SSC (Marathi Semi-English) 10thMaharashtra State Board
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# Formula for Solving a Quadratic Equation

#### notes

ax2 + bx + c, Divide the polynomial by a ( ∵a ≠ 0) to get x^2+b/ax+c/a.

Let us write the polynomial x^2+b/ax+c/a  in the form of difference of two square numbers. Now we can obtain roots or solutions of equation x^2+b/ax+c/a which is equivalent to ax^2 + bx + c = 0 .

ax^2 + bx + c = 0               .............(I)

x^2+b/ax+c/a=0               ..... dividing both sides by a

therefore x^2+b/ax+(b/(2a))^2-(b/(2a))^2+c/a=0

therefore (x+b/(2a))^2-b^2/(4a^2)+c/a=0

therefore(x+b/(2a))^2-(b^2-4ac)/(4a^2)=0       therefore(x+b/(2a))^2=(b^2-4ac)/(4a^2)

therefore(x+b/(2a))=sqrt((b^2-4ac)/(4a^2)) or (x+b/(2a))=-sqrt((b^2-4ac)/(4a^2))

therefore x=-b/(2a)+sqrt((b^2-4ac)/(4a^2)) or x=-b/(2a)-sqrt((b^2-4ac)/(4a^2))

therefore x=(-b+sqrt(b^2-4ac))/(2a) or x=(-b-sqrt(b^2-4ac))/(2a)

In short the solution is written as x=(-b+-sqrt(b^2-4ac))/(2a) and these values are denoted by alpha,beta.

therefore alpha=(-b+sqrt(b^2-4ac))/(2a) , beta=(-b-sqrt(b^2-4ac))/(2a) ............................(I)

The values of a, b, c from equation ax2 + bx + c = 0 are substituted in  (-b+-sqrt(b^2-4ac))/(2a) and further simplified to obtain the roots of the equation. So x=(-b+-sqrt(b^2-4ac))/(2a)  is the formula used to solve quadratic equation. Out of the two roots any one can be represented by α and the other by β.

That is, instead (I) we can write alpha=(-b+sqrt(b^2-4ac))/(2a) , beta=(-b-sqrt(b^2-4ac))/(2a) .......(II)

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