Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ 4 t^{2}+t+1=0 $$

Short answer

No real solutions.

Step-by-step solution

- Identify coefficients

The quadratic equation is of the form \(a t^{2} + b t + c = 0\). Identify the coefficients \(a\), \(b\), and \(c\) from the given equation \(4 t^{2} + t + 1 = 0\). Here, \(a = 4\), \(b = 1\), and \(c = 1\).

- Write the quadratic formula

The quadratic formula to find solutions to \(a t^{2} + b t + c = 0\) is \(t = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}\).

- Calculate the discriminant

The discriminant \(\triangle\) is given by \(\triangle = b^2 - 4ac\). Substitute the values \(a = 4\), \(b = 1\), and \(c = 1\) into the discriminant formula: \(\triangle = 1^2 - 4(4)(1)\) Simplify it: \(\triangle = 1 - 16 = -15\).

- Analyze the discriminant

The discriminant \(\triangle\) is \(-15\), which is less than \(0\). Since the discriminant is negative, the quadratic equation has no real solutions.

Key concepts

Discriminant

A key concept in solving quadratic equations using the quadratic formula is the discriminant. The discriminant \( \triangle \) is part of the quadratic formula, specifically found under the square root: \( \triangle = b^2 - 4ac \). This value helps us determine the nature of the roots of the quadratic equation.

Here’s what the discriminant can tell us:

• If \( \triangle > 0 \), there are two distinct real solutions.
• If \( \triangle = 0 \), there is exactly one real solution.
• If \( \triangle < 0 \), the equation has no real solutions and only complex solutions exist.

In this particular exercise, our discriminant was calculated as \( \triangle = -15 \), which is less than zero. Thus, we know the quadratic equation \( 4 t^{2}+t+1=0 \) has no real solutions.

Coefficients

To use the quadratic formula effectively, it’s crucial to correctly identify the coefficients in the quadratic equation. The general form of a quadratic equation is \( at^2 + bt + c = 0 \).

Here’s a breakdown of what each coefficient represents:

• \( a \): The coefficient of the \( t^2 \) term, which is 4 in this exercise.
• \( b \): The coefficient of the t term, which is 1.
• \( c \): The constant term, which is also 1.

These coefficients are substituted into the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which then allows us to solve for the roots of the equation. Identifying these values correctly is the first step to finding solutions.

Real Solutions

A quadratic equation can have different types of solutions based on the discriminant. In this exercise, we sought real solutions, which are the values of \( t \) that satisfy the equation.

We used the discriminant to determine if real solutions exist. For the equation \( 4 t^{2}+t+1=0 \), the discriminant \( \triangle \) was -15. Since it’s negative, there are no real solutions for this equation. In other words, the quadratic equation cannot be solved with real numbers; the solutions are complex.

It’s important to remember this when solving any quadratic equation using the quadratic formula. Always check the discriminant first to know the nature of the solutions:

• Positive discriminant means two real solutions.
• Zero discriminant means one real solution.
• Negative discriminant means no real solutions, but two complex solutions.