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How Can We Solve Quadratic Equations Using Different Methods?

How Can We Solve Quadratic Equations in Different Ways?

Quadratic equations are a big part of algebra. They usually look like this: ax2+bx+c=0ax^2 + bx + c = 0. In this formula, aa, bb, and cc are numbers, and aa cannot be zero. There are different ways to solve these equations, and each method has its own perks. Let's look at some of the most common methods.

1. Factoring

Factoring means rewriting the quadratic equation into two simpler parts called binomials. This method works best when the equation is easy to factor.

Example: Solve x25x+6=0x^2 - 5x + 6 = 0.

First, we need to find two numbers that multiply to 66 (the last number) and add up to 5-5 (the number in front of xx). The numbers we need are 2-2 and 3-3.

So, we can write the equation like this:

(x2)(x3)=0(x - 2)(x - 3) = 0

Now, we set each part equal to zero:

  1. x2=0x - 2 = 0 leads to x=2x = 2

  2. x3=0x - 3 = 0 leads to x=3x = 3

So, the answers are x=2x = 2 and x=3x = 3.

2. Completing the Square

Completing the square is a helpful method where we rewrite the equation so it looks like (xp)2=q(x - p)^2 = q.

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

First, we move the constant (the 55) to the other side:

x2+6x=5x^2 + 6x = -5

Next, to complete the square, we take half of the number in front of xx (which is 66), square it (which gives us 99), and add it to both sides:

x2+6x+9=4x^2 + 6x + 9 = 4

Now, we rewrite the left side:

(x+3)2=4(x + 3)^2 = 4

We take the square root of both sides:

x+3=±2x + 3 = \pm 2

So, we solve to get:

x=1andx=5x = -1 \quad \text{and} \quad x = -5

3. Using the Quadratic Formula

If factoring is tricky or doesn’t work, we can always use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula can be used for any quadratic equation.

Example: Solve 2x2+4x6=02x^2 + 4x - 6 = 0.

Here, a=2a = 2, b=4b = 4, and c=6c = -6. Let’s use the formula:

  1. First, calculate the discriminant:

    b24ac=424(2)(6)=16+48=64b^2 - 4ac = 4^2 - 4(2)(-6) = 16 + 48 = 64

  2. Now, apply the quadratic formula:

    x=4±642(2)x = \frac{-4 \pm \sqrt{64}}{2(2)} x=4±84x = \frac{-4 \pm 8}{4}

This gives us the solutions:

x=1andx=3x = 1 \quad \text{and} \quad x = -3

Conclusion

Each method has its benefits. Factoring is fast for simpler equations, completing the square helps us understand curves better, and the quadratic formula works for any quadratic equation. Choose the method you find easiest!

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How Can We Solve Quadratic Equations Using Different Methods?

How Can We Solve Quadratic Equations in Different Ways?

Quadratic equations are a big part of algebra. They usually look like this: ax2+bx+c=0ax^2 + bx + c = 0. In this formula, aa, bb, and cc are numbers, and aa cannot be zero. There are different ways to solve these equations, and each method has its own perks. Let's look at some of the most common methods.

1. Factoring

Factoring means rewriting the quadratic equation into two simpler parts called binomials. This method works best when the equation is easy to factor.

Example: Solve x25x+6=0x^2 - 5x + 6 = 0.

First, we need to find two numbers that multiply to 66 (the last number) and add up to 5-5 (the number in front of xx). The numbers we need are 2-2 and 3-3.

So, we can write the equation like this:

(x2)(x3)=0(x - 2)(x - 3) = 0

Now, we set each part equal to zero:

  1. x2=0x - 2 = 0 leads to x=2x = 2

  2. x3=0x - 3 = 0 leads to x=3x = 3

So, the answers are x=2x = 2 and x=3x = 3.

2. Completing the Square

Completing the square is a helpful method where we rewrite the equation so it looks like (xp)2=q(x - p)^2 = q.

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

First, we move the constant (the 55) to the other side:

x2+6x=5x^2 + 6x = -5

Next, to complete the square, we take half of the number in front of xx (which is 66), square it (which gives us 99), and add it to both sides:

x2+6x+9=4x^2 + 6x + 9 = 4

Now, we rewrite the left side:

(x+3)2=4(x + 3)^2 = 4

We take the square root of both sides:

x+3=±2x + 3 = \pm 2

So, we solve to get:

x=1andx=5x = -1 \quad \text{and} \quad x = -5

3. Using the Quadratic Formula

If factoring is tricky or doesn’t work, we can always use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula can be used for any quadratic equation.

Example: Solve 2x2+4x6=02x^2 + 4x - 6 = 0.

Here, a=2a = 2, b=4b = 4, and c=6c = -6. Let’s use the formula:

  1. First, calculate the discriminant:

    b24ac=424(2)(6)=16+48=64b^2 - 4ac = 4^2 - 4(2)(-6) = 16 + 48 = 64

  2. Now, apply the quadratic formula:

    x=4±642(2)x = \frac{-4 \pm \sqrt{64}}{2(2)} x=4±84x = \frac{-4 \pm 8}{4}

This gives us the solutions:

x=1andx=3x = 1 \quad \text{and} \quad x = -3

Conclusion

Each method has its benefits. Factoring is fast for simpler equations, completing the square helps us understand curves better, and the quadratic formula works for any quadratic equation. Choose the method you find easiest!

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