Math Problem Statement

Find the range of values of x for which the expression 2x^2 + 5x - 12 is positive and find the minimum value of the expression.

Solution

The given problem asks to find the range of values for xx such that the quadratic expression 2x2+5x122x^2 + 5x - 12 is positive, and also find the minimum value of this expression.

1. Finding the minimum value of the quadratic expression

A quadratic expression of the form ax2+bx+cax^2 + bx + c reaches its minimum (or maximum) at the vertex, where the xx-coordinate is given by:

x=b2ax = -\frac{b}{2a}

For 2x2+5x122x^2 + 5x - 12, the coefficients are a=2a = 2, b=5b = 5, and c=12c = -12.

xmin=52(2)=54x_{\text{min}} = -\frac{5}{2(2)} = -\frac{5}{4}

To find the minimum value, substitute x=54x = -\frac{5}{4} into the quadratic expression:

f(54)=2(54)2+5(54)12f\left(-\frac{5}{4}\right) = 2\left(-\frac{5}{4}\right)^2 + 5\left(-\frac{5}{4}\right) - 12

=2(2516)+5(54)12= 2\left(\frac{25}{16}\right) + 5\left(-\frac{5}{4}\right) - 12

=501625412= \frac{50}{16} - \frac{25}{4} - 12

=50161001619216= \frac{50}{16} - \frac{100}{16} - \frac{192}{16}

=5010019216=24216=15.125= \frac{50 - 100 - 192}{16} = \frac{-242}{16} = -15.125

Thus, the minimum value of the expression is 15.125-15.125.

2. Finding when the expression is positive

To determine when 2x2+5x12>02x^2 + 5x - 12 > 0, we first find the roots by solving the equation 2x2+5x12=02x^2 + 5x - 12 = 0 using the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute a=2a = 2, b=5b = 5, and c=12c = -12:

x=5±524(2)(12)2(2)x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-12)}}{2(2)}

x=5±25+964=5±1214x = \frac{-5 \pm \sqrt{25 + 96}}{4} = \frac{-5 \pm \sqrt{121}}{4}

x=5±114x = \frac{-5 \pm 11}{4}

The two roots are:

x1=5+114=64=1.5x_1 = \frac{-5 + 11}{4} = \frac{6}{4} = 1.5

x2=5114=164=4x_2 = \frac{-5 - 11}{4} = \frac{-16}{4} = -4

The quadratic expression changes signs at the roots. Since the coefficient of x2x^2 is positive, the quadratic expression is positive when x<4x < -4 or x>1.5x > 1.5.

Final Answer:

  • The expression 2x2+5x122x^2 + 5x - 12 is positive for x<4x < -4 or x>1.5x > 1.5.
  • The minimum value of the expression is 15.125-15.125.

Would you like more details on any part of the solution?


Here are 5 related questions:

  1. How do you determine whether a quadratic expression opens upwards or downwards?
  2. How can you find the vertex of a quadratic equation using the completing the square method?
  3. What does the discriminant of a quadratic equation tell you about its roots?
  4. How can you solve quadratic inequalities algebraically?
  5. How can you graph a quadratic function to visually identify the range of positive values?

Tip: Always check the sign of the leading coefficient to determine whether a parabola opens upwards (minimum) or downwards (maximum).

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Expressions
Inequalities

Formulas

Quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
Vertex formula: x = -b / 2a

Theorems

Quadratic Formula Theorem

Suitable Grade Level

Grades 9-10