PSAT Math Flashcards: Solving Nonlinear Functions

What this deck covers

This deck focuses on Solving Nonlinear Functions, giving you a quick way to review the definitions, rules, and examples that matter most for PSAT Math.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Solve for $x$: $(x-4)(x+1)=0$.

Answer: $x=4$ or $x=-1$. Set each factor to zero: $x-4=0$ or $x+1=0$.

Flashcard 2: What does $\Delta=0$ imply about the real solutions of $ax^2+bx+c=0$?

Answer: $1$ real solution (a repeated root). Zero discriminant makes $\pm\sqrt{0}=0$, giving one solution.

Flashcard 3: What is the standard method to solve $a(bx+c)^2=d$ when $a\ne 0$ and $d/a\ge 0$?

Answer: $(bx+c)=\pm\sqrt{\frac{d}{a}}$. Isolate the squared term, then take square root of both sides.

Flashcard 4: Identify the solutions of $x^4-16=0$.

Answer: $x=\pm 2$. Factor as $(x^2-4)(x^2+4)=0$; only $x^2=4$ has real solutions.

Flashcard 5: Identify the first step to solve $f(x)=g(x)$ when $f$ and $g$ are nonlinear functions.

Answer: Set $f(x)-g(x)=0$ and solve the resulting equation. Rearranging creates a single equation to solve.

Flashcard 6: State the quadratic formula for solutions of $ax^2+bx+c=0$.

Answer: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Derived by completing the square on the standard form.

Flashcard 7: What is the axis of symmetry for $y=a(x-h)^2+k$?

Answer: $x=h$. The vertical line through the vertex where the parabola is symmetric.

Flashcard 8: What is the solution set of $\sqrt{x-1}=3$?

Answer: $x=10$. Square both sides: $x-1=9$, so $x=10$.

Flashcard 9: Solve for $x$: $(x+1)^2=16$.

Answer: $x=3$ or $x=-5$. Take square root: $x+1=\pm 4$, so $x=3$ or $x=-5$.

Flashcard 10: State the vertex $x$-coordinate formula for $y=ax^2+bx+c$.

Answer: $x=-\frac{b}{2a}$. The parabola's vertex occurs where the derivative equals zero.

Flashcard 11: What are the solutions to $x^2+6x+9=0$?

Answer: $x=-3$. Perfect square $(x+3)^2=0$ has one repeated root.

Flashcard 12: What is the zero-product property used after factoring?

Answer: If $ab=0$, then $a=0$ or $b=0$. At least one factor must be zero for the product to be zero.

Flashcard 13: What is the quadratic formula for solutions to $ax^2+bx+c=0$?

Answer: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Derived from completing the square on the general quadratic equation.

Flashcard 14: What is the zero-product property used when solving factored equations?

Answer: If $ab=0$, then $a=0$ or $b=0$. At least one factor must be zero for the product to be zero.

Flashcard 15: What are the solutions to $x^2-9=0$?

Answer: $x=\pm 3$. Factor as $(x-3)(x+3)=0$, then apply zero-product property.

Flashcard 16: Solve for $x$: $x^4-5x^2+4=0$.

Answer: $x=\pm 1$ or $x=\pm 2$. Let $u=x^2$: $u^2-5u+4=0$ factors to $(u-1)(u-4)=0$.

Flashcard 17: What substitution turns a quadratic in $x^2$ into a linear equation in $u$?

Answer: Let $u=x^2$. This transforms equations like $x^4+3x^2-4=0$ into $u^2+3u-4=0$.

Flashcard 18: What are the solutions to $2x^2-8=0$?

Answer: $x=\pm 2$. Divide by 2: $x^2=4$, then take square roots.

Flashcard 19: What are the solutions to $x^2+5x=0$?

Answer: $x=0$ or $x=-5$. Factor out $x$: $x(x+5)=0$, then apply zero-product property.

Flashcard 20: What does $b^2-4ac=0$ tell you about the real solutions of $ax^2+bx+c=0$?

Answer: One real solution (a double root). Zero discriminant makes $\pm\sqrt{0}=0$, so both roots coincide.

Flashcard 21: What is the solution set of $\sqrt{x+1}=3$?

Answer: $x=8$. Square both sides: $x+1=9$, so $x=8$.

Flashcard 22: What is the square root property for $(x-h)^2=k$?

Answer: $x=h\pm\sqrt{k}$. Take square root of both sides, giving two solutions.

Flashcard 23: What does $\Delta>0$ indicate about the real solutions of $ax^2+bx+c=0$?

Answer: Two distinct real solutions. Positive discriminant means the square root yields two different values.

Flashcard 24: What is the solution set of $x^2=-16$ over the real numbers?

Answer: No real solution. Square of real number cannot be negative.

Flashcard 25: Solve for $x$: $\sqrt{x-1}=5$.

