The value of the root is, (b – d)/(c – a), If two equations x2 + ax + b = 0 and x2 + cx + d = 0, have a common root. The solution for the same is added below:
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
- 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).
Similar Examples
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.
