Solve quadratic equations

1. The equation has x^2 alone, so take the square root of both sides.
2. x = \pm\sqrt{64} = \pm 8
3. Both values work, so the solutions are x = 8 and x = -8.

Check: 8^2 = 8 \times 8 = 64 ✓ and (-8)^2 = (-8) \times (-8) = 64 ✓ (a negative times a negative is positive).

Tip: When you take the square root to solve an equation, always write \pm. A positive number has two square roots.

1. The equation has no x-term, so use the square-root method. First isolate x^2.
2. Divide both sides by 2: x^2 = \frac{72}{2} = 36
3. Take the square root of both sides. Both 6^2 = 36 and (-6)^2 = 36, so there are two solutions: x = \pm\sqrt{36} = \pm 6
4. The solutions are x = 6 and x = -6.

Check: 2(6)^2 = 2(36) = 72 ✓ and 2(-6)^2 = 2(36) = 72 ✓

Tip: When you take the square root to solve an equation, always write \pm. A positive number has two square roots, so x^2 = k (with k > 0) has two solutions.

1. Factor x^2 - 9x + 20. We need two numbers whose product is 20 and whose sum is -9.
2. Try factor pairs of 20 with both numbers negative (a positive product and a negative sum):
   • -1 and -20: sum -21 ✗
   • -2 and -10: sum -12 ✗
   • -4 and -5: sum -9 ✓
3. So x^2 - 9x + 20 = (x - 4)(x - 5) = 0.
4. Zero-product property: x - 4 = 0 or x - 5 = 0, so x = 4 or x = 5.

Check: 4^2 - 9(4) + 20 = 16 - 36 + 20 = 0 ✓ and 5^2 - 9(5) + 20 = 25 - 45 + 20 = 0 ✓

Tip: The numbers inside the factors have the opposite sign of the solutions: (x - 4) gives x = +4.

1. Start with x^2 + 6 = 42. To isolate x^2, undo "+ 6" by subtracting 6 from both sides: x^2 = 42 - 6 = 36
2. Jordan wrote x^2 = 48, which means he added 6 instead. So the first error is in Step 1.
3. Steps 2 and 3 are done correctly from his wrong value: \pm\sqrt{48} \approx \pm 6.93. But they start from a wrong number.
4. Correct solution: x = \pm\sqrt{36} = \pm 6.

Check: Jordan's answer fails: 6.93^2 + 6 \approx 48 + 6 = 54 \ne 42. The correct answer works: 6^2 + 6 = 42 ✓

Tip: In error questions, plug the final answer back into the original equation. If it fails, check each step from the top — the first wrong step is the answer.

1. The equation is in standard form with b = 3 and c = -10.
2. Find two numbers that multiply to -10 and add to 3: 5 \times (-2) = -10 and 5 + (-2) = 3.
3. Factor: (x + 5)(x - 2) = 0
4. Zero product property: x + 5 = 0 \Rightarrow x = -5; x - 2 = 0 \Rightarrow x = 2.
5. The solutions are x = -5 or x = 2.

Check: (-5)^2 + 3(-5) - 10 = 25 - 15 - 10 = 0 ✓; 2^2 + 3(2) - 10 = 4 + 6 - 10 = 0 ✓

Tip: Check your factors by multiplying back: (x + 5)(x - 2) = x^2 - 2x + 5x - 10 = x^2 + 3x - 10 ✓

1. Standard form ax^2 + bx + c = 0: a = 3, b = -5, c = -1.
2. Quadratic formula: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
3. -b = -(-5) = 5.
4. Discriminant: b^2 - 4ac = (-5)^2 - 4(3)(-1) = 25 - (-12) = 25 + 12 = 37.
5. Denominator: 2a = 2(3) = 6.
6. x = \frac{5 \pm \sqrt{37}}{6} So the answer is \dfrac{5 \pm \sqrt{37}}{6}.