Answer: $x=26$. Square both sides: $x-1=25$, so $x=26$.

Flashcard 26: What does $\Delta=b^2-4ac=0$ imply about the solutions of $ax^2+bx+c=0$?

Answer: $1$ real double solution. Zero discriminant means both roots coincide.

Flashcard 27: What are the real solutions to $(x-2)^2=9$?

Answer: $x=5$ or $x=-1$. Take square root: $x-2=\pm 3$, so $x=5$ or $x=-1$.

Flashcard 28: What are the solutions to $x^2-5x=0$?

Answer: $x=0$ or $x=5$. Factor as $x(x-5)=0$, giving two solutions.

Flashcard 29: What is the solution set of $\sqrt{x-1}=3$?

Answer: $x=10$. Square both sides: $x-1=9$, so $x=10$.

Flashcard 30: What is the axis of symmetry of $y=a(x-h)^2+k$?

Answer: $x=h$. In vertex form, the parabola is symmetric about the vertical line $x=h$.

Flashcard 31: What does $\Delta<0$ imply about the real solutions of $ax^2+bx+c=0$?

Answer: No real solutions. Negative discriminant means no real square root exists.

Flashcard 32: What is the zero-product property used when factoring nonlinear equations?

Answer: If $ab=0$, then $a=0$ or $b=0$. Allows solving factored equations by setting each factor to zero.

Flashcard 33: What step is required after squaring both sides of an equation with radicals?

Answer: Check for extraneous solutions. Squaring can introduce false solutions that don't satisfy the original.

Flashcard 34: Identify the key step to solve $x^2=k$ for real $x$ when $k\ge 0$.

Answer: $x=\pm\sqrt{k}$. Take square root of both sides, considering both positive and negative roots.

Flashcard 35: What are the solutions of $x^2-5x+6=0$?

Answer: $x=2$ and $x=3$. Factors to $(x-2)(x-3)=0$; apply zero-product property.

Flashcard 36: What is the discriminant of $ax^2+bx+c=0$?

Answer: $b^2-4ac$. The expression under the square root in the quadratic formula.

Flashcard 37: What are the solutions to $x^2=\frac{1}{4}$?

Answer: $x=\pm\frac{1}{2}$. Take square root of both sides: $x=\pm\sqrt{\frac{1}{4}}$.

Flashcard 38: What is the solution set of $|x-3|=5$?

Answer: $x=8$ and $x=-2$. Split into two cases: $x-3=5$ or $x-3=-5$.

Flashcard 39: What is the solution of $x^2=50$ in simplest radical form?

Answer: $x=\pm 5\sqrt{2}$. Take square root of both sides: $x=\pm\sqrt{50}=\pm\sqrt{25\cdot 2}=\pm 5\sqrt{2}$.

Flashcard 40: What is the standard form of a quadratic function?

Answer: $y=ax^2+bx+c$. General polynomial form with degree 2; coefficients determine shape.

Flashcard 41: Identify the vertex of $y=(x-2)^2+5$.

Answer: $(2,5)$. In vertex form $(x-h)^2+k$, the vertex is at $(h,k)$.

Flashcard 42: State the quadratic formula for solutions of $ax^2+bx+c=0$.

Answer: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Derived by completing the square on the standard form.

Flashcard 43: What is the key restriction when solving a rational equation with denominators in $x$?

Answer: Exclude values that make any denominator equal to $0$. Division by zero is undefined, so these values must be excluded.

Flashcard 44: What is the discriminant for $ax^2+bx+c=0$, and what does its sign determine?

Answer: $\Delta=b^2-4ac$; sign gives number of real solutions. Positive means 2 real roots, zero means 1, negative means none.

Flashcard 45: What is the square root property for solving $(x-h)^2=k$?

Answer: $x=h\pm\sqrt{k}$. Take square root of both sides, remembering $\pm$ for both solutions.

Flashcard 46: What are the solutions of $(x-4)(x+1)=0$?

Answer: $x=4$ and $x=-1$. Apply zero-product property: each factor equals zero.

Flashcard 47: What is the standard form of a circle centered at $(h,k)$ with radius $r$?

Answer: $(x-h)^2+(y-k)^2=r^2$. Distance formula squared: all points exactly $r$ units from $(h,k)$.

Flashcard 48: What is the solution set of $x^2+5x=0$?

Answer: $x=0$ and $x=-5$. Factor out $x$: $x(x+5)=0$, so $x=0$ or $x=-5$.

Flashcard 49: What is the inverse operation used to solve $x^2=k$ for real $x$?

Answer: $x=\pm\sqrt{k}$ (requires $k\ge 0$). Take square root of both sides; negative $k$ has no real solutions.

Flashcard 50: What is the substitution that turns a biquadratic $x^4+bx^2+c=0$ into a quadratic?

