GMAT / GRE Mathematics Tutorial Series: Quadratic Equation
In this tutorial, we cover quadratic equations – the definitions, formats, solved problems, and sample practice questions.
A quadratic equation is a polynomial equation of degree 2 or an equation in which the highest power is the square of a variable (x2, y2, etc.)
A monomial is an algebraic expression or a polynomial that consists of only a single term.
Example: x3, 2x, y2, 3xyz etc.
A polynomial is an algebraic expression or a polynomial that consists of more than one term.
In other words,
A polynomial is formed by the addition/subtraction of multiple monomials.
Example: x3+6y2+7x+16, 2x2+7x-1, 9y-5 etc.
A polynomial consisting of two terms is called a binomial expression.
A polynomial consisting of three terms is called a trinomial expression.
A standard GMAT quadratic equation tutorials formula look like this:
ax2+bx+c = 0
Where a, b, c are numbers and a≥1.
a, b are called the coefficients of x2 and x respectively and c is called the constant.
The following are examples of some quadratic equations:
1) 5x2+x+6 = 0 where a=5, b=1 and c=6.
2) x2+11x-3 = 0 where a=1, b=11 and c= -3
3) 5x2+2x = 1
→ 5x2+2x-1 = 0 where a=5, b=2 and c= -1
4) 11x2 = 4
→ 11x2-4 = 0 where a=11, b=0 and c= -4
For every quadratic equation, there can be one or more than one solution. These solutions are referred to as the roots of the quadratic equation.
For a quadratic equation ax2+bx+c = 0,
the sum of its roots = –b/a and the product of its roots = c/a.
A quadratic equation may be expressed as a product of two binomials.
For example, in the following equation,
x2-(a+b)x+ab = 0
x2-ax-bx+ab = 0
x(x-a)-b(x-a) = 0
(x-a)(x-b) = 0
x-a = 0 or x-b = 0
x = a or x=b
Here, a and b are called the roots of the given quadratic equation.
Now, let’s calculate the roots of an equation x2+5x+6 = 0.
We have to take two numbers, which on the addition will add up to 5 and upon multiplication will give 6. They are 2 and 3.
Let us express the middle term as an addition of 2x and 3x.
→ x2+2x+3x+6 = 0
→ x(x+2)+3(x+2) = 0
→ (x+2)(x+3) = 0
→ x+2 = 0 or x+3 = 0
→ x = -2 or x = -3
This method is known as factoring.
We know the sum of the roots to be –b/a and the product of the roots to be c/a. Let’s verify the same in the above equation.
Sum of the roots for the equation x2+5x+6 = 0 is -5 and the product of the roots is 6.
The roots of this equation are -2 and -3 which upon addition gives -5 and upon multiplication gives 6.
Also Check GMAT / GRE Tutorial Series Quadrilaterals
Solved examples of Quadratic equations
Here are a few solved examples to understand this method.
Problem 1: Solve for x: x2-3x-10 = 0
Let us express -3x as a sum of -5x and +2x.
→ x2-5x+2x-10 = 0
→ x(x-5)+2(x-5) = 0
→ (x-5)(x+2) = 0
→ x-5 = 0 or x+2 = 0
→ x = 5 or x = -2
Problem 2: Solve for x: x2-18x+45 = 0
The numbers which add up to -18 and give +45 when multiplied are -15 and -3.
Rewriting the equation,
→ x2-15x-3x+45 = 0
→ x(x-15)-3(x-15) = 0
→ (x-15) (x-3) = 0
→ x-15 = 0 or x-3 = 0
→ x = 15 or x = 3
Till now we solved examples in which the coefficient of x2 was 1. We’ll now move on to equations where the coefficient of x2 is greater than 1.
Problem 3: Solve for x: 3x2+2x =1
Rewriting our equation, we get 3x2+2x-1= 0
Here, the coefficient of x2 is 3. In these cases, we multiply the constant c with the coefficient of x2. Therefore, the product of the numbers we choose should be equal to -3 (-1*3).
