How do you find the nature of the roots using the discriminant given $x^{2} - 7 x + 12 = 0$ $x^2 - 7x + 12 = 0$ ?

Question

10363 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-25T17:21:45

Roots are rational. These are $x = 3$ $x=3$ or $x = 4$ $x=4$ .

If the equation is $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , nature of roots is decided by discriminant $b^{2} - 4 a c$ $b^2-4ac$

If $a$ $a$ , $b$ $b$ and $c$ $c$ are rational and $b^{2} - 4 a c$ $b^2-4ac$ is square of a rational number, roots are rational.

If $b^{2} - 4 a c > 0$ $b^2-4ac>0$ but is not a square of a rational number, roots are real but not rational.

If $b^{2} - 4 a c > 0 - 0$ $b^2-4ac>0-0$ we have equal roots.

If $b^{2} - 4 a c < 0$ $b^2-4ac<0$ roots are complex

In $x^{2} - 7 x + 12 = 0$ $x^2-7x+12=0$ , discriminant is ${(- 7)}^{2} - 4 \times 1 \times 12 = 49 - 48 = 1 = 1^{2}$ $(-7)^2-4xx1xx12=49-48=1=1^2$

hence roots are rational. In fact

$x^{2} - 7 x + 12 = 0$ $x^2-7x+12=0$

$\Leftrightarrow x^{2} - 4 x - 3 x + 12 = 0$ $hArrx^2-4x-3x+12=0$

or $x (x - 4) - 3 (x - 4) = 0$ $x(x-4)-3(x-4)=0$

or $(x - 3) (x - 4) = 0$ $(x-3)(x-4)=0$ i.e. $x = 3$ $x=3$ or $x = 4$ $x=4$

How do you find the nature of the roots using the discriminant given x^2 - 7x + 12 = 0x2−7x+12=0x^2 - 7x + 12 = 0?