Solve For $y$. $y^2 + Y - 12 = 0$ If There Is More Than One Solution, Separate Them With Commas. If There Is No Solution, Click On No Solution.

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Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a quadratic equation of the form y2+yβˆ’12=0y^2 + y - 12 = 0. We will use the quadratic formula to find the solutions, and we will also explore the concept of factoring and the discriminant.

What is a Quadratic Equation?

A quadratic equation is a polynomial equation of degree two, which means that the highest power of the variable (in this case, yy) is two. The general form of a quadratic equation is:

ax2+bx+c=0ax^2 + bx + c = 0

where aa, bb, and cc are constants, and xx is the variable. In our case, the equation is y2+yβˆ’12=0y^2 + y - 12 = 0, where a=1a = 1, b=1b = 1, and c=βˆ’12c = -12.

The Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where aa, bb, and cc are the coefficients of the quadratic equation. In our case, we have:

y=βˆ’1Β±12βˆ’4(1)(βˆ’12)2(1)y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)}

Simplifying the Quadratic Formula

To simplify the quadratic formula, we need to evaluate the expression inside the square root. We have:

b2βˆ’4ac=12βˆ’4(1)(βˆ’12)=1+48=49b^2 - 4ac = 1^2 - 4(1)(-12) = 1 + 48 = 49

So, the quadratic formula becomes:

y=βˆ’1Β±492(1)y = \frac{-1 \pm \sqrt{49}}{2(1)}

Evaluating the Square Root

The square root of 49 is 7, so we have:

y=βˆ’1Β±72(1)y = \frac{-1 \pm 7}{2(1)}

Solving for y

Now, we can solve for yy by evaluating the two possible values:

y1=βˆ’1+72(1)=62=3y_1 = \frac{-1 + 7}{2(1)} = \frac{6}{2} = 3

y2=βˆ’1βˆ’72(1)=βˆ’82=βˆ’4y_2 = \frac{-1 - 7}{2(1)} = \frac{-8}{2} = -4

Conclusion

In this article, we have solved the quadratic equation y2+yβˆ’12=0y^2 + y - 12 = 0 using the quadratic formula. We have found two solutions, y=3y = 3 and y=βˆ’4y = -4, which are separated by a comma. This is a common way to present the solutions to a quadratic equation.

Factoring the Quadratic Equation

Another way to solve the quadratic equation is to factor it. We can write the equation as:

(y+4)(yβˆ’3)=0(y + 4)(y - 3) = 0

This tells us that either (y+4)=0(y + 4) = 0 or (yβˆ’3)=0(y - 3) = 0. Solving for yy, we get:

y+4=0β‡’y=βˆ’4y + 4 = 0 \Rightarrow y = -4

yβˆ’3=0β‡’y=3y - 3 = 0 \Rightarrow y = 3

The Discriminant

The discriminant is a value that can be calculated from the coefficients of the quadratic equation. It is given by:

Ξ”=b2βˆ’4ac\Delta = b^2 - 4ac

In our case, we have:

Ξ”=12βˆ’4(1)(βˆ’12)=49\Delta = 1^2 - 4(1)(-12) = 49

The discriminant is positive, which means that the quadratic equation has two distinct solutions.

No Solution

If the discriminant is negative, then the quadratic equation has no solution. In this case, we would click on "No solution".

Conclusion

In this article, we have solved the quadratic equation y2+yβˆ’12=0y^2 + y - 12 = 0 using the quadratic formula and factoring. We have found two solutions, y=3y = 3 and y=βˆ’4y = -4, which are separated by a comma. We have also explored the concept of the discriminant and how it can be used to determine the number of solutions to a quadratic equation.

Final Answer

The final answer is: 3,βˆ’4\boxed{3, -4}

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Introduction

In our previous article, we solved the quadratic equation y2+yβˆ’12=0y^2 + y - 12 = 0 using the quadratic formula and factoring. In this article, we will answer some common questions that students and professionals may have when it comes to solving quadratic equations.

Q&A

Q: What is a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means that the highest power of the variable (in this case, yy) is two. The general form of a quadratic equation is:

ax2+bx+c=0ax^2 + bx + c = 0

where aa, bb, and cc are constants, and xx is the variable.

Q: How do I solve a quadratic equation?

A: There are several ways to solve a quadratic equation, including:

  • Using the quadratic formula: x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • Factoring: If the quadratic equation can be written as a product of two binomials, then we can factor it and solve for the variable.
  • Graphing: We can graph the quadratic equation and find the x-intercepts, which are the solutions to the equation.

Q: What is the quadratic formula?

A: The quadratic formula is a powerful tool for solving quadratic equations. It is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where aa, bb, and cc are the coefficients of the quadratic equation.

Q: What is the discriminant?

A: The discriminant is a value that can be calculated from the coefficients of the quadratic equation. It is given by:

Ξ”=b2βˆ’4ac\Delta = b^2 - 4ac

The discriminant can be used to determine the number of solutions to a quadratic equation. If the discriminant is positive, then the equation has two distinct solutions. If the discriminant is zero, then the equation has one solution. If the discriminant is negative, then the equation has no solution.

Q: How do I determine the number of solutions to a quadratic equation?

A: To determine the number of solutions to a quadratic equation, we need to calculate the discriminant. If the discriminant is positive, then the equation has two distinct solutions. If the discriminant is zero, then the equation has one solution. If the discriminant is negative, then the equation has no solution.

Q: What is the difference between a quadratic equation and a linear equation?

A: A quadratic equation is a polynomial equation of degree two, while a linear equation is a polynomial equation of degree one. In other words, a quadratic equation has a highest power of two, while a linear equation has a highest power of one.

Q: Can I solve a quadratic equation by graphing?

A: Yes, we can solve a quadratic equation by graphing. We can graph the quadratic equation and find the x-intercepts, which are the solutions to the equation.

Conclusion

In this article, we have answered some common questions that students and professionals may have when it comes to solving quadratic equations. We have discussed the quadratic formula, the discriminant, and how to determine the number of solutions to a quadratic equation. We have also explored the difference between a quadratic equation and a linear equation.

Final Answer

The final answer is: 3,βˆ’4\boxed{3, -4}