 / x = 10 / 3
Multiply all terms in the equation by 3 x and simplify.
3 x^{2} + 3 = 10x
Write the above equation with right hand term equal to 0, factor and solve.
3 x^{2}  10 x + 3 = 0
(3x  1)(x  3) = 0
Two solutions to the equation
x = 1 / 3 and x = 3
We are looking for the solution that is greater than 1. Hence the solution to the problem is
x = 3
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