Check: \sqrt{37} \approx 6.083, so x \approx \tfrac{11.083}{6} \approx 1.847. Then 3(1.847)^2 - 5(1.847) - 1 \approx 10.235 - 9.235 - 1 \approx 0 ✓

Tip: Write a, b, c with their signs before you substitute. A negative c makes -4ac positive.

1. Take the square root of both sides: x + 2 = \pm\sqrt{25} = \pm 5
2. Case 1: x + 2 = 5 \Rightarrow x = 5 - 2 = 3.
3. Case 2: x + 2 = -5 \Rightarrow x = -5 - 2 = -7.
4. The solutions are x = 3 or x = -7.

Check: (3 + 2)^2 = 5^2 = 25 ✓ and (-7 + 2)^2 = (-5)^2 = 25 ✓

Tip: With a squared group, there's no need to multiply it out — take the square root (with \pm) and solve two simple equations.

1. Move every term to one side so the other side is 0: x^2 - 7x = 0
2. Both terms share a factor of x. Factor it out (the GCF): x(x - 7) = 0
3. Zero-product property: x = 0 or x - 7 = 0.
4. So the solutions are x = 0 or x = 7.

Check: 0^2 = 7(0) gives 0 = 0 ✓. 7^2 = 49 and 7(7) = 49 ✓

Tip: Never divide both sides by x — x might be 0, and dividing throws that solution away. Instead, move everything to one side and factor out x.

1. Identify the coefficients: a = 3, b = -5, c = 4.
2. Compute the discriminant: b^2 - 4ac = (-5)^2 - 4(3)(4) = 25 - 48 = -23
3. Interpret it:
   • positive → two real solutions
   • zero → exactly one real solution
   • negative → no real solutions
4. Since -23 < 0, the answer is -23; no real solutions.

Check: The quadratic formula would need \sqrt{-23}. There is no real number whose square is negative, so no real solution exists ✓

Tip: (-5)^2 is +25 — squaring a negative gives a positive. And keep the full 4ac: here 4 \cdot 3 \cdot 4 = 48.

1. Match 3x^2 + 2x - 5 = 0 to ax^2 + bx + c = 0: a = 3, b = 2, c = -5.
2. Substitute into x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}: x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-5)}}{2(3)}
3. Simplify carefully: 2^2 = 4 and -4(3)(-5) = +60, and 2(3) = 6: x = \frac{-2 \pm \sqrt{4 + 60}}{6}
4. So the correct setup is x = \dfrac{-2 \pm \sqrt{4 + 60}}{6}.

Check: Finish it: \sqrt{64} = 8, so x = \tfrac{6}{6} = 1 or x = \tfrac{-10}{6} = -\tfrac{5}{3}. Test x = 1: 3 + 2 - 5 = 0 ✓

Tip: When c is negative, -4ac becomes positive (a negative times a negative). The formula is on the GED formula sheet, but the signs are up to you.

1. Each solution r gives a factor (x - r):
   • x = 5 → (x - 5)
   • x = -3 → (x - (-3)) = (x + 3)
2. So the equation is (x + 3)(x - 5) = 0.
3. Multiply out (FOIL): x^2 - 5x + 3x - 15 = x^2 - 2x - 15
4. The equation is x^2 - 2x - 15 = 0.

Check: x = -3: 9 + 6 - 15 = 0 ✓. x = 5: 25 - 10 - 15 = 0 ✓

Tip: You can also test each answer choice: substitute both solutions. The right equation gives 0 for both.

1. Rewrite in standard form by subtracting 5x and 2 from both sides: 3x^2 - 5x - 2 = 0
2. a = 3, b = -5, c = -2. Discriminant: (-5)^2 - 4(3)(-2) = 25 + 24 = 49.
3. x = \frac{5 \pm \sqrt{49}}{2(3)} = \frac{5 \pm 7}{6}
4. x = \dfrac{5 + 7}{6} = 2 or x = \dfrac{5 - 7}{6} = -\dfrac{2}{6} = -\dfrac{1}{3}.
5. The solutions are x = 2 or x = -\tfrac{1}{3}. (By factoring: (3x + 1)(x - 2) = 0.)

