In this article we cover quadratic equations – definitions, formats, solved problems and sample questions for practice.
A quadratic equation is a polynomial whose highest power is the square of a variable (x2, y2 etc.)
Definitions
A monomial is an algebraic expression with only one term in it.
Example: x3, 2x, y2, 3xyz etc.
A polynomial is an algebraic expression with more than one term in it.
Alternatively it can be stated as –
A polynomial is formed by adding/subtracting multiple monomials.
Example: x3+2y2+6x+10, 3x2+2x-1, 7y-2 etc.
A polynomial that contains two terms is called a binomial expression.
A polynomial that contains three terms is called a trinomial expression.
A standard quadratic equation looks 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) x2+5x+6 = 0 where a=1, b=5 and c=6.
2) x2+2x-3 = 0 where a=1, b=2 and c= -3
3) 3x2+2x = 1
→ 3x2+2x-1 = 0 where a=3, b=2 and c= -1
4) 9x2 = 4
→ 9x2-4 = 0 where a=9, b=0 and c= -4
For every quadratic equation, there can be one or more than one solution. These are called 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, consider 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 adding which we get 5 and multiplying which we get 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 called factoring.
We saw earlier that the sum of the roots is –b/a and the product of the roots is c/a. Let us verify that.
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 -2 and -3 when added give -5 and when multiplied give 6.
Solved examples of Quadratic equations
Let us solve some more examples using this method.
Problem 1: Solve for x: x2-3x-10 = 0
Solution:
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
Solution:
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, the coefficient of x2 was 1. Let us see how to solve the equations where the coefficient of x2 is greater than 1.
Problem 3: Solve for x: 3x2+2x =1
Solution:
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
Solution:
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. But 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.
For this kind of equations, 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.
Let us consider another example.
Solve for x: x2 = 24 – 10x
Solution:
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.
So this can be solved by the factoring method. But let’s solve it using the new method, 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.
Discriminant
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.
Solution:
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.
Hence verified.
Quadratic Equations Quiz: Solve the following
Problem 1: Click here
Answer 1: Click here
Problem 2: Click here
Answer 2: Click here
Problem 3: Click here
Answer 3: Click here
36(x)^2-24(2)^1/2 x+8 = 0 solve it
Very good question
Roots: [x – {2(2^1/2)}/3] [x – {2(2^1/2)}/3]
Credit: Pro. Po-Shen Loh
(1/2)^1/2
36x^2 – 24(√2)x + 8=0
x =24√2±[√(1152 – 4*36*8)]/2*36
=24√2±[√1152-1152]/72
=24√2±0/72
so, x =24√2/72
you can simplify ahead by dividing further
x=√2 / 3
Yo how on earth is 0/72 = 72. It is zero. What nonsense. So ans is 24√2
There should be a bracket [(24 √(2)±0)/72] and hence the answer is (√(2)/3).
do not say something is nonsense..even if you have second opinions about them
there are some students who learn from this site…please be mindful of your words
If p and q are the root of the equation ax2+bx+c=0,than the equation whose roots are p+1/q and q+1 ?
acx^2 + b(a+c)x + a^2 + b^2 +2ac=0
is it p + (1/q) or (p+1)/q?
It will be ,
X^2-{p+(1/q) +q+1}x+pq+P+(1/q) +1
Find the valu of x and y. Root x +y=7 and x+ root y =11 were x=2
u may use elimination method to solve for x any y
X2-10×+9=0
9, 1
2x-10x+9=0
Here, a=2,b= -10,c=9
X=[(-b)±√ (b² – 4ac)]/2a
= [-(-10)± √{(-10)² – 4×2×9}]/2×2
={10±√(100 – 72)}/4
={10± √28}/4
={10±2√7}/4
=(10+2√7)/4 or (10 – 2√7)/4
={2(5+√7)}/4 or {2(5 – √7)}/4
X=(5+√7)/2 or (5 – √7)/2
Solve it x2-20x+a2-b2=0
show that x²+2x=(2a+2b+1)(2a+2b-1) are the root of the equation are integers
If a+b is a root of square of X+ax+b equals 0 then what is the maximum value of square of b?
In 3/(x+2)(2x-1) = a/(2x+1) + b/(x+2), what is the value of a and b
The sum of the roots of a quadratic equation is 5/2 and the product of its root is 4. The quadratic equation is_______
Equation is 2x^2-5x+4=0 with roots (5+i7)/4 & (5-7i)/4
The sum of a quadratic equation is k+2 and the product if its roots is 6-k^2.the quadratic equation is ___________?
what is the quadratic equation
15-2x-x=0
Using a graph of the equation y=2x³+5x²-x-6 for -4≤x≥2. Use 2cm to represent 1 unit on the x-axis and 1cm to represent 5 units on the y-axis. Then solve the equations:
1) 2x³+5x²+x-4
2) 2x³ +5x²-x+2
Please tell solution of x^2-4ax+4a^2-9=0
find the limit between which k must lie in order that kx square -6x+4 over4x square -6x+x
may be capable of all values when x is real
p+1/q =0 q+1=0 q=-1 p+1/-1=0 then p=1 to find the equation we use (x-x1) (x-x2)=0 , (x-(-1)) (x-1)=0, (x+1) (x-1)=0 the quadratic equation x^2 -x+x -1=0, x^2-1=0
Thanks but I got a problem with this number
Find the possible values of k if the equation
2kx^2-8x+1=2k(x-2) has equal roots.
In the question find for balie of x,
X² + 10X – 24,
X = +2 or -12
If x = -2 or 12,
The equation would be X² – 10X – 24
Solve 2ײ-11×+14
It’s a good platform and I’ve learnt a lot. However, i request that you include question and solutions for questions like” show that the equation X2+(2-k)x+k=3 has equal roots for all k values?” I would really appreciate for the help. Thanks you God bless you!!!
Solve the quadratic equation by finding I.F
( x-x^2y)dy + (y+xy^2)dx=0