Answer: Let $u=x^2$. Reduces the degree from 4 to 2, making it solvable by quadratic formula.

Flashcard 51: What domain restriction must be stated for $\frac{1}{x-3}$ before solving equations?

Answer: $x\ne 3$. The denominator cannot be zero, so exclude $x=3$.

Flashcard 52: Identify the method: What is the first step to solve $(x-3)(x+5)=0$?

Answer: Set each factor equal to $0$. Zero product property: if $AB=0$, then $A=0$ or $B=0$.

Flashcard 53: What is the quadratic formula for solving $ax^2+bx+c=0$ in terms of $a$, $b$, and $c$?

Answer: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Derived using completing the square on the general quadratic equation.

Flashcard 54: Identify the solutions of $\sqrt{x-1}=3$.

Answer: $x=10$. Square both sides: $x-1=9$, so $x=10$.

Flashcard 55: What is the axis of symmetry for $y=ax^2+bx+c$?

Answer: $x=-\frac{b}{2a}$. The $x$-coordinate of the vertex, halfway between roots.

Flashcard 56: What does $\Delta<0$ indicate about the real solutions of $ax^2+bx+c=0$?

Answer: No real solutions. Negative discriminant means $\sqrt{\Delta}$ is imaginary.

Flashcard 57: What is the standard form of a quadratic equation to solve for its zeros?

Answer: $ax^2+bx+c=0$ with $a\ne 0$. Standard form requires the equation to equal zero with $a \ne 0$.

Flashcard 58: What is the first step to solve an equation with a rational expression in $x$?

Answer: Multiply both sides by the LCD. Clears denominators to create a polynomial equation.

Flashcard 59: What is the solution set of $\frac{1}{x}=2$?

Answer: $x=\frac{1}{2}$. Multiply both sides by $x$: $1=2x$, so $x=\frac{1}{2}$.

Flashcard 60: What does $b^2-4ac<0$ tell you about the real solutions of $ax^2+bx+c=0$?

Answer: No real solutions. Negative discriminant gives imaginary square roots, no real values.

Flashcard 61: What does $\Delta<0$ tell you about the real solutions of $ax^2+bx+c=0$?

Answer: No real solutions. Negative discriminant means square root of negative number, which isn't real.

Flashcard 62: What is the vertex $x$-coordinate of $y=ax^2+bx+c$ in terms of $a$ and $b$?

Answer: $x=-\frac{b}{2a}$. The vertex lies on the axis of symmetry.

Flashcard 63: Solve for $x$: $x^2-5x+6=0$.

Answer: $x=2$ or $x=3$. Factors to $(x-2)(x-3)=0$ using zero-product property.

Flashcard 64: Identify the substitution that makes $x^4-5x^2+4=0$ quadratic in one variable.

Answer: Let $u=x^2$. Transforms to $u^2-5u+4=0$, a standard quadratic form.

Flashcard 65: What is the solution to the system $y=x^2$ and $y=9$?

Answer: $(x,y)=(3,9)$ and $(x,y)=(-3,9)$. Substitute $y=9$ into $y=x^2$: $x^2=9$, so $x=\pm 3$.

Flashcard 66: Solve for $x$: $x^2-9=0$.

Answer: $x=\pm 3$. Factor as $(x-3)(x+3)=0$, giving $x=3$ or $x=-3$.

Flashcard 67: State the factored form of a quadratic with zeros $r_1$ and $r_2$.

Answer: $y=a(x-r_1)(x-r_2)$. Each factor $(x-r)$ equals zero when $x$ equals that root.

Flashcard 68: What does $\Delta=b^2-4ac>0$ imply about the solutions of $ax^2+bx+c=0$?

Answer: $2$ distinct real solutions. Positive discriminant means the square root is real.

Flashcard 69: Solve $x^2=49$ for $x$.

Answer: $x=\pm 7$. Take square root of both sides: $x=\pm\sqrt{49}$.

Flashcard 70: What is the quadratic formula for solving $ax^2+bx+c=0$?

Answer: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Derived using completing the square on the general quadratic equation.

Flashcard 71: What are the solutions of $(x-2)(x+5)=0$?

Answer: $x=2$ or $x=-5$. Apply zero-product property: each factor equals zero.

Flashcard 72: What is the quadratic formula for solving $ax^2+bx+c=0$?

Answer: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Derived from completing the square on the standard form.

Flashcard 73: What are the solutions to $x^2=49$?

Answer: $x=\pm 7$. $\sqrt{49}=7$, so $x=\pm 7$.

Flashcard 74: Solve for $x$: $x^2=49$.

Answer: $x=\pm 7$. Take square roots of both sides: $x=\pm\sqrt{49}=\pm 7$.

Flashcard 75: What is the vertex form of a quadratic function?

Answer: $y=a(x-h)^2+k$. Shows parabola with vertex at $(h,k)$ and vertical stretch factor $a$.