Check: x = 2: 3(2)^2 = 12 and 5(2) + 2 = 12 ✓. x = -\tfrac{1}{3}: 3 \cdot \tfrac{1}{9} = \tfrac{1}{3} and 5(-\tfrac{1}{3}) + 2 = -\tfrac{5}{3} + \tfrac{6}{3} = \tfrac{1}{3} ✓

Tip: Always get "= 0" first. When moving a term across the equals sign, its sign changes.

1. Rewrite in standard form by subtracting 7x from both sides: 2x^2 - 7x + 3 = 0
2. Factor. We need (2x \;\_\;)(x \;\_\;) with a product of +3 and a middle term of -7x. Try (2x - 1)(x - 3): 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3 \;✓
3. Zero-product property:
   • 2x - 1 = 0 → 2x = 1 → x = \tfrac{1}{2}
   • x - 3 = 0 → x = 3
4. The solutions are x = \tfrac{1}{2} or x = 3.

Check: x = \tfrac{1}{2}: 2(\tfrac{1}{4}) + 3 = 3.5 and 7(\tfrac{1}{2}) = 3.5 ✓. x = 3: 18 + 3 = 21 = 7(3) ✓

Tip: When a \ne 1, always multiply your factors back out to test the middle term. Two different factor pairs can give the right first and last terms.

1. Isolate the squared group. Divide both sides by 4: (x - 1)^2 = 9
2. Take the square root of both sides with \pm: x - 1 = \pm 3
3. Solve each equation by adding 1: x = 1 + 3 = 4 \quad \text{or} \quad x = 1 - 3 = -2
4. The solutions are x = 4 or x = -2.

Check: 4(4 - 1)^2 = 4(9) = 36 ✓ and 4(-2 - 1)^2 = 4(9) = 36 ✓

Tip: Undo operations in reverse order: the 4 multiplies the square, so divide by 4 first, then take the root, then move the 1.

1. Identify a = 1, b = 4, c = -7.
2. Discriminant: 4^2 - 4(1)(-7) = 16 + 28 = 44.
3. Quadratic formula: x = \frac{-4 \pm \sqrt{44}}{2(1)} = \frac{-4 \pm \sqrt{44}}{2}
4. \sqrt{44} \approx 6.633.
   • Plus: \dfrac{-4 + 6.633}{2} = \dfrac{2.633}{2} \approx 1.3
   • Minus: \dfrac{-4 - 6.633}{2} = \dfrac{-10.633}{2} \approx -5.3
5. The solutions are x \approx 1.3 and x \approx -5.3.

Check: The two solutions should add to -\tfrac{b}{a} = -4: 1.3 + (-5.3) = -4 ✓. And 1.3^2 + 4(1.3) - 7 = 1.69 + 5.2 - 7 \approx -0.1 (close to 0 because of rounding) ✓

Tip: Divide the whole numerator by 2a, not just the square root.

1. Solving x^2 + x - 12 = 0 means finding the x-values that make y = 0.
2. Scan the y row for 0. It appears in two columns: x = -4 and x = 3.
3. So the solutions are x = -4 and x = 3.
4. A quadratic has at most two solutions, so we have found them all.

Check: Factor: x^2 + x - 12 = (x + 4)(x - 3), which gives x = -4 or x = 3 ✓. Also (-4)^2 + (-4) - 12 = 16 - 4 - 12 = 0 ✓

Tip: In a table, the solutions are the x-values in the columns where y = 0 — never the y-values themselves, and not where y is smallest.

1. Since x = 5 is a solution, substituting it makes the equation true: 2(5)^2 + b(5) - 15 = 0
2. Simplify: 2(25) = 50, so 50 + 5b - 15 = 0 \;\Rightarrow\; 35 + 5b = 0
3. Subtract 35: 5b = -35. Divide by 5: b = -7.
4. The equation is 2x^2 - 7x - 15 = 0.

