Quadratic Formula

Introduction

The quadratic equation is a one-variable polynomial equation with degree two. The highest power of the unknown variable in a quadratic equation is two. The general form of a quadratic equation f(x) with variable x is f(x)=ax²+bx+c=0, in which a≠0 and a,b,c ϵ R. Every quadratic equation has two roots which can be real or imaginary. The discriminant of a quadratic equation determines the nature of the roots. The roots can be calculated using the quadratic formula.

Quadratic Equations

A Quadratic equation is a polynomial equation of one variable with a degree of two. The general form of a quadratic equation f(x)=ax²+bx+c=0, in which x is the unknown variable, a≠0, and a,b,c ϵ R. a is the leading coefficient of the quadratic equation, c is the absolute term of the quadratic equation. The values of unknown variables satisfying the quadratic equation are called the roots. The roots can be real or imaginary. There are various methods to solve a quadratic equation and find the roots of the quadratic equation.

Example: 3x²+5x+6=0, -x²+2x-1=0, etc… are a few quadratic equations.

Quadratic Formula

The roots of a quadratic equation can be found using the quadratic formula. Consider the quadratic equation f(x)=ax²+bx+c=0, in which x is the unknown variable, a≠0, and a,b,c ϵ R. Now, the roots of the quadratic equation using the quadratic formula are −

$\mathrm{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$, a,b,c are the values of coefficients of a quadratic equation ax²+bx+c=0.

Discriminant

• The discriminant (D) of a quadratic equation f(x)=ax^2+bx+c=0 is equal to b²-4ac. The value of the discriminant determines the nature of roots in a quadratic equation.

• If the value of D > 0, there exist two different real roots.

• If the value of D − 0 there exist two different imaginary roots.

• If the value of D = 0 there exists a single real root (both the roots are equal).

• If the discriminant value of a quadratic equation ax²+bx+c=0 is equal to zero that is b²=4ac, then the roots of the quadratic equation are $\mathrm{x=\frac{-b}{2a}, \frac{-b}{2a}}$

• If the discriminant value of a quadratic equation ax²+bx+c=0 is less than zero that is b²<4ac, then the roots of the quadratic equation are always a conjugate pair. If one root is p+iq, then the other root is p-iq.

Derivation of Quadratic Formula

Consider the quadratic equation f(x)=ax²+bx+c=0, in which x is the unknown variable, a≠0, and a,b,c ϵ R. Now to derive the roots the quadratic formula send c to the other side of the equation.

$$\mathrm{ax^2+bx=-c}$$

Divide both sides of the equation by $\mathrm{\frac{1}{a}}$

$$\mathrm{x^2+\frac{b}{a} x=-\frac{c}{a}}$$

Now, add $\mathrm{(\frac{b}{2a} )^2}$ to both sides of the equation to make a perfect square on the left-hand side.

$$\mathrm{x^2+\frac{b}{a} x+ (\frac{b}{2a} )^2=-\frac{c}{a}+ (\frac{b}{2a} )^2}$$

$$\mathrm{(x+\frac{b}{2a})^2=-\frac{c}{a}+ (\frac{b}{2a})^2}$$

$$\mathrm{(x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2}}$$

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

Now, taking the square root on both sides of the equation

$$\mathrm{x+\frac{b}{2a}=\sqrt{\frac{b^2-4ac}{4a^2}}}$$

$$\mathrm{x+\frac{b}{2a}=\frac{±\sqrt{b^2-4ac}}{2a}}$$

$$\mathrm{x=-\frac{b}{2a}±\frac{\sqrt{b^2-4ac}}{2a}}$$

$$\mathrm{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$$

The roots of the quadratic equation f(x)=ax²+bx+c=0 are $\mathrm{x=\frac{-b+\sqrt{b^2-4ac}}{2a},x=\frac{-b-\sqrt{b^2-4ac}}{2a}}$

Solved Examples

1) Solve the quadratic equation x²+3x+2=0 using the quadratic formula?

Given, that the quadratic equation x²+3x+2=0, in the equation the values of a,b,c are 1, 3, 2 respectively.

