Question

The quadratic equation ax^2+bx+c=0 has two real roots if. (a) b^2=4ac (b) b^2>4ac (c) b^2<4ac (d) b^2=ac
If done correctly with some explanation I will mark you as brainliest... MCQ......

Answer 1

AnsWer :

( a ) b² = 4ac.

Solution :

We have, Quadratic Equation.

★ General Equation.

$\tt \dagger \: \: \: \: \: a {x}^{2} + bx + c =0 .$

$\tt \dagger \: \: \: \: \: \red{a \neq0 . }$

Condition Given.

• Has both roots are Equal.
• Real roots.

So,

• When discriminate become 0, then our both roots are real and equal.

$\tt \dagger \: \: \: \: \: D = {b}^{2} - 4ac.$

$\tt : \implies 0 = {b}^{2} - 4ac.$

$\tt : \implies {b}^{2} = 4ac.$

When, Putting the value of D in Quadratic Formula, We get.

$\tt \dagger \: \: \: \: \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\tt : \implies x = \dfrac{ - b \pm \sqrt{ D} }{2a}$

$\tt : \implies x = \dfrac{ - b \pm \sqrt{ 0} }{2a}$

$\tt : \implies x = \dfrac{ - b }{2a}$

Therefore, correct option is ( a )