 As we have established in the previous sections, complex numbers are an addition to the real number system. Does the addition of complex numbers to mathematics make any difference? How does it help in the solution of quadratic equations by using the Quadratic formula? Let us start the quest of finding the answers to these questions here!

### Suggested Videos For You        Concept of Complex Numbers Modulus and it's Properties Complex Numbers Problem and its Solution In earlier classes, you read about the quadratic equations. We say that any equation that has the form ax2 + bx + c = 0, or an equation that we can reduce to this form is a quadratic equation. The equation has two solutions which may be identical or different. The most effective way to solve a quadratic equation is to use the quadratic formula.

### Browse more Topics under Complex Numbers And Quadratic Equations

A quadratic equation is an equation of degree 2. How many solutions must it have? How many solutions does any equation have? The answer is provided by the Fundamental Theorem of Algebra:

“A polynomial equation has at least one solution”

Any equation of the form a1x(n-1) + a2x(n-2) + a3x(n-3) + … + an = 0 is a polynomial equation of degree n. The fundamental theorem of algebra, when applied to polynomial equations of this form, provide an important result, which is:

“A polynomial equation of degree n, has n roots”

So a quadratic equation, which is a polynomial equation of order 2, must have 2 solutions. Let us see further. The general form of the quadratic equation can be written as ax2 + bx + c  = 0. Solving it for x, we get the following two solutions:

x = $$\frac{-b ± {(b^2 – 4ac)^{1/2}} }{2a}$$

This is known as the quadratic formula and gives two values for ‘x’. One for the + sign and the other for the – sign. The quantity in the square root is called the discriminant D. We have three conditions:

• If D > 0, the quantity in the square root is positive and thus real. In this case, there are two different and real solutions.
• In case D = 0, the ± sign no longer plays a part and we have a single, real solution, repeated twice.
• However, if D < 0, the quantity in the square root is negative and the solutions become complex or imaginary numbers.

Let’s discuss the third case. Now that we have a firm idea of complex numbers, we can use them to solve all the quadratic equations with D < 0. The three cases of the solution of a quadratic equation in the Argand plane

### The Complex Case

The discriminant of the solution of a quadratic equation is given as = $$\sqrt[]{(b^2 – 4ac)}$$. The solution will have a non-zero imaginary part if 4ac > b2 . Hence we can write $$\sqrt[]{(-(4ac – b^2) )}$$ = $$\sqrt[]{( 4ac – b^2 )}$$ . i

The square root $$\sqrt[]{( 4ac – b^2 )}$$ is now positive and real. And thus  we get two solutions from the Quadratic formula again: x = $$\frac{-b ± {i. ( 4ac – b^2)^{1/2}} }{2a}$$

Now let us use this result to solve quadratic equations with complex roots.

## Solved Examples For You

Question 1: Solve x2 + x + 1 = 0.

Answer : Here, D = $$\sqrt[]{(b^2 – 4ac )}$$ =  $$\sqrt[]{-3}$$

Hence the equation has complex roots given by:

x = $$\frac{-b ± {i. ( 4ac – b^2)^{1/2}} }{2a}$$ = $$\frac{-1 ± {i. ( 3 )^{1/2}} }{2}$$

Question 2: Find the value of k for the following quadratic equation, so that it has two real and equal roots. x22(k+1)x+k2=0

A) 1/2                              B) 2                                  C) -1/2                                 D) -2

Answer : C) The discrimminant has to  be = 0. Therefore, {-2(k + 1)}2 – 4 (1)(k2) = 0

Thus, 4(k2 + 1 + 2k) – 4k2 = 0

4 + 8k = 0 or k = -1/2

Question 3: How should one solve the quadratic equation by factoring?

• Put all terms on one side of the equal sign such that zero is left on the other side.
• Factor.
• Factor equal must be set to zero.
• Solve each equation.
• One can check by inserting answer in the original equation.

Question 4: In maths, what is meant by quadratic equations?

Answer: In maths, a quadratic equation is an equation that one can rearrange in the standard form. Here, x is representative of an unknown. Moreover, a, b, and c are representative of the known numbers.

Question 5: How does a problem become quadratic?

Answer: A quadratic refers to a type of problem that deals with a variable multiplied by itself. Furthermore, such an operation is known as squaring.

Question 6: Why the quadratic equation is known by the name quadratic?

Answer: Quadratum is the Latin word for square. Furthermore, the area of a square of side length x is such that it is given by x2, a polynomial equation consisting of an exponent. This is known as a quadratic equation or the square-like equation.

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