Free math worksheets

- Formula coefficients come from an equation with zero on one side. - Expand the product before collecting like terms. - Keep each sign attached when identifying the coefficients.

1. Expand: \(3x^2-6x+5=2x\). 2. Move all terms to one side: \(3x^2-8x+5=0\). 3. Thus, \(a=3\), \(b=-8\), and \(c=5\).

\(3x^2-8x+5=0\); \(a=3\), \(b=-8\), \(c=5\).

- Identify the coefficients only after the equation is in the standard form you intend to use. - A common factor may be removed before formula substitution if every term is divided by it. - Check the sign of the constant term inside the discriminant.

1. a) With \(a=1\), \(b=-10\), and \(c=21\), \(x=\frac{10\pm\sqrt{100-84}}{2}=\frac{10\pm4}{2}\), so \(x=3\) or \(x=7\). 2. b) Divide by \(2\): \(x^2+4x-12=0\). Then \(x=\frac{-4\pm\sqrt{16+48}}{2}=\frac{-4\pm8}{2}\), so \(x=-6\) or \(x=2\).

a) \(x=\frac{-(-10)\pm\sqrt{(-10)^2-4\cdot1\cdot21}}{2\cdot1}\); \(x\in\{3,7\}\) b) After dividing by \(2\): \(x=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot(-12)}}{2\cdot1}\); \(x\in\{-6,2\}\)

- Clear the decimal in part a) before identifying the formula coefficients. - Use the discriminant sign to decide whether a real square root exists. - Only apply the full formula in cases where real roots are present.

1. a) Multiply by \(2\) to get \(x^2-4x-12=0\). Its discriminant is \(64>0\). The formula gives \(x=\frac{4\pm8}{2}\), so \(x=-2\) or \(x=6\). 2. b) The discriminant is \(D=4^2-4\cdot1\cdot7=-12<0\), so there are no real zeros.

a) \(D=(-4)^2-4\cdot1\cdot(-12)=64\); \(x=\frac{-(-4)\pm\sqrt{64}}{2\cdot1}\), so the zeros are \(-2\) and \(6\). b) \(D=4^2-4\cdot1\cdot7=-12\); no real zeros.

- Treat the negative values of \(b\) and \(c\) as part of the coefficients. - Do not simplify pieces of the formula until the complete substitution is written. - Factor the radicand before reducing the final fraction.

1. Here \(a=3\), \(b=-4\), and \(c=-2\). 2. Substitute: \(x=\frac{4\pm\sqrt{16+24}}{6}=\frac{4\pm\sqrt{40}}{6}\). 3. Simplify: \(x=\frac{2\pm\sqrt{10}}{3}\).

\(x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot3\cdot(-2)}}{2\cdot3}\) Therefore, \(x=\frac{2-\sqrt{10}}{3}\) or \(x=\frac{2+\sqrt{10}}{3}\).

- Expand the product before identifying the three formula coefficients. - Keep the negative constant attached to \(c\) inside \(-4ac\). - Simplify the discriminant before evaluating the two signs.

1. Expand and rearrange: \(2x^2+5x-12=0\). 2. Substitute \(a=2\), \(b=5\), and \(c=-12\): \(x=\frac{-5\pm\sqrt{25+96}}{4}\). 3. Since \(\sqrt{121}=11\), the roots are \(x=\frac32\) and \(x=-4\).

Standard form: \(2x^2+5x-12=0\) Formula substitution: \(x=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot(-12)}}{2\cdot2}\) \(x\in\left\{-4,\frac32\right\}\)

- Set the function expressions equal. - Clear the decimal coefficient before solving. - Keep irrational coordinates in exact radical form.

1. Set the functions equal: \(0.5x^2 = x + 0.5\). 2. Multiply by \(2\) and rearrange: \(x^2 - 2x - 1 = 0\). 3. Apply the quadratic formula: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}\). 4. Using \(g(x) = x + 0.5\), the y-values are \(\frac{3}{2} + \sqrt{2}\) and \(\frac{3}{2} - \sqrt{2}\). 5. The intersection points are \(\left(1 + \sqrt{2}, \frac{3}{2} + \sqrt{2}\right)\) and \(\left(1 - \sqrt{2}, \frac{3}{2} - \sqrt{2}\right)\).

\(\left(1 + \sqrt{2}, \frac{3}{2} + \sqrt{2}\right)\) and \(\left(1 - \sqrt{2}, \frac{3}{2} - \sqrt{2}\right)\)

- Write the coefficient triple with signs before putting anything into the formula. - Simplify the radical after the full substitution is correct. - For the check, keep the radical exact so the cancellation is visible.

