Problem 95
Question
Solve equation by the method of your choice. $$ 3 x^{2}-12 x+12=0 $$
Step-by-Step Solution
Verified Answer
The solution to the equation \(3x^2 - 12x + 12 = 0\) is \(x = 2\).
1Step 1: Identify Coefficients
In the quadratic equation \(3x^2 - 12x + 12 = 0\), the coefficients are \(a = 3\), \(b = -12\), and \(c = 12\).
2Step 2: Plug Coefficients into Quadratic Formula
Plug the coefficients into the Quadratic Formula, yielding \(x = \frac{{-(-12) \pm \sqrt{{(-12)^2 - 4*3*12}}}}{2*3}\).
3Step 3: Simplify Inside the Square Root
First calculate the value inside the square root, known as the discriminant. Here, \((-12)^2 - 4*3*12 = 144 - 144= 0\). This leaves the formula as \(x = \frac{{12 \pm \sqrt{0}}}{6}\).
4Step 4: Simplify Further
The square root of 0 is 0, so the formula simplifies to \(x = \frac{12}{6}\).
5Step 5: Final simplifying
Finally, performing the division gives \(x = 2\).
Key Concepts
Quadratic FormulaDiscriminant in QuadraticsCoefficients of a Quadratic Equation
Quadratic Formula
When facing a quadratic equation like the one we have, which is in the standard form of \( ax^2 + bx + c = 0 \), your go-to method for finding its solutions is the quadratic formula. The quadratic formula looks like this: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \].
Using this formula, we are guaranteed to find the value(s) of \(x\) that satisfy the equation, no matter if the solutions are real or complex numbers. This powerful tool requires you to first identify the coefficients \(a\), \(b\), and \(c\) from the equation, then plug these values into the formula and carry out the arithmetic. In the given exercise, the quadratic formula quickly revealed the solution, showing how straightforward and invaluable this method can be in your mathematical toolkit.
Using this formula, we are guaranteed to find the value(s) of \(x\) that satisfy the equation, no matter if the solutions are real or complex numbers. This powerful tool requires you to first identify the coefficients \(a\), \(b\), and \(c\) from the equation, then plug these values into the formula and carry out the arithmetic. In the given exercise, the quadratic formula quickly revealed the solution, showing how straightforward and invaluable this method can be in your mathematical toolkit.
Discriminant in Quadratics
The discriminant in quadratics is a key player within the quadratic formula. Simply put, it's the part under the square root: \( b^2 - 4ac \). The value of the discriminant tells us about the nature of the roots of the quadratic equation.
- If it's positive, there are two distinct real roots.
- If it's zero, there's exactly one real root (also known as a repeated or double root).
- If it's negative, there are no real roots; instead, there are two complex roots.
Coefficients of a Quadratic Equation
In every quadratic equation, the terms \(a\), \(b\), and \(c\) are the coefficients. Specifically, \(a\) is the coefficient of the \(x^2\) term, \(b\) is the coefficient of the \(x\) term, and \(c\) is the constant term. It's vital to accurately identify these coefficients since they directly affect the solutions of the equation.
In the worked exercise, the coefficients were \(a = 3\), \(b = -12\), and \(c = 12\). These values were then used in the quadratic formula. A common mistake is misidentifying these coefficients, which can lead to incorrect solutions. Always pay attention to the signs of the coefficients as well; a negative coefficient can significantly change the discriminant and the equation's solutions.
In the worked exercise, the coefficients were \(a = 3\), \(b = -12\), and \(c = 12\). These values were then used in the quadratic formula. A common mistake is misidentifying these coefficients, which can lead to incorrect solutions. Always pay attention to the signs of the coefficients as well; a negative coefficient can significantly change the discriminant and the equation's solutions.
Other exercises in this chapter
Problem 94
Find all values of \(x\) satisfying the given conditions. $$y_{1}=6\left(\frac{2 x}{x-3}\right)^{2}, y_{2}=5\left(\frac{2 x}{x-3}\right), \text { and } y_{1} \t
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Solve each equation. $$0.5(x+2)=0.1+3(0.1 x+0.3)$$
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Solve each equation. $$\left|x^{2}+2 x-36\right|=12$$
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Solve each equation. $$4 x+13-\\{2 x-[4(x-3)-5]\\}=2(x-6)$$
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