Question

Solve the quadratic equation by extracting square roots. List both the exact answer and a decimal answer that has been rounded to two decimal places. $$ x^{2}+3\left(x^{2}-5\right)=10 $$

Short answer

The exact solutions are \(x = \pm \frac{5}{2}\) and the decimal solutions are \(x = \pm 2.5\)

Step-by-step solution

Simplify the Equation

First, simplify the equation, combine like terms: \(x^{2}+3x^{2}-5*3 = 10\), which simplifies to \(4x^{2} - 15 = 10\)

Move Constants to one side

Rearrange the equation to shift -15 to right side by adding both sides by 15 to get \(4x^2 = 25\)

Divide By the Coefficient of \(x^{2}\)

Divide Equation by 4 on both sides to isolate \(x^2\). So we get, \(x^{2} = \frac{25}{4}\)

Extract Square Roots

Take the square root of both sides. Since \(x^{2}\) can be either positive or negative, we get \(x = \pm \sqrt{\frac{25}{4}}\) which simplifies to \(x = \pm \frac{5}{2}\)

Find Decimal Value

Convert the fractional roots to decimal to nearest 2 decimal places. So, we get \(x = \pm 2.5\)

Key concepts

Extracting Square Roots

When solving a quadratic equation by extracting square roots, the goal is to make it simple so we can find the root of the variable squared. This is possible when the quadratic equation is in the form of:

• \(x^2 = k\), where \(k\) is some number.

To solve this equation, take the square root of both sides. Remember, taking a square root involves both the positive and negative roots. This is because both \(+\sqrt{k}\) and \(-\sqrt{k}\) will satisfy the equation when squared again.
For example, in the solution above, we had \(x^2 = \frac{25}{4}\), and by taking the square roots, we concluded that:

• \(x = \pm \frac{5}{2}\)

This means there are two potential values for \(x\), an essential part of understanding quadratics!

Combining Like Terms

Combining like terms is a useful technique for simplifying algebraic equations. It helps in bundling similar items together to make the equation clearer and more manageable. When you have terms with the same variable raised to the same power, you can add or subtract them.
In the example at hand, we see the equation:

• \(x^2 + 3(x^2 - 5)\)

By distributing and then combining, it becomes:

• \(x^2 + 3x^2 - 15\), which simplifies further to \(4x^2 - 15\)

Combining like terms reduces the complexity of the equation, making it much easier to solve. It’s like cleaning up a messy room, getting similar items into one box!

Decimal Approximation

Decimal approximation is a way to represent numbers that can't be expressed exactly as simple fractions. It is beneficial when you require an answer in a more universal numerical form. To convert a fraction into a decimal, you divide the numerator by the denominator and round it as needed.
In mathematics, especially when interpreting square roots or fractions, you often prefer decimals because they are easier to understand and use in calculations. For instance, from our exact value of \(\pm \frac{5}{2}\), we derived:

• \(\pm 2.5\)

This decimal form is exact in this instance due to the clean division, but rounding usually handles longer or repeating decimals smoothly, keeping calculations handy and precise, up to required places.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, usually expressed in the standard form:

• \(ax^2 + bx + c = 0\)

Here, \(x\) is the variable, and \(a\), \(b\), and \(c\) are constants. The solutions to quadratic equations can be found using various methods such as factoring, completing the square, extracting square roots, or using the quadratic formula.
In the given example:

• By simplifying \(x^2 + 3(x^2 - 5) = 10\) into \(4x^2 = 25\), we use extracting square roots as an approach to finding \(x\).

Quadratics often have two solutions since the "squaring" process in these equations inherently involves two possible roots. This dual solution is crucial for calculations in many fields, including physics, engineering, and economics.