**Solving Quadratic Equations:**

Standard form of a quadratic equation is \(ax^2~+~bx~+~c~=~0\), where \(a\), \(b\) and \(c\) are constants and \(a~≠~0\). In algebra, there are generally two types of problems which come up:

I. Flexible Questions

II.Fixed Questions

The first type of problems, i.e. flexible problems are variable based. In such questions all the required information like roots, coefficients are provided in the form of variables like \(α,β,a,b\) etc. The solution to such problems could be easily found out using various methods such as substitution.

In fixed questions of quadratic equations, the basic types of problems that are encountered are as follows:

- Range of \(x\)
- Nature of roots of quadratic equation
- Roots and coefficients
- Finding roots of quadratic equation using the quadratic formula

i) In problems involving range of \(x\), graphs are used mostly for solving quadratic equations. We plot the graph of given quadratic equation, which is a parabola and range of \(x\) is evaluated using this graph. The domain is represented by all possible values on \(x\), which satisfy the given equation.

ii) In problems related to the nature of the roots of quadratic equation, discriminant(\(D\)) is found out and depending upon that, nature of roots is decided.

Quadratic equations in standard form are represented as, \(ax^2~+~bx~+~c~=~0\), where \(a~≠~0\) and \(a,b,c\) are constants. The solutions or the roots of the equation are given as:

\(x\) = \(\frac{-b~±~√{b^2~-~4ac}}{2a}\)

The term \(b^2~-~4ac\) is known as discriminant of the quadratic equation \(ax^2~+~bx~+~c~=~0\) and is denoted by \(D\) or \(Δ\).

Depending on value of \(D\), nature of roots are:

- \(D\gt 0\) , roots of the equation are real and distinct.
- \(D~=~0\), roots of the equation are real and equal.
- \(D\lt 0 \), roots are unreal and complex conjugates of each other.

iii) In problems involving roots and coefficients, the relationship between the coefficients of equation and its root is established and required variable is figured out using these relations.

For a function

if the roots are given by \(α_{n-1},α_{n-2}……α_0\) ,

then,

\(f(x)\) = \(α_n (x~-~α_{n-1})(x~-~α_{n-2})~………~(x~-~α_0)\)

\(\Rightarrow~a_n x^n~+~a_{n-1}~x^{n~-~1}~+~⋯..~+~a_1 ~x~+~a_0\) = \(α_n (x-α_{n-1})(x-α_{n-2})~………~(x~-~α_0)\)

On comparing coefficients of \(x^{n~-~1}\) ,we get:

On comparing coefficients of \(x^{n~-~2}\), we get:

On comparing coefficients of \(x^{n~-~3}\), we get:

\(S_2\) = \((α_{n~-~1})(α_{n~-~2})(α_{n~-~3})~+~(α_{n~-~2})(α_{n~-~3})(α_{n~-~4})~+~⋯~=~∑_(i≠j≠k)α_i α_j α_k~=~-\frac{a_{n~-~3}}{a_n}\)=\(~(-1)^3~\frac{coefficient~ of~x^{n~-~3}}{coefficient~of~x^n}\)

Similarly,

For quadratic equation of the form \(ax^2~+~bx~+~c~=~0\), having roots α and β,

\(α~+~β\) = \(-\frac{b}{a}\)

\(αβ\) = \(\frac{c}{a}\)

For cubic equation of the form

having roots \(α,β\) and \(γ\),

\(α~+~β~+~γ\) = \(-\frac{b}{a}\)

\(αβ~+~βγ~+~γα\) = \(\frac{c}{a}\)

\(αβγ\) = \(-\frac{d}{a}\)

Using these relations between roots and coefficients, problems can be solved easily.

iv) In problems which require to finding the roots of quadratic equation directly, quadratic formula is used directly to determine the roots.

If \(α\) and \(β\) are the roots or solutions of quadratic equation \(ax^2~+~bx~+~c~=~0\), then it can be expressed as:

\(ax^2~+~bx~+~c~=~(x~-~α)(x~-~β)~=~0\)

The solutions or the roots are given:

\(x\) = \(\frac{-b~±~√{b^2~-~4ac}}{2a}\)

\(x~=~-\frac{b}{2a}~±~\frac{√{b^2~-~4ac}}{2a}\)

Solve the model solutions to the chapter titled Quadratic Equations that are provided in detail through simple step-step solutions to all questions in an NCERT textbooks only at Byju’s.