**Solving Quadratic Equations:**

Standard form of a quadratic equation is \(ax^2~+~bx~+~c~=~0\)

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 \(Î±,Î²,a,b\)

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 \(x\)

ii) In problems related to the nature of the roots of quadratic equation, discriminant(\(D\)

Quadratic equations in standard form are represented as, \(ax^2~+~bx~+~c~=~0\)

\(x\)

The term \(b^2~-~4ac\)

Depending on value of \(D\)

- \(D\gtÂ 0\)
, 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 \(Î±_{n-1},Î±_{n-2}â€¦â€¦Î±_0\)

then,

\(\Rightarrow~a_n x^n~+~a_{n-1}~x^{n~-~1}~+~â‹¯..~+~a_1 ~x~+~a_0\)

On comparing coefficients of \(x^{n~-~1}\)

On comparing coefficients of \(x^{n~-~2}\)

On comparing coefficients of \(x^{n~-~3}\)

=\(~(-1)^3~\frac{coefficient~ of~x^{n~-~3}}{coefficient~of~x^n}\)

Similarly,

For quadratic equation of the form \(ax^2~+~bx~+~c~=~0\)

\(Î±~+~Î²\)

\(Î±Î²\)

For cubic equation of the form

having roots \(Î±,Î²\)

\(Î±~+~Î²~+~Î³\)

\(Î±Î²~+~Î²Î³~+~Î³Î±\)

\(Î±Î²Î³\)

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 \(Î±\)

\(ax^2~+~bx~+~c~=~(x~-~Î±)(x~-~Î²)~=~0\)

The solutions or the roots are given:

\(x\)

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

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.