Expressing 2x as a sum of +3x and –x
→ 3x2+3x-x-1 = 0
→ 3x(x+1)-1(x+1) = 0
→ (3x-1)(x+1) = 0
→ 3x-1 = 0 or x+1 = 0
→ x = 1/3 or x = -1
Problem 4: Solve for x: 11x2+18x+7 = 0
In this case, the sum of the numbers we choose should equal to 18 and the product of the numbers should equal 11*7 = 77.
This can be done by expressing 18x as the sum of 11x and 7x.
→ 11x2+11x+7x+7 = 0
→ 11x(x+1) +7(x+1) = 0
→ (x+1)(11x+7) = 0
→ x+1 = 0 or 11x+7 = 0
→ x = -1 or x = -7/11.
The factoring method is an easy way of finding the roots. However, this method can be applied only to equations that can be factored.
For example, consider the equation x2+2x-6=0.
If we take +3 and -2, multiplying them gives -6 but adding them doesn’t give +2. Hence this quadratic equation cannot be factored.
Therefore we apply the quadratic formula to find the roots.
The quadratic formula to find the roots,
x = [-b ± √(b2-4ac)] / 2a
Now, let us find the roots of the equation above.
x2+2x-6 = 0
Here, a = 1, b=2 and c= -6.
Substituting these values in the formula,
x = [-2 ± √(4 – (4*1*-6))] / 2*1
→ x = [-2 ± √(4+24)] / 2
→ x = [-2 ± √28] / 2
When we get a non-perfect square in a square root, we usually try to express it as a product of two numbers in which one is a perfect square. This is for simplification purpose. Here 28 can be expressed as a product of 4 and 7.
→ x = [-2 ± √(4*7)] / 2
→ x = [-2 ± 2√7] / 2
→ x = 2[ -1 ± √7] / 2
→ x = -1 ± √7
Hence, √7-1 and -√7-1 are the roots of this equation.
Take a look at another example.
Problem: Solve for x: x2 = 24 – 10x
Rewriting the equation into the standard quadratic form,
x2 +10x-24 = 0
What are the two numbers which when added give +10 and when multiplied give -24? 12 and -2.
This equation can be solved by using the factoring method. But let’s solve it by applying the quadratic formula.
Here, a = 1, b = 10 and c = -24.
x = [-10 ± √(100 – 4*1*-24)] / 2*1
x = [-10 ± √(100-(-96))] / 2
x = [-10 ± √196] / 2
x = [-10 ± 14] / 2
x = 2 or x= -12 are the roots.
For an equation ax2+bx+c = 0, b2-4ac is called the discriminant and helps in determining the nature of the roots of a quadratic equation.
If b2-4ac > 0, the roots are real and distinct.
If b2-4ac = 0, the roots are real and equal.
If b2-4ac < 0, the roots are not real (they are complex).
Consider the following example:
Problem: Find the nature of roots for the equation x2+x+12 = 0.
b2-4ac = -47 for this equation. So it has complex roots. Let us verify this.
→ [ -1±√(1-48)] / 2(1)
→ [-1±√-47] / 2
√-47 is usually written as i √47 indicating it’s an imaginary number.
Quadratic Equations Quiz: Solve the following
1) Solve for x: x2-15x+56 = 0
- x = 14 or x = 4
B. x = 8 or x = 7
C. x = 28 or x = 2
D. All of the above
Only 8 and 7 satisfy the conditions of adding up to 15 and giving a product of 56.
2) Find x if 2x2+7x+4 = 0
A. -7 ± √17 / 4
B. -7 ± √7 / 4
C. [-7 ± √17] / 4
D. [-7 ± √17] / 2
Applying the quadratic formula and substituting a=2, b=7 and c=4, we get the answer as C.
3) For what value of k does the equation x2-12x+k = 0 have real and equal roots?
b2-4ac = 0 for the equation to have real and equal roots.
144-4k = 0 → k = 36
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