Check: Factor: 2x^2 - 7x - 15 = (2x + 3)(x - 5), so x = 5 is a solution ✓ (the other is -\tfrac{3}{2}). Also 2(25) - 7(5) - 15 = 50 - 35 - 15 = 0 ✓

Tip: "One solution is…" means you can plug it in. Square first (5^2 = 25), then multiply by 2.

1. Step 1 is correct: x^2 - 4x - 5 has a = 1, b = -4, c = -5.
2. Step 2: The formula starts with -b. Since b = -4, -b = -(-4) = +4. Lena wrote -4, so Step 2 has the first error. (Her square root part is right: (-4)^2 - 4(1)(-5) = 16 + 20 = 36.)
3. Correct work: x = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}
4. So x = \tfrac{10}{2} = 5 or x = \tfrac{-2}{2} = -1.

Check: 5^2 - 4(5) - 5 = 25 - 20 - 5 = 0 ✓. Lena's x = 1: 1 - 4 - 5 = -8 \ne 0 ✗ — her answers do not work. (Factoring also gives (x - 5)(x + 1) = 0.)

Tip: -b means "the opposite of b." When b is negative, -b is positive.

1. Add 15x to both sides so the right side is 0: 6x^2 + 15x = 0
2. The GCF of 6x^2 and 15x is 3x. Factor it out: 3x(2x + 5) = 0
3. Zero-product property:
   • 3x = 0 → x = 0
   • 2x + 5 = 0 → 2x = -5 → x = -\tfrac{5}{2}
4. The solutions are x = 0 or x = -\tfrac{5}{2}.

Check: x = -\tfrac{5}{2}: 6(\tfrac{25}{4}) = 37.5 and -15(-\tfrac{5}{2}) = 37.5 ✓. x = 0: 0 = 0 ✓

Tip: Moving a term across the equal sign changes its sign: -15x on the right becomes +15x on the left. And keep x = 0 — don't divide by x.

1. Substitute x = 4: 4^2 + k(4) - 20 = 0 \Rightarrow 16 + 4k - 20 = 0 \Rightarrow 4k - 4 = 0
2. Solve: 4k = 4, so k = 1.
3. The equation is x^2 + x - 20 = 0.
4. Factor: two numbers that multiply to -20 and add to 1 are 5 and -4: (x + 5)(x - 4) = 0
5. Solutions: x = -5 and x = 4. The other solution is -5.

Check: (-5)^2 + (-5) - 20 = 25 - 5 - 20 = 0 ✓. Shortcut: when the x^2 coefficient is 1, the two solutions multiply to the constant: 4 \times (-5) = -20 ✓

Tip: In "work backwards" problems, use the given solution to find the missing constant first, then solve normally.

1. Move all terms to the left: subtract 6x and add 4: x^2 - 6x + 4 = 0 \quad (a = 1,\ b = -6,\ c = 4)
2. Discriminant: (-6)^2 - 4(1)(4) = 36 - 16 = 20.
3. Formula: x = \frac{6 \pm \sqrt{20}}{2}
4. Simplify \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}: x = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}
5. The answer is 3 \pm \sqrt{5}.

Check: 3 + \sqrt{5} \approx 5.236: 5.236^2 \approx 27.42 and 6(5.236) - 4 \approx 27.42 ✓

Tip: When you simplify \dfrac{6 \pm 2\sqrt{5}}{2}, divide both terms of the numerator by 2.

1. The right side is 7, not 0, so you cannot set each factor equal to 7. Multiply out first: (x + 5)(x - 1) = x^2 - x + 5x - 5 = x^2 + 4x - 5
2. Set up and subtract 7: x^2 + 4x - 5 = 7 \;\Rightarrow\; x^2 + 4x - 12 = 0
3. Factor: two numbers with product -12 and sum 4 are 6 and -2: (x + 6)(x - 2) = 0
4. So x = -6 or x = 2.

Check: x = 2: (7)(1) = 7 ✓. x = -6: (-1)(-7) = 7 ✓

Tip: If a product equals a nonzero number, many pairs of factors are possible, so there is no shortcut. Expand, move everything to one side to get 0, then factor.