1. With \(a=1\), \(b=-4\), and \(c=-1\), the formula gives \(x=\frac{4\pm\sqrt{20}}{2}=2\pm\sqrt5\). 2. The smaller root is \(2-\sqrt5\). 3. Substitution gives \((2-\sqrt5)^2-4(2-\sqrt5)-1=9-4\sqrt5-8+4\sqrt5-1=0\).

Formula substitution: \(x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot(-1)}}{2\cdot1}\) Solutions: \(x=2-\sqrt5\) and \(x=2+\sqrt5\) Check: \((2-\sqrt5)^2-4(2-\sqrt5)-1=0\).

- Compare the student's expression with every part of the quadratic formula, not just the discriminant. - Check how the leading coefficient affects the denominator. - After correcting the setup, simplify only if there is a common factor in the entire numerator and denominator.

1. For \(2x^2 + 3x - 1 = 0\), \(a = 2\), \(b = 3\), and \(c = -1\). The discriminant is \(3^2 - 4 \cdot 2 \cdot (-1) = 17\), so the numerator is correct. 2. The denominator of the quadratic formula is \(2a\), not \(2\). Since \(a = 2\), the denominator must be \(4\). 3. Therefore, the correct solutions are \(x = \frac{-3 \pm \sqrt{17}}{4}\).

The student used \(2\) instead of \(2a = 4\) in the denominator. The correct solutions are \(x = \frac{-3 - \sqrt{17}}{4}\) and \(x = \frac{-3 + \sqrt{17}}{4}\).

- Read the x-intercepts before doing algebra so the graph estimate is independent. - Keep the coefficient signs visible when substituting into the quadratic formula. - Compare decimal approximations only after obtaining the exact roots.

1. The graph crosses the x-axis at about \(x=-0.4\) and \(x=2.4\). 2. With \(a=1\), \(b=-2\), and \(c=-1\), the formula gives \(x=\frac{2\pm\sqrt8}{2}=1\pm\sqrt2\). 3. Since \(1-\sqrt2\approx-0.41\) and \(1+\sqrt2\approx2.41\), the exact roots agree with the graph estimates to the nearest tenth.

a) \(x\approx-0.4\) and \(x\approx2.4\) b) \(x=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot1\cdot(-1)}}{2\cdot1}=1\pm\sqrt2\) c) The exact roots are approximately \(-0.41\) and \(2.41\), matching the graph estimates.

- Multiply every term by the common denominator before identifying coefficients. - Keep the negative signs on \(b\) and \(c\) when substituting. - Simplify the radical and fraction only after the full formula is correct.

1. Multiply by \(8\): \(4x^2-6x-1=0\). 2. With \(a=4\), \(b=-6\), and \(c=-1\), \(x=\frac{6\pm\sqrt{36+16}}{8}=\frac{6\pm\sqrt{52}}{8}\). 3. Simplify: \(x=\frac{3\pm\sqrt{13}}{4}\).

Standard form: \(4x^2-6x-1=0\) Formula substitution: \(x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot4\cdot(-1)}}{2\cdot4}\) Roots: \(x=\frac{3-\sqrt{13}}{4}\) or \(x=\frac{3+\sqrt{13}}{4}\)

- Expand and move all terms to one side before removing decimals. - Reduce a common factor after scaling the equation if possible. - Use the signed coefficients from the simplified standard form in the formula.

1. Expand and collect terms: \(0.4x^2-1.8x-2=0\). 2. Multiply by \(10\) and divide by \(2\): \(2x^2-9x-10=0\). 3. The formula gives \(x=\frac{9\pm\sqrt{81+80}}{4}=\frac{9\pm\sqrt{161}}{4}\).

Standard form: \(2x^2-9x-10=0\) Formula substitution: \(x=\frac{-(-9)\pm\sqrt{(-9)^2-4\cdot2\cdot(-10)}}{2\cdot2}\) Roots: \(x=\frac{9-\sqrt{161}}{4}\) or \(x=\frac{9+\sqrt{161}}{4}\)

- Identify \(a\), \(b\), and \(c\) with their signs before inspecting the proposed formula. - Compare the first numerator term with \(-b\), not with \(b\). - After correcting the substitution, evaluate both signs of \(\pm\).

