Question

find the roots of ax²+bx+c=0​

Answer 1

Given: The quadratic equation $ax^2+bx+c=0$.

Reuired answer: The solutions of the quadratic equation.

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Things to Know

• Complete the Square

→ It is used because the highest degree is 2. After completing the square, we take the square root to find the solutions.

• Square Root

→ The number to find when the square of the number is given.

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Solution

$ax^2+bx+c=0$

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

After completing the square,

$\implies a(x^2+\dfrac{b}{a} x+\dfrac{b^2}{4a^2} )-\dfrac{b^2}{4a} +c=0$

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

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

This is equivalent to,

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

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

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

Hence, we derived the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}$.

Answer 2

Given :-

• Quadratic equation is –
• $\sf\boxed{\sf {{ ax²  + bx + c = 0}}}$

Need To Find Out :-

• Roots of the quadratic polynomial

Solution :-

We are given :-

• $\sf\boxed{\sf{{ ax²  + bx + c = 0}}}$

Were,

• a = Coefficient of x².
• b = Coefficient of x.
• c = Constant term.

Formula to find out roots:-

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

• Discriminant = b² - 4ac.The discriminant tells us about the nature of roots.

• If Discriminant is greater than 0, roots are real and distinct.

• If the Discriminant is equal to 0, roots are equal and real.

• If the Discriminant is less than 0, no real roots exists.

Now, derivation of Quadratic Formula:-

$\sf :\implies ax^{2}+bx+c=0$

Multiplying both sides by 4a :-

$\sf:\implies 4a^{2}x^{2}+4abx+4ac=0$

Adding b² to both sides :-

$\sf:\implies 4a^{2}x^{2}+4abx+b^{2}+4ac=b^{2}$

$\sf:\implies (2ax)^{2}+2\times (2ax)\times (b)+b^{2}=b^{2}-4ac$

$\sf:\implies (2ax + b)^{2}=b^{2}-4ac$

$\sf:\implies 2ax + b=\pm\sqrt{b^{2}-4ac}$

$\sf:\implies 2ax =-b\pm\sqrt{b^{2}-4ac}$

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

• Hence,the roots are :-

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

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