We will discuss that a quadratic equation has only two roots or in other words we can say that a quadratic equation cannot have more than two roots.
We will prove this one-by-one.
A quadratic equation has only two roots.
Proof:
Let us, consider the quadratic equation of the general form
ax2 + bx + c = 0, (a ≠ 0) ............... (i)
Now divide each term by a (since, a ≠ 0), we get
x2 + bax + ca = 0
⇒ x2 + 2 * x * b2a + (b2a)2 – (b2a)2 + ca = 0
⇒ (x + b2a)2 - b2−4ac4a2 = 0
⇒ (x + b2a)2 – (√b2−4ac2a)2 = 0
⇒ (x + b2a + √b2−4ac2a)(x + b2a - √b2−4ac2a) = 0
⇒ [x - (−b−√b2−4ac2a)][x - (−b+√b2−4ac2a)] = 0
⇒ (x - α)(x - β) = 0, where α = −b−√b2−4ac2a and β = −b+√b2−4ac2a
Now we can clearly see that the equation ax2 + bx + c = 0 reduces to (x - α)(x - β) = 0 and the equation ax2 + bx + c = 0 is only satisfied by the values x = α and x = β.
Except α and β no other values of x satisfies the equation ax2 + bx + c = 0.
Hence, we can say that the equation ax2 + bx + c = 0 has two and only two roots.
Therefore, a quadratic equation has two and only two roots.
Solved example on quadratic equation:
Solve the quadratic equation x2 - 4x + 13 = 0
Solution:
The given quadratic equation is x2 - 4x + 13 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get
a = 1, b = -4 and c = 13
Therefore, x = −b±√b2−4ac2a
⇒ x = −(−4)±√(−4)2−4(1)(13)2(1)
⇒ x = 4±√16−522
⇒ x = 4±√−362
⇒ x = 4±6i2, [Since i = √-1]
⇒ x = 2 ± 3i
Hence, the given quadratic equation has two and only two roots.
The roots are 2 + 3i and 2 - 3i.
11 and 12 Grade Math
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