Question

prove that the quadratic equations x square +ax-4=0 has distinct real roots​

Answer 1

Given Question :-

• Prove that the quadratic equations x² +ax-4=0 has distinct real roots

Answer

Given :-

• A quadratic equation x² + ax - 4 = 0.

To Prove :-

• Equation x² + ax - 4 = 0 has distinct real roots.

Concept Used :-

Concept Used :- Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal or distinct real roots.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

CALCULATION :-

Given

• A quadratic equation x² + ax - 4 = 0

Now,

• On comparing the given quadratic equation with ax² + bx + c = 0,

we get

• a = 1

• b = a

• c = - 4

Now,

• Discriminant (D) of quadratic equation is evaluated as

$\rm :\longmapsto\:Discriminant \: (D) = {b}^{2} - 4ac$

$\rm :\longmapsto\:Discriminant \: (D) = {a}^{2} - 4 \times (1) \times ( - 4)$

$\rm :\implies\:Discriminant \: (D) = {a}^{2} + 16$

$\bf\implies \:Discriminant \: (D) > 0$

$\boxed{ \because \: \bf \: as \: a \in \: real \: number \: so \: {a}^{2} \geqslant 0\implies\: {a}^{2} + 16 > 0 }$

• Hence, roots of x² + ax - 4 = 0 are real and distinct.

${\boxed{\boxed{\bf{Hence, Proved}}}}$