$\mathrm{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$, substitute the respective values of a,b,c in the formula to get the roots of the equation.

$$\mathrm{x=\frac{-3±\sqrt{3^2-8}}{2}=\frac{-3±\sqrt{1}}{2}=-2,-1}$$

-2,-1 are the roots of the quadratic equation x²+3x+2=0.

The roots are real and distinct because the discriminant value is greater than zero.

2) Solve the quadratic equation -x²+4x-5=0 using the quadratic formula?

Given, that the quadratic equation -x²+4x-5=0, in the equation the values of a,b,c are -1, 4, -5 respectively.

$\mathrm{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$, substitute the respective values of a,b,c in the formula to get the roots of the equation.

$$\mathrm{x=\frac{-4±\sqrt{4^2-20}}{-2}=\frac{-4±\sqrt{-4}}{-2}=\frac{-4±i2}{-2}}$$

2+i,2-i are the roots of the quadratic equation -x²+4x-5=0.

The roots are imaginary and a conjugate pair because the discriminant value is less than zero.

3)Solve the quadratic equation x²+2x+2=0 using the quadratic formula?

Given, that the quadratic equation x²+2x+2=0, in the equation the values of a,b,c are 1, 2, 2 respectively.

$\mathrm{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$, substitute the respective values of a,b,c in the formula to get the roots of the equation.

$$\mathrm{x=\frac{-2±\sqrt{2^2-8}}{2}=\frac{-2±\sqrt{-4}}{2}=\frac{-2±i2}{2}}$$

-1+i,-1-i are the roots of the quadratic equation x²+2x+4=0.

The roots are imaginary and a conjugate pair because the discriminant value is less than zero.

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4)Solve the quadratic equation x²+5x+4=0 using the quadratic formula?

Given, that the quadratic equation x²+5x+4=0, in the equation the values of a,b,c are 1, 5, 4 respectively.

$\mathrm{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$, substitute the respective values of a,b,c in the formula to get the roots of the equation.

$$\mathrm{x=\frac{-5±\sqrt{5^2-16}}{2}=\frac{-5±\sqrt{9}}{2}=-1,-4}$$

-1,-4 are the roots of the quadratic equation x²+5x+4=0.

The roots are distinct and real because the discriminant value is greater than zero.

Conclusion

In this tutorial, we learned about quadratic equations, quadratic formula, discriminant, how the discriminant determines the nature of the roots, derivation of quadratic formula, and a few solved examples.

FAQs

1.How to determine the nature of roots in a quadratic equation ax²+bx+c=0?

The discriminant of a quadratic equation is D =b²-4ac determines the nature of the roots of the quadratic equation whether they are real or imaginary or equal.

2.What if the discriminant of a quadratic equation is greater than zero?

Then the roots are real and distinct.

3.What is the value of the discriminant in a quadratic equation ax²+bx+c=0?

The value of the discriminant D =b²-4ac.

4.What is the quadratic formula for the quadratic equation ax²+bx+c=0?

$\mathrm{x=\frac{-b±\sqrt{b^2-4ac}}{2a}}$, substitute the respective values of a,b,c in the formula. Where if the part under the square root i.e., b²-4ac is positive i.e., < 0 then the equation has distinct real root. Whereas if the same part under the square root i.e., b²-4ac is 0 then the equation has equal real roots, and if it is negative then the equation no real roots but distinct complex roots.

5.Determine if the roots of the quadratic equation 2x²+3x+5=0 are real or imaginary?

To determine the nature of the roots, calculate the value of the discriminant of the quadratic equation. The values of a,b,c of the quadratic equation are 2, 3, 5 respectively.

D =b²-4ac=3²-40=-31.

D < 0, therefore the roots are imaginary.

6.If one root of a quadratic equation is 2-5i, what is the value of the other root?

If the discriminant value of a quadratic equation ax²+bx+c=0 is less than zero that is b²<4ac, then the roots of the quadratic equation are always a conjugate pair. If one root is p+iq, then the other root is p-iq therefore the other root is equal to 2+5i.

7.Determine if the roots of the quadratic equation x²-2x+1=0 are real or imaginary?

To determine the nature of the roots, calculate the value of the discriminant of the quadratic equation. The values of a,b,c of the quadratic equation are 1, -2, 1 respectively.

D =b²-4ac=(-2)²-4=0.

D = 0, therefore the roots are equal and real.