GMAT Problems on Quadratic Equations
with Solutions and Explanations Sample 3
Solutions and detailed explanations to quadratic equations problems in sample 3 similar to the problems in the GMAT test.
Solution to Question 1
Write the given equation with right hand side equal to 0 and factor
x2 + 5x + 6 = 0
(x + 2)(x + 3)
Solve for x
x + 2 = 0 or x + 3 = 0
Solutions: x = - 2 and x = - 3
Solution to Question 2
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
Solution to Question 3
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
Solution to Question 4
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 = ± √20 = ± 2 √5
solutions: x = 2 + 2 √5 and x = 2 - 2 √5
Solution to Question 5
For the given equation to have a solution, the left hand side must be a square. Hence
x2 + k x + 4 = (x + 2)2
Expand the right hand side
x2 + k x + 4 = x2 + 4x + 4
Comparing the left hand side and the right hand sides, the value of k is 4
k = 4
Solution to Question 6
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)x2 = 16
Substitute m by 2 in the given equation and solve it.
4 x2 = 16
x2 = 4
x = ± 2
The second solution is
x = - 2
Solution to Question 7
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) x2 + 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
Solution to Question 8
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
x2 - x - 2 = 0
If we now compare the given equation x2 + bx + c = 0 with the obtained equation x2 - x - 2 = 0, we can identify constants b and c by their values as follows
b = -1 and c = -2.
Solution to Question 9
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 X2 + 2 = 11 x
5 X2 - 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
Solution to Question 10
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 x2 + 3 = 10x
Write the above equation with right hand term equal to 0, factor and solve.
3 x2 - 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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