Rewrite The Equation In Vertex Form. { (t-\square \nabla)^2-\square=0$} M U L T I P L Y T H E E Q U A T I O N B Y − 1 A N D T H E N B Y 16. Multiply The Equation By -1 And Then By 16. M U Lt I Pl Y T H Ee Q U A T I O Nb Y − 1 An D T H E Nb Y 16. { -16(t-\square)^2+\square\$}

Introduction

In mathematics, the vertex form of a quadratic equation is a powerful tool for analyzing and solving quadratic equations. The vertex form is a way of expressing a quadratic equation in the form of $(x-h)^2+k=0$, where $(h,k)$ is the vertex of the parabola. In this article, we will explore how to rewrite the equation in vertex form and then multiply it by -1 and 16.

Understanding the Vertex Form

The vertex form of a quadratic equation is given by:

$$y = a(x-h)^2 + k$$

where $(h,k)$ is the vertex of the parabola. The vertex form is useful because it allows us to easily identify the vertex of the parabola, which is the point where the parabola changes direction.

Rewriting the Equation in Vertex Form

To rewrite the equation in vertex form, we need to complete the square. The given equation is:

$$(t-\square \nabla)^2-\square=0$$

We can start by expanding the squared term:

$$(t-\square \nabla)^2 = t^2 - 2t\square \nabla + (\square \nabla)^2$$

Now, we can rewrite the equation as:

$$t^2 - 2t\square \nabla + (\square \nabla)^2 - \square = 0$$

To complete the square, we need to add and subtract $(\square \nabla)^2$:

$$t^2 - 2t\square \nabla + (\square \nabla)^2 - (\square \nabla)^2 - \square = 0$$

Now, we can factor the perfect square trinomial:

$$(t - \square \nabla)^2 - (\square \nabla)^2 - \square = 0$$

Simplifying the equation, we get:

$$(t - \square \nabla)^2 = (\square \nabla)^2 + \square$$

Multiplying the Equation by -1 and 16

Now that we have rewritten the equation in vertex form, we can multiply it by -1 and 16:

$$-16(t - \square \nabla)^2 = -16((\square \nabla)^2 + \square)$$

Distributing the -16, we get:

$$-16(t - \square \nabla)^2 = -16(\square \nabla)^2 - 16\square$$

Simplifying the equation, we get:

$$-16(t - \square \nabla)^2 + 16\square = -16(\square \nabla)^2$$

Conclusion

In this article, we have rewritten the equation in vertex form and then multiplied it by -1 and 16. The vertex form is a powerful tool for analyzing and solving quadratic equations, and it allows us to easily identify the vertex of the parabola. By completing the square and factoring the perfect square trinomial, we were able to rewrite the equation in vertex form. Multiplying the equation by -1 and 16 allowed us to simplify the equation and make it easier to solve.

Example Problems

Problem 1

Rewrite the equation in vertex form:

$$(x-3)^2-4=0$$

Solution

To rewrite the equation in vertex form, we need to complete the square. Expanding the squared term, we get:

$$(x-3)^2 = x^2 - 6x + 9$$

Now, we can rewrite the equation as:

$$x^2 - 6x + 9 - 4 = 0$$

To complete the square, we need to add and subtract 9:

$$x^2 - 6x + 9 - 9 - 4 = 0$$

Now, we can factor the perfect square trinomial:

$$(x-3)^2 - 9 - 4 = 0$$

Simplifying the equation, we get:

$$(x-3)^2 = 13$$

Problem 2

Multiply the equation by -1 and 16:

$$(t-2)^2-3=0$$

Solution

To multiply the equation by -1 and 16, we need to distribute the -16:

$$-16(t-2)^2 + 48 = 0$$

Simplifying the equation, we get:

$$-16(t-2)^2 = -48$$

Final Answer

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Introduction

In our previous article, we explored the vertex form of a quadratic equation and how to rewrite an equation in vertex form. In this article, we will answer some frequently asked questions about the vertex form of a quadratic equation.

Q: What is the vertex form of a quadratic equation?

A: The vertex form of a quadratic equation is a way of expressing a quadratic equation in the form of $(x-h)^2+k=0$, where $(h,k)$ is the vertex of the parabola.

Q: How do I rewrite an equation in vertex form?

A: To rewrite an equation in vertex form, you need to complete the square. This involves expanding the squared term, adding and subtracting the same value, and then factoring the perfect square trinomial.

Q: What is the vertex of a parabola?

A: The vertex of a parabola is the point where the parabola changes direction. It is the minimum or maximum point of the parabola, depending on whether the parabola opens upward or downward.

Q: How do I find the vertex of a parabola?

A: To find the vertex of a parabola, you need to rewrite the equation in vertex form. The vertex is then given by the values of $h$ and $k$ in the equation $(x-h)^2+k=0$.

Q: What is the significance of the vertex form of a quadratic equation?

A: The vertex form of a quadratic equation is significant because it allows us to easily identify the vertex of the parabola. This is useful in a variety of applications, including physics, engineering, and economics.

Q: Can I use the vertex form of a quadratic equation to solve quadratic equations?

A: Yes, you can use the vertex form of a quadratic equation to solve quadratic equations. By rewriting the equation in vertex form, you can easily identify the vertex of the parabola and then use the equation to solve for the variable.

Q: How do I multiply an equation by -1 and 16?

A: To multiply an equation by -1 and 16, you need to distribute the -16. This involves multiplying each term in the equation by -16 and then simplifying the resulting equation.

Q: What is the final answer to the equation $(t-2)^2-3=0$?

A: To find the final answer to the equation $(t-2)^2-3=0$, you need to multiply the equation by -1 and 16. This gives:

$$-16(t-2)^2 + 48 = 0$$

Simplifying the equation, you get:

$$-16(t-2)^2 = -48$$

Dividing both sides by -16, you get:

$$(t-2)^2 = 3$$

Taking the square root of both sides, you get:

$$t-2 = \pm\sqrt{3}$$

Adding 2 to both sides, you get:

$$t = 2 \pm \sqrt{3}$$

Conclusion

In this article, we have answered some frequently asked questions about the vertex form of a quadratic equation. We have also provided examples of how to rewrite an equation in vertex form and how to multiply an equation by -1 and 16. By understanding the vertex form of a quadratic equation, you can easily identify the vertex of a parabola and use the equation to solve quadratic equations.

Example Problems

Problem 1

Rewrite the equation in vertex form:

$$(x-3)^2-4=0$$

Solution

To rewrite the equation in vertex form, we need to complete the square. Expanding the squared term, we get:

$$(x-3)^2 = x^2 - 6x + 9$$

Now, we can rewrite the equation as:

$$x^2 - 6x + 9 - 4 = 0$$

To complete the square, we need to add and subtract 9:

$$x^2 - 6x + 9 - 9 - 4 = 0$$

Now, we can factor the perfect square trinomial:

$$(x-3)^2 - 9 - 4 = 0$$

Simplifying the equation, we get:

$$(x-3)^2 = 13$$

Problem 2

Multiply the equation by -1 and 16:

$$(t-2)^2-3=0$$

Solution

To multiply the equation by -1 and 16, we need to distribute the -16:

$$-16(t-2)^2 + 48 = 0$$

Simplifying the equation, we get:

$$-16(t-2)^2 = -48$$

Dividing both sides by -16, we get:

$$(t-2)^2 = 3$$

Taking the square root of both sides, we get:

$$t-2 = \pm\sqrt{3}$$

Adding 2 to both sides, we get:

$$t = 2 \pm \sqrt{3}$$

Final Answer

The final answer is: $\boxed{2 \pm \sqrt{3}}$