GRE Quadratic Equation: Methods to Solve & Practice Samples

In the GRE, Quadratic equation questions test your understanding and ability to solve, apply, and analyze a quadratic equation. They fall under the GRE Algebra, under the quant section. A quadratic equation has the highest exponent of 2 and is a polynomial equation. A quadratic equation is represented in the general form ax²+bx+c=0, Where a,b, and c are constants, and a is not equal to ‘0’. And x is a variable. Usually, a quadratic equation is formed as a parabola when represented in a graph. 

In the below sections, we will explain the GRE Quadratic Equation concepts and a few practice problems with solutions and explanations. 

What are the Methods to Solve GRE Quadratic Equations?

GRE Quadratic Equations can appear in various forms such as standard form, factored form, vertex, intercept, and expanded form. These questions appear in the GRE Quantitative Reasoning section. The quadratic equations can be solved using three main methods namely Factorization, Completing the Square Method, and Quadratic Equation Formula methods. Let’s see these methods in detail with the examples below.

Method 1: Factorization Method

You’ll have to follow a few strategies to solve a quadratic equation in the factorization method. Before solving the equation, express the equation as a product of two linear factors. Then set the equation and each factor to zero. Setting the equation to zero is called the “Zero Product Property”.

For eg:
Solve the below equation using the factorization method.
x² +5x = 6

First represent this equation to a standardized form, which is 
x²+5x-6 = 0

Step 2:
Multiply to a.c = 1. (-6) = -6
Add up to b = 5

The two numbers are 6 and -1.

Step 3:
Upon expanding the equation to x²+6x−x−6=0
And grouping, we get

(x²+6x)−(x+6)=0

Step 4:

Then we have to factorize each group, we get
x(x+6)−1(x+6)=0, which is further reduced to (x+6)(x−1)=0

Step 5: 
When setting each factor to zero, we get
x+6=0 or x−1=0
𝑥=−6 or 𝑥=1

Hence the answer is x = -6 and 1.

Method 2: Completing the Square Method

The completing the square method is the second technique for solving quadratic equations. This method involves changing a quadratic equation into a perfect square trinomial, then by taking square roots. The formula is ax² + bx + c ⇒ a(x + m)² + n, where m=b/2a and n=c -(b² /4a)

To explain this method, let's see an example.

Solve the below quadratic equation.
x² +6x+5=0

First we should isolate the quadratic and linear term, which is x²+6x=−5

As a next step, take the half the coefficient of x(6), square it and add to both the sides, which is 
(6/2)² which is equal to 9.
Then substituting this on both the sides,
x²+6x+9=−5+9
(x+3)² = 4

Taking the square root on both the sides
X+3 = ±√4 
X+3 = ±2

While solving for x, substituting ±2 in the equation, we get
X = -3+2 = -1 or X = -3-2 = -5

So, X = -1 and X = -5

Method 3: Quadratic Equation Formula

Quadratic formula is the most universal technique to solve the quadratic equation. The solution for a quadratic equation is called roots. The roots of the quadratic equation can be found using the formula 

X = −b±√b2 -4ac / 2a

Here b² -4ac is called the discriminant represented by Δ. 

The equation will not have any solution if the b² <4ac 
The equation has one solution if b² =4ac 
The equation has 2 solution if the b² >4ac 

This can be explained with an example

Solve the quadratic equation. 2x² +3x−2=0
By identifying the coefficients a = 2, b=3 and c=-3
With this coefficients, we can substitute in the formula 
X = −b±√b² -4ac / 2a
X = -3±√3² -4(2)(-2) / 2(2)

The next step is to simplify the discriminant 

Δ = b² -4ac = 3²-4(2)(-2) = 9+16 = 25

So, x = -3±√25/4

The next step is to solve for x, which is 
x=-3+5 /4 or x = -3-5/4
x= 2/4 = 1/2 or x = -8/4 = -2

Hence, the solutions are
X = ½ or x = -2

GRE Quadratic Equations Practice Problems

Problem 1
1. Find the value of X.

x² +7x+10=0

Steps:
Since the equation is already in standard form, expand the equation.
x²+5x+2x+10=0

On grouping and factorizing we get,

(x²+5x) + (2x+10)=0
Which can be represented as 
x(x+5) + 2(x+5), which can be simplified as 

(x+5)(x+2)=0

When solving for x, we get
Answer = x=-2 and x=-5

Problem 2:
Solve x for the equation 2x²−3x−5=0

Steps:
Since the equation is already in standard form, expand the equation.
2x²-5x+2x-5=0

On grouping and factorizing we get,

(2x²-5x) + (2x-5)=0
Which can be represented as 
x(2x-5) + 1(2x-5), which can be simplified as 

(2x-5)(x+1)=0

When solving for x, we get
Answer = x=1 and x=-5/2

Problem 3:
Solve for x in the following equation x²+6x+5=0.
Steps:
Move the constant to the other side.
x²+6x= -5

Next step is to complete the square
(6/2)² = 3² = 9
x²+6x+9 = -5+9
(x+3)² = 4

Solve by taking the square root
X+3 = ±√4
X+3 = ±2
x= -3+2 = -1 or x = -3-2 = -5
Hence, Answer = x=−1 and x=−5

Problem 4: Solve 2x2+8x−6=0

Steps:
Since already in standard form divide through by 2 to make the coefficient of x² to 1
x²+4x−3=0
Move the constant to the other side,
x²+4x=3
While completing the square we get
(4/2)² = 2² = 4
x²+4x+4 = 3+4
(x+2)² = 7

X+2 = ±√7

Hence, Answer x= -2+√7 or x=-2-√7

Problem 5
Solve for x² -4x-5 = 0

Steps:
Solving using for quadratic formula, first we need to identify the coefficients
The coefficients for a=1,b=−4,c=−5.

While substituting the coefficients to the above formula

X = -(-4)±√(-4)² - 4(1)(-5)/2(1)
X = 4±√16+20 /2
X = 4±√36 /2 
X = 10/2 = 5 or x = 4-6/2 = -2 / 2 =-1

Hence, Answer= x=5 and x=-1

Problem 6:
Solve for X 2x² −3x−2=0

Steps:
Solving by the quadratic formula, we need to find the coefficients
The coefficients for a=2,b=−3,c=−2

X = -(-3)±√(-3)² - 4(2)(-2)/2(2)
X = 3±√9+16/4
X = 3±√25/4
X = 3±5/4
X = 8/4 = 2 or x = -2/4 = -½

Hence, Answer = x=2 and x=-½

Top 5 tips to solve the GRE Quadratic Equations Effectively

1. As a first step, convert any equation to a standardized equation. ax² +bx+c =0. Then simplify the equation or eliminate common factors to work on the equation easily.
2. By seeing the given question, immediately identify which type of method (Factorization, Completing the square or quadratic formula) to be used, unless the method is stated explicitly.
3. To quickly identify the nature of roots, simplify the discriminant. As stated in the discriminant rule, you can identify whether it has 2 real roots, or 1 real root or no root. This can save time when solving the equation.
4. You can identify the common patterns in the given equation. It can be a perfect square of trinomials like x²+2ab+b² = (x+b)²  or difference of squares a² -b² = (a+b)(a-b).
5. The questions can be asked either as a word problem or given as a graphical representation. You should be able to understand the tricky words given and represented in the simple equation. 

As we have come to the end of the article, by now you might have understood about the GRE Quadratic Equation concepts and a few tips to solve it effectively in the GRE main exam. If you wish to know more about the GRE concepts and join the best coaching for GRE, then book a free demo with us!. Our representatives will get in touch with you.