If two equations x^{2}+ax+b = 0 and x^{2}+cx+d = 0, have a common root, then what is the value of that root?
An algebraic equation having the highest degree of two is defined as a quadratic equation. The word “quadratic” comes from the Latin word “quadrature”, which means “square”.
The standard form of a quadratic equation in x is given as follows:
ax^{2}+bx+c = 0
Where a ≠ 0, a, b, and c ∈ R
In the equation given above, x is the variable, a and b are the coefficients of x^{2} and x, and c is a constant. The values of the unknown variable x that satisfy the given quadratic equation are called solutions of the quadratic equation, where these solutions have termed the roots (or) zeros of the quadratic equation. Since the highest degree of the quadratic equation is two, it has two solutions.
The Quadratic formula for finding the roots of a quadratic equation, ax^{2}+bx+c = 0, is
x = [–b ± √(b^{2} – 4ac)]/2a, a ≠ 0
 The plus or minus sign in the formula represents that there will be two solutions for x.
Here, b^{2} – 4ac is called discriminant (D) which helps to predict the nature of the roots.

Discriminant value (D) D = b^{2} – 4ac 
Nature of Roots 

Case 1 
D = 0 
Two real and equal roots 
Case 2 
D > 0 
Two real and unequal roots 
Case 3 
D < 0 
Imaginary roots 
Sum and Product of Roots of Quadratic Equation
If α and β be the roots of the quadratic equation ax^{2}+bx+c = 0, then
 The sum of roots of the quadratic equation (α + β) = –b/a = – coefficient of x/ coefficient of x^{2}
 Product of roots of the quadratic equation (αβ) = c/a = constant/coefficient of x^{2}
 The formula for the quadratic equation whose roots are α and β is given as follows:
x^{2} – (α + β)x + αβ = 0
If two equations x^{2}+ax+b =0 and x^{2}+cx+d =0, have a common root, then what is the value of that root?
Solution:
Given equations:
x^{2}+ax+b =0 and x^{2}+cx+d =0
Let “α” be the common root of both equations.
Now, substitute α in both equations.
As α is the root of the given quadratic equations, their values will be zero at α.
So,
α^{2} + aα + b = 0 ————— (1)
α^{2} + cα + d = 0 ————— (2)
By subtracting equation (1) from equation (2), we get
α^{2} + cα + d – (α^{2} + aα + b) = 0
α^{2} + cα + d – α^{2} – aα – b = 0
cα + d – aα – b = 0
(c – a)α + (d – b) = 0
(c – a)α = b – d
α = (b – d)/(c – a)
Hence, the common root is (b – d)/(c – a).
Solved Example on Quadratic equations
Example 1: Solve the following equations: a) 3x^{2} – 21x = 0 and b) 25x^{2} – 36 = 0.
Solution:
a) 3x^{2} – 21x = 0
Given Equation: 3x^{2} – 21x = 0
⇒ 3x(x – 7) = 0
⇒ 3x = 0 ⇒ x = 0
(or)
⇒ x – 7 = 0 ⇒ x = 7
Hence, the solutions to the given equation are: x = 0 and x = 7.
b) 25x^{2} – 36 = 0
Given Equation: 25x^{2} – 36 = 0
⇒ 25x^{2} = 36
⇒ x^{2} = 36/25
⇒ x = √(36/25) = ±6/5
⇒ x = 6/5 (or) –6/5
Hence, the solutions to the given equation are: x = 6/5 and x = –6/5.
Example 2: Solve: 5x^{2} + x – 4 = 0.
Solution:
Given equation: 5x^{2} – x – 4 = 0
By comparing the given equation with the standard equation,
we get a = 5, b = 1, c = –4
x = [–b ± √(b^{2}–4ac)]/2a
⇒ x = [–1 ± √((1)^{2} –4(5)(–4)]/2(5)
⇒ x = [–1 ± √(1–80)]/10
⇒ x = [–1 ± √81]/10
⇒ x = (–1 ± 9)10 = (–1–9)/10 (or) (–1+9)/10
⇒ x = (–10)/10 (or) x = 8/10
⇒ x = –1 (or) 4/5
Hence, the solutions to the given equation are –1 and 4/5.
Example 3: Find the sum and product of roots of the quadratic equation 9x^{2} – 19x + 10 = 0.
Solution:
Given equation: 9x^{2} – 19x + 10 = 0
By comparing the given equation with the standard equation ax^{2}+bx+c = 0,
we get have a = 9, b= –19 and c = 10
Now, substitute the values in the formulae of sum and product of roots.
Sum of roots = –b/a = –(–19/9) = 19/9
Product of roots = c/a = 10/9.
Example 4: Solve: 12m^{2} + 13m + 1 = 0.
Solution:
Given Equation: 12m^{2} + 13m + 1 = 0
⇒ 12m^{2} + 12m + m + 1 = 0
⇒ 12m(m + 1) + 1(m + 1) = 0
⇒ (12m + 1) (m + 1) = 0
⇒ 12m + 1 = 0 or m + 1 = 0
⇒ m = –1/12 or m = –1
Hence, the solutions of the given quadratic equation are m = –1/12 and m = –1.
Example 5: Find the nature of roots of the quadratic equation x^{2} + 3x + 5 = 0.
Solution:
Given equation: x^{2} + 3x + 5 = 0
Now compare the equation with the standard form ax^{2}+bx+ c =0
So, we have a = 1, b = 3, c = 5
Now, calculate the discriminant,
D = b^{2} – 4ac
= 3^{2} – 4(1)(5)
= 9 –20 = –11 < 0
As the discriminant (D) is less than zero, the equation has two imaginary solutions.
FAQs on Quadratic Equations
Question 1: What is a Quadratic equation and write its standard form.
Answer:
A quadratic equation is a seconddegree algebraic equation. The standard form of the quadratic equation is
ax^{2}+bx+c = 0, where a ≠ 0, a, b, and c ∈ R
In the equation given above, x is the variable, a and b are the coefficients of x^{2} and x, and c is a constant.
Question 2: How to determine the nature of the roots of a quadratic equation?
Answer:
The discriminant of a quadratic equation helps to predict the nature of its roots. The formula for the discriminant of a quadratic equation ax^{2}+bx+c=0 is b^{2} – 4ac. If D > 0, then the roots are real and distinct. If D = 0, then the roots are real and equal. If D < 0, then the roots are imaginary complex numbers.
Question 3: How Many Roots does Quadratic Equation Have?
Answer:
A quadratic equation is a seconddegree algebraic equation, hence, it has two roots. These roots can be obtained using the quadratic formula. One root can be obtained using the positive sign, while the other is obtained using the negative sign in the formula.
Question 4: How to apply the Quadratic Formula?
Answer:
The formula for finding the roots of a quadratic equation, ax^{2}+bx+c = 0, is x = [–b ± √(b^{2} – 4ac)]/2a, a ≠ 0. Here, the plus or minus sign in the formula represents that there will be two solutions for x. To find the roots of a quadratic equation substitute the values of a, b, and c in the quadratic formula. Do not forget to apply positive (+) and negative (–) signs separately.
Question 5: When will a Quadratic Equation have equal roots?
Answer:
A given quadratic equation will have equal roots if the value of the discriminant is equal to zero. For a quadratic equation of the form ax^{2}+bx+c=0, the formula to find the discriminant is D = (b^{2} – 4ac). So, if D = 0, the given equation will have equal roots, and the value of each root will be x = –b/2a.
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