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Chapter 1: Problem 50

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

Short Answer

Expert verified
No real solutions.

Step by step solution

01

- 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 \(c = 1\).
02

- Write the quadratic formula

The quadratic formula to find solutions to \(a t^{2} + b t + c = 0\) is \(t = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}\).
03

- 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: \(\triangle = 1 - 16 = -15\).
04

- 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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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鈥檚 crucial to correctly identify the coefficients in the quadratic equation. The general form of a quadratic equation is \( at^2 + bt + c = 0 \).

Here鈥檚 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鈥檚 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鈥檚 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.

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