Question

What are the possible values of \(x\) if \(6 x^{2}-10=11 x ?\)

Short answer

The possible values of \(x\) are \( x = \frac{5}{2} \) and \( x = \frac{-2}{3} \).

Step-by-step solution

- Move All Terms to One Side

First, to solve the quadratic equation, move all terms to one side so that it is in the standard form of a quadratic equation. Subtract 11x from both sides: 6x^2 - 11x - 10 = 0

- Identify Coefficients

Identify the coefficients from the quadratic equation. Here, a = 6 b = -11 c = -10

- Apply the Quadratic Formula

Use the quadratic formula to find the values of x: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plug in the values of a, b, and c: \[ x = \frac{11 \pm \sqrt{(-11)^2 - 4 \times 6 \times (-10)}}{2 \times 6} \]

- Simplify Under the Square Root

Calculate the value under the square root: \[ (-11)^2 = 121 \] \[ 4 \times 6 \times (-10) = -240 \] \[ b^2 - 4ac = 121 + 240 = 361 \] This gives: \[ x = \frac{11 \pm \sqrt{361}}{12} \]

- Calculate the Roots

Simplify further by finding the square root of 361: \[ \sqrt{361} = 19 \] So, the equation becomes \[ x = \frac{11 \pm 19}{12} \] This results in two possible solutions: \[ x = \frac{11 + 19}{12} = \frac{30}{12} = \frac{5}{2} \] \[ x = \frac{11 - 19}{12} = \frac{-8}{12} = \frac{-2}{3} \]

- State the Solution

The solutions to the quadratic equation \(6 x^{2} - 10 = 11 x\) are: \( x = \frac{5}{2} \) \( x = \frac{-2}{3} \)

Key concepts

quadratic formula

The quadratic formula is a powerful tool for finding the roots of any quadratic equation. It is especially useful when the equation cannot be easily factored. The formula is:
\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \.
Here, 'a', 'b', and 'c' are the coefficients from the quadratic equation in its standard form.
\[ ax^2 + bx + c = 0 \]
The term
\ b^2 - 4ac \
under the square root is known as the discriminant. It tells us how many real roots the equation has:

• If the discriminant is positive, there are two real and distinct roots.
• If it is zero, there is one real root (sometimes known as a repeated root).
• If it is negative, there are no real roots, but two complex roots.

Remember, once you compute the roots using the quadratic formula, simplify the results to find the final solutions.

standard form of quadratic equations

To use the quadratic formula, the equation must be in standard form. The standard form of a quadratic equation is:
\ ax^2 + bx + c = 0 \.
Here, 'a', 'b', and 'c' are constants, and 'x' is the variable.
The term 'a' is the coefficient of the squared term, 'b' is the coefficient of the linear term, and 'c' is the constant term.
For example, in the given equation \(6x^2 - 10 = 11x\), we need to rearrange it to the standard form. We do this by moving all terms to one side:
\ 6x^2 - 11x - 10 = 0 \.
Now, it matches the standard form with:

• 'a' = 6
• 'b' = -11
• 'c' = -10

Once in the standard form, the coefficients 'a', 'b', and 'c' can be identified and used in the quadratic formula.

roots of quadratic equations

The roots of a quadratic equation are the values of 'x' that make the equation true. Another term for roots is 'solutions'.
When you use the quadratic formula, the solutions are found by inserting the coefficients of the equation into the formula.
For the equation \(6x^2 - 11x - 10 = 0\), after using the quadratic formula:
We have:
\ x = \frac{11 \pm \sqrt{361}}{12} \
The calculation under the square root (the discriminant) was simplified to 361. Taking the square root of 361 gives us 19.
Therefore, we end up with two solutions:

• \( x = \frac{11 + 19}{12} = \frac{30}{12} = \frac{5}{2} \)
• \( x = \frac{11 - 19}{12} = \frac{-8}{12} = \frac{-2}{3} \)

These roots (\( \frac{5}{2} \) and \( \frac{-2}{3} \)) are the values of 'x' which satisfy the original quadratic equation.