Flashcard 76: Solve $x^2-5x=0$ for $x$.

Answer: $x=0$ or $x=5$. Factor out $x$: $x(x-5)=0$, then use zero-product property.

Flashcard 77: What are the solutions to $(x-4)(x+1)=0$?

Answer: $x=4$ or $x=-1$. Apply zero-product property to each factor.

Flashcard 78: What is the solution set of $\frac{2}{x-1}=1$?

Answer: $x=3$. Cross-multiply: $2=x-1$, so $x=3$.

Flashcard 79: What are the solutions to $(x-4)^2=25$?

Answer: $x=9$ or $x=-1$. Take square root: $x-4=\pm 5$, so $x=4\pm 5$.

Flashcard 80: Solve for $x$: $x^2+4x+5=0$.

Answer: No real solutions. Discriminant is $16-20=-4<0$, so no real solutions exist.

Flashcard 81: What is the solution set of $\sqrt{x+2}=x$?

Answer: $x=2$. Square both sides: $x+2=x^2$; solve $x^2-x-2=0$ to get $x=2$ (reject $x=-1$).

Flashcard 82: What are the solutions to $(x-4)(x+1)=0$?

Answer: $x=4$ or $x=-1$. Apply zero-product property: each factor can equal zero.

Flashcard 83: What is the axis of symmetry of $y=ax^2+bx+c$ written in terms of $a$ and $b$?

Answer: $x=-\frac{b}{2a}$. The parabola's line of symmetry passes through the vertex.

Flashcard 84: Solve for $x$: $x^2-5x=0$.

Answer: $x=0$ or $x=5$. Factor out $x$: $x(x-5)=0$, so $x=0$ or $x=5$.

Flashcard 85: What are the solutions to $x^2=16$?

Answer: $x=\pm 4$. Take square root of both sides: $x=\pm\sqrt{16}$.

Flashcard 86: What is the solution of $\sqrt{x-1}=4$?

Answer: $x=17$. Square both sides: $x-1=16$, so $x=17$.

Flashcard 87: Solve for $x$: $3x^2-12=0$.

Answer: $x=\pm 2$. Add 12 to both sides: $3x^2=12$, so $x^2=4$, thus $x=\pm 2$.

Flashcard 88: Solve for $x$: $x^{\frac{1}{2}}=3$.

Answer: $x=9$. Square both sides: $(x^{\frac{1}{2}})^2=3^2$.

Flashcard 89: What does $\Delta<0$ imply about the solutions of $ax^2+bx+c=0$?

Answer: No real solutions (two complex solutions). Negative discriminant requires imaginary numbers for the square root.

Flashcard 90: What is the solution to $x^2-6x+9=0$?

Answer: $x=3$. Perfect square: $(x-3)^2=0$, so $x=3$ (double root).

Flashcard 91: What does $b^2-4ac>0$ tell you about the real solutions of $ax^2+bx+c=0$?

Answer: Two distinct real solutions. Positive discriminant means the square root is real, yielding two values.

Flashcard 92: Solve for $x$: $\sqrt{x+1}=3$.

Answer: $x=8$. Square both sides: $x+1=9$, so $x=8$.

Flashcard 93: What does it mean if $\Delta<0$ for $ax^2+bx+c=0$?

Answer: No real solutions. Negative discriminant means the parabola doesn't cross the x-axis.

Flashcard 94: What does $b^2-4ac=0$ tell you about solutions of $ax^2+bx+c=0$?

Answer: $1$ real solution (a double root). Zero discriminant means $\pm\sqrt{0}=0$, giving one repeated root.

Flashcard 95: Solve $\frac{1}{x-2}=3$ for $x$.

Answer: $x=\frac{7}{3}$. Cross-multiply: $1=3(x-2)$, so $1=3x-6$.

Flashcard 96: Identify the $x$-intercepts of $y=a(x-r_1)(x-r_2)$.

Answer: $x=r_1$ and $x=r_2$. The parabola crosses the $x$-axis where each factor equals zero.

Flashcard 97: What does $\Delta=0$ indicate about the real solutions of $ax^2+bx+c=0$?

Answer: One real solution (a double root). Zero discriminant makes $\pm\sqrt{0}=0$, giving one repeated root.

Flashcard 98: What does $b^2-4ac<0$ tell you about solutions of $ax^2+bx+c=0$?

Answer: No real solutions. Negative discriminant means square root of negative number is not real.

Flashcard 99: What is the zero-product property used after factoring an equation like $(x-3)(x+2)=0$?

Answer: If $ab=0$, then $a=0$ or $b=0$. A product equals zero only when at least one factor is zero.

Flashcard 100: What is the standard form of a quadratic equation in $x$?

Answer: $ax^2+bx+c=0$ with $a\ne 0$. The general form with $a$ as the coefficient of $x^2$.