1. Here \(a=2\), \(b=-5\), and \(c=-3\). The numerator begins with \(-b\), so it should begin with \(5\), not \(-5\). 2. The correct substitution is \(x=\frac{5\pm\sqrt{(-5)^2-4\cdot2\cdot(-3)}}{4}=\frac{5\pm7}{4}\). 3. Therefore, \(x=3\) or \(x=-\frac12\).

Jordan used \(b\) instead of \(-b\) in the numerator. Correct substitution: \(x=\frac{5\pm\sqrt{(-5)^2-4\cdot2\cdot(-3)}}{4}\) Solutions: \(x=3\) and \(x=-\frac12\).

- Inspect the displayed structure before doing algebra; the equation's current form is often the strongest clue. - A method may be the best choice more than once, so do not assign methods by elimination. - Your structural reason and first step should agree with the method name you choose.

1. a) Factoring/zero-product is immediate because the equation is already a product equal to zero: \(x-4=0\) or \(x+3=0\). Thus, \(x=4,-3\). 2. b) The square-root method is direct because a squared binomial is isolated after division: \((x+5)^2=9\). Thus, \(x=-2,-8\). 3. c) Completing the square is efficient because the equation is already in the form \(x^2+bx=c\): \(x^2+8x+16=19\), so \((x+4)^2=19\) and \(x=-4\pm\sqrt{19}\). 4. d) The quadratic formula is efficient because the standard-form quadratic has no convenient integer factorization: \(x=\frac{-2\pm\sqrt{2^2-4\cdot3\cdot(-7)}}{6}=\frac{-1\pm\sqrt{22}}{3}\). 5. e) Factoring is efficient: \((2x-1)(x-3)=0\), so \(x=\frac12,3\).

a) Factoring; already a product equal to zero. First step: \(x-4=0\) or \(x+3=0\). Solutions: \(-3,4\). b) Square-root method; a squared binomial is isolated after division. First step: \((x+5)^2=9\). Solutions: \(-8,-2\). c) Completing the square; the equation has form \(x^2+bx=c\). First step: \(x^2+8x+16=19\). Solutions: \(x=-4\pm\sqrt{19}\). d) Quadratic formula; it is in standard form and has no convenient integer factorization. First step: \(x=\frac{-2\pm\sqrt{2^2-4\cdot3\cdot(-7)}}{2\cdot3}\). Solutions: \(x=\frac{-1\pm\sqrt{22}}{3}\). e) Factoring; the trinomial factors over the integers. First step: \((2x-1)(x-3)=0\). Solutions: \(\frac12,3\).

- Express the longer dimension in terms of the width before using the area. - Check whether the resulting standard-form trinomial factors conveniently before seeing why the general formula is useful here. - Interpret the two formula roots in the physical context before rounding.

1. Let the width be \(w\). Then the length is \(w+3\), so \(w(w+3)=52\). 2. Standard form is \(w^2+3w-52=0\). 3. The quadratic formula gives \(w=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-52)}}{2}=\frac{-3\pm\sqrt{217}}{2}\). 4. The negative root is not a physical width. The positive root is approximately \(5.865\), so the width is \(5.9\,\text{ft}\) and the length is \(8.9\,\text{ft}\).

Standard form: \(w^2+3w-52=0\) Formula substitution: \(w=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-52)}}{2\cdot1}\) Width: \(5.9\,\text{ft}\) Length: \(8.9\,\text{ft}\)

- Keep the sign of \(b=-1\) attached when substituting into the formula. - Multiply the conjugate exact roots before converting anything to decimals. - Exactly one real root corresponds to a zero discriminant.

1. With \(a=1\), \(b=-1\), and \(c=-5\), the quadratic formula gives \(x=\frac{1\pm\sqrt{21}}{2}\). 2. Their product is \(\frac{(1+\sqrt{21})(1-\sqrt{21})}{4}=\frac{1-21}{4}=-5\). 3. For \(x^2-x+c=0\), exactly one real solution requires \(D=(-1)^2-4c=0\). 4. Thus, \(c=\frac14\).

a) \(x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-5)}}{2\cdot1}=\frac{1\pm\sqrt{21}}{2}\) b) \(\frac{1+\sqrt{21}}{2}\cdot\frac{1-\sqrt{21}}{2}=-5\) c) \(D=1-4c=0\), so \(c=\frac14\).

- Express all three side lengths in terms of the width before using the Pythagorean theorem. - Expand the squares and collect the equation into standard quadratic form. - After using the formula, reject any value that cannot represent a positive length.

