Free Math Worksheets

- Clear the decimal coefficient in part a) before applying the formula. - Substitute the coefficients into the quadratic formula. - Check the sign of the discriminant before taking a square root.

1. For a), set \(0.5x^2 - 2x - 6 = 0\). Multiply by \(2\) to obtain \(x^2 - 4x - 12 = 0\). 2. The quadratic formula gives \(x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2} = \frac{4 \pm \sqrt{64}}{2}\). Thus, \(x = -2\) or \(x = 6\). 3. For b), the discriminant is \(D = 4^2 - 4 \cdot 1 \cdot 7 = -12\). Since \(D < 0\), the function has no real zeros.

a) Two real zeros: \(x = -2\) and \(x = 6\) b) No real zeros

- Identify \(a\), \(b\), and \(c\) carefully, including their signs. - Divide every term by a common factor before using the formula when that simplifies the coefficients. - Simplify the discriminant before taking its square root.

1. For part a, \(a = 1\), \(b = -10\), and \(c = 21\). Then \(x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 21}}{2} = \frac{10 \pm 4}{2}\). Thus, \(x = 7\) or \(x = 3\). 2. For part b, divide by \(2\): \(x^2 + 4x - 12 = 0\). 3. Use \(a = 1\), \(b = 4\), and \(c = -12\): \(x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2} = \frac{-4 \pm 8}{2}\). Thus, \(x = 2\) or \(x = -6\).

a) \(x \in \{3, 7\}\) b) \(x \in \{-6, 2\}\)

- Expand the product and write the equation in standard form. - Identify \(a\), \(b\), and \(c\), including their signs. - Simplify the discriminant before calculating the two solutions.

1. Expand and move all terms to one side: \(2x^2 + 5x - 12 = 0\). 2. Use \(a = 2\), \(b = 5\), and \(c = -12\): \(x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-12)}}{2 \cdot 2}\). 3. Simplify the discriminant: \(25 + 96 = 121\), so \(\sqrt{121} = 11\). 4. Therefore, \(x = \frac{-5 + 11}{4} = \frac{3}{2}\) or \(x = \frac{-5 - 11}{4} = -4\).

\(x \in \left\{-4, \frac{3}{2}\right\}\)

- Identify \(a\), \(b\), and \(c\) before using the quadratic formula. - Simplify the radical in the formula before writing the final roots. - When checking the smaller root, expand the squared binomial and combine like radical terms.

1. Use \(a = 1\), \(b = -4\), and \(c = -1\): \(x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}\). 2. The smaller solution is \(x = 2 - \sqrt{5}\). 3. Substitute it into the left side: \((2 - \sqrt{5})^2 - 4(2 - \sqrt{5}) - 1\). 4. Simplify: \(9 - 4\sqrt{5} - 8 + 4\sqrt{5} - 1 = 0\). Therefore, \(2 - \sqrt{5}\) is a solution.

The solutions are \(x = 2 + \sqrt{5}\) and \(x = 2 - \sqrt{5}\). Substitution confirms that \(2 - \sqrt{5}\) makes the equation true.

- Write the diagonal in terms of the side length. - Combine that relationship with \(d^2=2a^2\). - Solve the resulting quadratic and reject the negative value.

1. The diagonal is \(d=a+5\), and for a square, \(d^2=2a^2\). 2. Substitute: \((a+5)^2=2a^2\), which simplifies to \(a^2-10a-25=0\). 3. Use the quadratic formula: \(a=\frac{10\pm\sqrt{100+100}}{2}=5\pm 5\sqrt{2}\). 4. A side length must be positive, so \(a=5+5\sqrt{2}\,\text{cm}\approx 12.07\,\text{cm}\).

\(a\approx 12.07\,\text{cm}\)