If two equations x2+ax+b = 0 and x2+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:

ax²+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 x² 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, ax²+bx+c = 0, is

x = [–b ± √(b² – 4ac)]/2a, a ≠ 0

• The plus or minus sign in the formula represents that there will be two solutions for x.

Here, b² – 4ac is called discriminant (D) which helps to predict the nature of the roots.

	Discriminant value (D)   D = b2 – 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 ax²+bx+c = 0, then

• The sum of roots of the quadratic equation (α + β) = –b/a = – coefficient of x/ coefficient of x²
• Product of roots of the quadratic equation (αβ) = c/a = constant/coefficient of x²
• The formula for the quadratic equation whose roots are α and β is given as follows:

x² – (α + β)x + αβ = 0

If two equations x²+ax+b =0 and x²+cx+d =0, have a common root, then what is the value of that root?

Solution:

Given equations: 

x²+ax+b =0 and x²+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,

α² + aα + b = 0       ————— (1)

α² + cα + d = 0       ————— (2)

By subtracting equation (1) from equation (2), we get

α² + cα + d – (α² + aα + b) = 0

α² + cα + d – α² – 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) 3x² – 21x = 0 and b) 25x² – 36 = 0.

Solution:

a) 3x² – 21x = 0

Given Equation: 3x² – 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) 25x² – 36 = 0

Given Equation: 25x² – 36 = 0

⇒ 25x² = 36

⇒ x² = 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: 5x² + x – 4 = 0.

Solution:

Given equation: 5x² – x – 4 = 0

By comparing the given equation with the standard equation,

we get a = 5, b = 1, c = –4

x = [–b ± √(b²–4ac)]/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 9x² – 19x + 10 = 0.

Solution:

Given equation: 9x² – 19x + 10 = 0

By comparing the given equation with the standard equation ax²+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: 12m² + 13m + 1 = 0.

Solution:

Given Equation: 12m² + 13m + 1 = 0

⇒ 12m² + 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 x² + 3x + 5 = 0.

Solution:

Given equation: x² + 3x + 5 = 0

Now compare the equation with the standard form ax²+bx+ c =0

So, we have a = 1, b = 3, c = 5

Now, calculate the discriminant,

D = b² – 4ac

= 3² – 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 second-degree algebraic equation. The standard form of the quadratic equation is 

ax²+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 x² 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 ax²+bx+c=0 is b² – 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 second-degree 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, ax²+bx+c = 0, is x = [–b ± √(b² – 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 ax²+bx+c=0, the formula to find the discriminant is D = (b² – 4ac). So, if D = 0, the given equation will have equal roots, and the value of each root will be x = –b/2a.