1. The Pythagorean theorem gives \(a^2+(a+5)^2=(a+9)^2\). 2. Expanding and simplifying yields \(a^2-8a-56=0\). 3. The quadratic formula gives \(a=\frac{8\pm\sqrt{(-8)^2-4\cdot1\cdot(-56)}}{2}=\frac{8\pm\sqrt{288}}{2}=4\pm6\sqrt2\). 4. A width must be positive, so \(a=4+6\sqrt2\approx12.49\,\text{cm}\). 5. The length is \(a+5\approx17.49\,\text{cm}\).

Standard form: \(a^2-8a-56=0\) Formula substitution: \(a=\frac{-(-8)\pm\sqrt{(-8)^2-4\cdot1\cdot(-56)}}{2\cdot1}\) Width: \(12.49\,\text{cm}\) Length: \(17.49\,\text{cm}\)

- Use the vertex together with one other point to determine the parabola. - Use the line's slope and y-intercept to write its equation. - After forming the quadratic equation, check whether integer factoring is actually possible before choosing a solving method.

1. The parabola has vertex \((1, 4)\) and passes through \((2, 3)\), so \(f(x)=-(x-1)^2+4=-x^2+2x+3\). 2. The line has slope \(1\) and y-intercept \(2\), so \(g(x)=x+2\). 3. Set the functions equal: \(-x^2+2x+3=x+2\). 4. Rearrange: \(x^2-x-1=0\). 5. By the quadratic formula, \(x=\frac{1\pm\sqrt{5}}{2}\). 6. Substituting into \(g\) gives the intersections \(\left(\frac{1-\sqrt{5}}{2}, \frac{5-\sqrt{5}}{2}\right)\) and \(\left(\frac{1+\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right)\), which agree with the graph.

a) \(f(x)=-x^2+2x+3\), \(g(x)=x+2\) b) \(x^2-x-1=0\) c) \(\left(\frac{1-\sqrt{5}}{2}, \frac{5-\sqrt{5}}{2}\right)\) and \(\left(\frac{1+\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right)\)

- Use each parabola's vertex and one additional point to determine its equation. - At an intersection, the two functions have the same output. - After simplifying the quadratic equation, check whether integer factoring is possible before choosing a solving method. - Substitute each x-value into either function to find the corresponding y-value.

1. The graph of \(f\) has vertex \((0,-4)\) and passes through \((2,0)\). Therefore, \(f(x)=x^2-4\). 2. The graph of \(g\) has vertex \((1,2)\) and opens downward. Write \(g(x)=a(x-1)^2+2\). Since it passes through \((0,1)\), \(1=a+2\), so \(a=-1\). Thus, \(g(x)=-(x-1)^2+2=-x^2+2x+1\). 3. Set the functions equal: \(x^2-4=-x^2+2x+1\). This simplifies to \(2x^2-2x-5=0\). 4. By the quadratic formula, \(x=\frac{1\pm\sqrt{11}}{2}\). 5. Substitute into either function. The intersections are \(\left(\frac{1-\sqrt{11}}{2}, \frac{-2-\sqrt{11}}{2}\right)\) and \(\left(\frac{1+\sqrt{11}}{2}, \frac{-2+\sqrt{11}}{2}\right)\).

a) \(f(x)=x^2-4\) and \(g(x)=-x^2+2x+1\) b) \(\left(\frac{1-\sqrt{11}}{2}, \frac{-2-\sqrt{11}}{2}\right)\) and \(\left(\frac{1+\sqrt{11}}{2}, \frac{-2+\sqrt{11}}{2}\right)\)

- Use vertex form for the parabola. - Use the y-intercept and slope of the line. - Set the function expressions equal to find their common points.

1. The parabola has vertex \((1,0)\) and passes through \((0,1)\), so \(p(x)=(x-1)^2\). 2. The line passes through \((0,0)\) and \((1,1)\), so its slope is \(1\) and \(g(x)=x\). 3. Set the functions equal: \((x-1)^2=x\). 4. Expand and rearrange: \(x^2-3x+1=0\). By the quadratic formula, \(x=\frac{3\pm\sqrt5}{2}\). 5. Since \(g(x)=x\), the intersections are \(\left(\frac{3-\sqrt5}{2}, \frac{3-\sqrt5}{2}\right)\) and \(\left(\frac{3+\sqrt5}{2}, \frac{3+\sqrt5}{2}\right)\).

\(p(x)=(x-1)^2\); \(g(x)=x\) Intersections: \(\left(\frac{3-\sqrt5}{2}, \frac{3-\sqrt5}{2}\right)\) and \(\left(\frac{3+\sqrt5}{2}, \frac{3+\sqrt5}{2}\right)\)