Problem 59
Question
Solve the quadratic equation using any method. Find only real solutions. $$-2 x^{2}-1=3 x$$
Step-by-Step Solution
Verified Answer
The real solutions to the equation '-2x^2 - 1 = 3x' are x = -1 and x = 1/2.
1Step 1: Standard Quadratic Equation
Bring all terms to one side to form the quadratic equation in the form of 'ax^2 + bx + c = 0'. Our equation becomes '-2x^2 - 3x - 1 = 0'
2Step 2: Determining Values of a, b, c
We see that \( a = -2, b = -3, c = -1\) in our quadratic equation '-2x^2 - 3x - 1 = 0'.
3Step 3: Identifying the Discriminant
The discriminant \( D \) is given by the formula \( D = b^2 - 4ac \). Substituting our values of a, b, and c, we get \( D = (-3)^2 -4*(-2)*(-1) = 9 - 8 = 1 \)
4Step 4: Applying Quadratic Formula
When \( D \geq 0 \), we have real solutions. The roots of the quadratic equation can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \). On calculation, the roots are \( x = \frac{-(-3) + \sqrt{1}}{2*(-2)}, x = \frac{-(-3) - \sqrt{1}}{2*(-2)} \) which simplifies to \( x = -1, x = 1/2 \)
Key Concepts
Standard Quadratic EquationDiscriminantQuadratic Formula
Standard Quadratic Equation
In mathematics, a standard quadratic equation is a fundamental form of algebraic expression that represents a parabola when graphed. This equation is typically written as \( ax^2 + bx + c = 0 \), where \( a \) is the coefficient of the squared term, \( b \) is the coefficient of the linear term, and \( c \) is the constant.
To solve a quadratic equation, you must rearrange the terms so that the equation takes the form of the standard quadratic equation. This involves combining like terms and moving them to one side of the equation, setting it equal to zero. Once you have the equation in standard form, you can solve it using various methods, including factoring, completing the square, or applying the quadratic formula.
To solve a quadratic equation, you must rearrange the terms so that the equation takes the form of the standard quadratic equation. This involves combining like terms and moving them to one side of the equation, setting it equal to zero. Once you have the equation in standard form, you can solve it using various methods, including factoring, completing the square, or applying the quadratic formula.
Discriminant
When solving quadratic equations, the discriminant plays a crucial role in determining the types and number of solutions. The discriminant is the part of the quadratic formula under the square root sign. It is represented by the symbol \( D \) and calculated with the formula \( D = b^2 - 4ac \).
If the discriminant is greater than zero, \( D > 0 \), there are two distinct real solutions. A discriminant of zero, \( D = 0 \), means there is one real solution, and if the discriminant is negative, \( D < 0 \), there are no real solutions – instead, there are two complex solutions. Knowing the value of the discriminant can thus instantly give you an insight into the nature of the solutions for the quadratic equation without actually solving it.
If the discriminant is greater than zero, \( D > 0 \), there are two distinct real solutions. A discriminant of zero, \( D = 0 \), means there is one real solution, and if the discriminant is negative, \( D < 0 \), there are no real solutions – instead, there are two complex solutions. Knowing the value of the discriminant can thus instantly give you an insight into the nature of the solutions for the quadratic equation without actually solving it.
Quadratic Formula
The quadratic formula is a powerful tool for solving standard quadratic equations, and it guarantees a solution when dealing with real or complex numbers. The formula is \( x = \frac{-b \pm \sqrt{D}}{2a} \), where \( x \) represents the variable being solved for, and \( a \) , \( b \) , and \( c \) are the coefficients of the standard quadratic equation \( ax^2 + bx + c = 0 \).
The symbol \( \pm \) indicates that the quadratic equation may have two solutions. To find these solutions, you substitute the values of \( a \) , \( b \) , and \( c \) into the formula and solve for \( x \). This use of the quadratic formula is particularly useful when factoring is difficult or impossible, ensuring that even the most complex quadratic equations can be solved.
The symbol \( \pm \) indicates that the quadratic equation may have two solutions. To find these solutions, you substitute the values of \( a \) , \( b \) , and \( c \) into the formula and solve for \( x \). This use of the quadratic formula is particularly useful when factoring is difficult or impossible, ensuring that even the most complex quadratic equations can be solved.
Other exercises in this chapter
Problem 59
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In Exercises \(49-66,\) let \(f(x)=x^{2}+x, g(x)=\sqrt{x},\) and \(h(x)=-3 x\) Evaluate each of the following. $$(g \circ f)(1)$$
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Use a graphing utility to find all real solutions. You may need to adjust the window size manually or use the ZOOMFIT feature to get a clear graph. Graphically
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