‘Quadratic Equations’ are very useful in solving questions from the algebra and arithmetic part in Quantitative Aptitude. Quadratic Equations are having degree two with one variable. These can be represented as $$ay^{2}+by+c = 0$$ where a, b and c are the real numbers and ‘a’ is not equal to zero. Generally, we can solve any quadratic equation by either formula or with factors. In factors, the product of ‘a’ and ‘c’ in the equation is equal to the product of the factors ‘b’. For example, here in equation $$y^{2}-5y+6 = 0$$, 3 and 2 are the factors. The formula is also available for solving this which is applicable in any quadratic equation mentioned below.
For a quadratic equation $$ay^{2}+by+c = 0$$, to obtain the value of ‘y’ following formula can be used.
$$y = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$
Here D = $$b^{2}-4ac$$
On behalf of the value of D, there are three possible conditions which are given below.
Condition(i) D=0, then both the value of ‘y’ will be the same.
Condition(ii) D>0, then two distinct values of ‘y’ will be there.
Condition(iii) D<0, then there will be no real value for ‘y’
As we know that the two real roots of a quadratic equation are called $$\alpha$$ and $$\beta$$.
$$\alpha+\beta = -\frac{b}{a}$$
$$\alpha \beta = \frac{c}{a}$$
Q) Find out the values of M, for the equation $$M^{2} - 14M + 48 = 0$$.
Sol. $$M^{2} - 14M + 48 = 0$$
$$M^{2} - (8+6)M + 48 = 0$$
$$M^{2} - 8M - 6M + 48 = 0$$
$$M(M-8)-6(M-8) = 0$$
(M-8) (M-6) = 0
M = 6, 8
Q) For the equation $$5M^{2} + 31M + 48 = 0$$, find out the value of $$\alpha \beta$$.
Sol. $$\alpha \beta$$ = $$\frac{c}{a}$$
= $$\frac{48}{5}$$
= 9.6
