Quadratic Formula How To

Quadratic Formula How To, The quadratic formula is a powerful tool in algebra that allows you to solve quadratic equations of the form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0. This formula provides a systematic way to find the values of $x$ that satisfy the equation, regardless of the specific coefficients $a$, $b$, and $c$. In this article, we’ll explore the quadratic formula, its derivation, and how to apply it to solve quadratic equations.

What is the Quadratic Formula?

The quadratic formula states that the solutions to the equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 can be found using the following formula:

x=−b±b2−4ac2ax = \frac{{-b \pm \sqrt{{b^2 – 4ac}}}}{{2a}}x=2a−b±b2−4ac​​

In this formula:

• $a$ is the coefficient of x2x^2x2,
• $b$ is the coefficient of $x$,
• $c$ is the constant term,
• $\pm$ indicates that there are typically two solutions: one for addition and one for subtraction,
• b2−4acb^2 – 4acb2−4ac is called the discriminant.

Step-by-Step Guide to Using the Quadratic Formula

1. Identify the Coefficients

Start by identifying the coefficients $a$, $b$, and $c$ in your quadratic equation. For example, in the equation 2×2−4x−6=02x^2 – 4x – 6 = 02x2−4x−6=0:

• $a=2$
• $b=-4$
• $c=-6$

2. Calculate the Discriminant

The discriminant is crucial for determining the nature of the roots. Calculate it using the formula:

D=b2−4acD = b^2 – 4acD=b2−4ac

For our example:

D=(−4)2−4(2)(−6)=16+48=64D = (-4)^2 – 4(2)(-6) = 16 + 48 = 64D=(−4)2−4(2)(−6)=16+48=64

3. Evaluate the Discriminant

The value of the discriminant tells you about the solutions:

• If $D>0$: Two distinct real roots.
• If $D=0$: One real root (a repeated root).
• If $D<0$: No real roots (two complex roots).

In our example, since $D=64>0$, there will be two distinct real roots.

4. Plug the Values into the Quadratic Formula

Now, substitute $a$, $b$, and $D$ into the quadratic formula:

x=−b±D2ax = \frac{{-b \pm \sqrt{D}}}{{2a}}x=2a−b±D​​

For our example:

x=−(−4)±642(2)x = \frac{{-(-4) \pm \sqrt{64}}}{{2(2)}}x=2(2)−(−4)±64​​

5. Calculate the Roots

Now perform the calculations:

x=4±84x = \frac{{4 \pm 8}}{{4}}x=44±8​

This results in two calculations:

1. x1=4+84=124=3x_1 = \frac{{4 + 8}}{4} = \frac{12}{4} = 3x1​=44+8​=412​=3
2. x2=4−84=−44=−1x_2 = \frac{{4 – 8}}{4} = \frac{-4}{4} = -1x2​=44−8​=4−4​=−1

Thus, the solutions to the equation 2×2−4x−6=02x^2 – 4x – 6 = 02x2−4x−6=0 are $x=3$ and $x=-1$.

Example Problems

Example 1: Solve x2+5x+6=0x^2 + 5x + 6 = 0x2+5x+6=0

1. Identify coefficients: $a=1$, $b=5$, $c=6$.
2. Calculate the discriminant: D=52−4(1)(6)=25−24=1D = 5^2 – 4(1)(6) = 25 – 24 = 1D=52−4(1)(6)=25−24=1 (one real root).
3. Use the quadratic formula:

x=−5±12=−5±12x = \frac{{-5 \pm \sqrt{1}}}{2} = \frac{{-5 \pm 1}}{2}x=2−5±1​​=2−5±1​

Calculating gives:

• x1=−5+12=−2x_1 = \frac{{-5 + 1}}{2} = -2x1​=2−5+1​=−2
• x2=−5−12=−3x_2 = \frac{{-5 – 1}}{2} = -3x2​=2−5−1​=−3

So, the solutions are $x=-2$ and $x=-3$.

Example 2: Solve 3×2−2x+1=03x^2 – 2x + 1 = 03x2−2x+1=0

1. Coefficients: $a=3$, $b=-2$, $c=1$.
2. Discriminant: D=(−2)2−4(3)(1)=4−12=−8D = (-2)^2 – 4(3)(1) = 4 – 12 = -8D=(−2)2−4(3)(1)=4−12=−8 (no real roots).
3. Use the quadratic formula:

Since $D<0$, the roots will be complex:

x=2±−86=2±2i26=1±i23x = \frac{{2 \pm \sqrt{-8}}}{6} = \frac{{2 \pm 2i\sqrt{2}}}{6} = \frac{1 \pm i\sqrt{2}}{3}x=62±−8​​=62±2i2​​=31±i2​​

Thus, the solutions are x=1+i23x = \frac{1 + i\sqrt{2}}{3}x=31+i2​​ and x=1−i23x = \frac{1 – i\sqrt{2}}{3}x=31−i2​​.

Conclusion

The quadratic formula is an essential tool for solving quadratic equations. By following these steps, you can confidently tackle any quadratic equation and determine its solutions, whether they are real or complex. With practice, using the quadratic formula will become second nature, helping you excel in algebra and beyond.