Question

Prove that any quadratic equation has maximum two roots

Answer 1

This is an impossible situation because α and γ are both distinct roots of the quadratic equation and a≠0. Therefore, there is a maximum of 2 solutions. Show that Quadratic Equations cannot have more than 2 Roots.

Answer 2

$\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}$

To prove:

$\sf\green{\implies x = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

Proof:

We will start with the the standard form of a quadratic equation,

$\bold\green{\implies ax^{2}+bx+c=0}$

Now, divide both sides of the equation by 'a' so you can complete the square,

$\sf{\implies x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=\dfrac{0}{a}}$

$\sf{\implies x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$

Now, Subtract c/a from both sides.

$\sf{\implies x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}-\dfrac{c}{a}=0-\dfrac{c}{a}}$

$\tt{\implies x^{2}+\dfrac{b}{a}x=-\dfrac{c}{a}}$

Now, by completing the square method.

The coefficient of the second term is b/a . Divide this coefficient by 2 and square the result to get (b/2a)², add (b/2a)² to both sides:

$\sf{\implies x^{2}+\dfrac{b}{a}x+\Bigg(\dfrac{b}{2a}\Bigg)^{2}=-\dfrac{c}{a}+\Bigg(\dfrac{b}{2a}\Bigg)^{2}}$

Since the left side of the equation right above is a perfect square, you can factor the left side by using the coefficient of the first term (x) and the base of the last term(b/2a).

Then, square the right side to get (b²)/(4a²).

$\sf{\implies \Bigg(x+\dfrac{b}{2a}\Bigg)^{2}=-\dfrac{c}{a}+\dfrac{b^{2}}{4a^{2}}}$

Get the same denominator on the right side:

$\sf{\implies \Bigg(x+\dfrac{b}{2a}\Bigg)^{2} =-\dfrac{4ac}{4a^{2}}+\dfrac{b^{2}}{4a^{2}}}$

$\sf{\implies \Bigg(x+\dfrac{b}{2a}\Bigg)^{2} =\dfrac{b^{2}-4ac}{4a^{2}}}$

Now, take the square root of each side,

$\sf{\implies \sqrt{\bigg(x+\dfrac{b}{2a}\bigg)^{2}}=\sqrt{\dfrac{b^{2}-4ac}{4a^{2}}}}$

Now, Simplify the left side,

$\sf{\implies x+\dfrac{b}{2a}=\pm\sqrt{\dfrac{b^{2}-4ac}{4a^{2}}}}$

$\sf{\implies x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^{2}-4ac}}{\sqrt{4a^{2}}}}$

Rewrite the right sides,

$\sf{\implies x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^{2}-4ac}}{2a}}$

Now, Subtract b/2a from both sides,

$\sf{\implies x+\dfrac{b}{2a}-\dfrac{b}{2a}=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^{2}-4ac}}{2a}}$

Adding the numerator and keeping the same denominator, we get the quadratic formula,

$\sf{\implies x = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

_________________(HENCE PROVED)