Learn how to use the discriminant (\( \Delta \)), the complete formula and the reduced formula with step-by-step solutions. Includes analysis of real, double or impossible roots, simplification of radicals and handling of the standard form.
Exercise 1— level ★☆☆☆☆
\[ x^2 - 4x + 3 = 0 \]
Answer
\[ x_1 = 3 \qquad x_2 = 1 \]
Solution
Standard form and coefficients
The equation is already in standard form \(ax^2 + bx + c = 0\): \[ a = 1, \quad b = -4, \quad c = 3 \]
Discriminant
\[ \Delta = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 3 = 16 - 12 = 4 \]
Since \(\Delta > 0\), the equation has two distinct real solutions.
Complete formula
\[ x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2} \]
\[ x_1 = \frac{4 + 2}{2} = 3 \qquad x_2 = \frac{4 - 2}{2} = 1 \]
Reduced formula
Since \(b = -4 = 2k\), we have \(k = -2\). The reduced formula is: \[ x_{1,2} = \frac{-k \pm \sqrt{k^2 - ac}}{a} \]
\[ k^2 - ac = 4 - 3 = 1 \]
\[ x_{1,2} = \frac{2 \pm \sqrt{1}}{1} = 2 \pm 1 \]
\[ x_1 = 3 \qquad x_2 = 1 \]
Result
\[ \boxed{x_1 = 3 \qquad x_2 = 1} \]
Exercise 2 — level ★☆☆☆☆
\[ x^2 - 6x + 8 = 0 \]
Answer
\[ x_1 = 4 \qquad x_2 = 2 \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = -6, \quad c = 8 \]
Discriminant
\[ \Delta = (-6)^2 - 4 \cdot 1 \cdot 8 = 36 - 32 = 4 \]
Since \(\Delta > 0\), the equation has two distinct real solutions.
Complete formula
\[ x_{1,2} = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2} \]
\[ x_1 = \frac{6 + 2}{2} = 4 \qquad x_2 = \frac{6 - 2}{2} = 2 \]
Reduced formula
\(b = -6 = 2k\), so \(k = -3\): \[ k^2 - ac = 9 - 8 = 1 \]
\[ x_{1,2} = \frac{3 \pm \sqrt{1}}{1} = 3 \pm 1 \]
\[ x_1 = 4 \qquad x_2 = 2 \]
Result
\[ \boxed{x_1 = 4 \qquad x_2 = 2} \]
Exercise 3 — level ★☆☆☆☆
\[ x^2 + 2x - 3 = 0 \]
Answer
\[ x_1 = 1 \qquad x_2 = -3 \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = 2, \quad c = -3 \]
Discriminant
\[ \Delta = 2^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16 \]
Since \(\Delta > 0\), the equation has two distinct real solutions.
Complete formula
\[ x_{1,2} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2} \]
\[ x_1 = \frac{-2 + 4}{2} = 1 \qquad x_2 = \frac{-2 - 4}{2} = -3 \]
Reduced formula
\(b = 2 = 2k\), so \(k = 1\): \[ k^2 - ac = 1 - 1 \cdot (-3) = 1 + 3 = 4 \]
\[ x_{1,2} = \frac{-1 \pm \sqrt{4}}{1} = -1 \pm 2 \]
\[ x_1 = 1 \qquad x_2 = -3 \]
Result
\[ \boxed{x_1 = 1 \qquad x_2 = -3} \]
Exercise 4 — level ★★☆☆☆
\[ x^2 - 2x - 8 = 0 \]
Answer
\[ x_1 = 4 \qquad x_2 = -2 \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = -2, \quad c = -8 \]
Discriminant
\[ \Delta = (-2)^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \]
Since \(\Delta > 0\), the equation has two distinct real solutions.
Complete formula
\[ x_{1,2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \]
\[ x_1 = \frac{2 + 6}{2} = 4 \qquad x_2 = \frac{2 - 6}{2} = -2 \]
Reduced formula
\(b = -2 = 2k\), so \(k = -1\): \[ k^2 - ac = 1 - 1 \cdot (-8) = 1 + 8 = 9 \]
\[ x_{1,2} = \frac{1 \pm \sqrt{9}}{1} = 1 \pm 3 \]
\[ x_1 = 4 \qquad x_2 = -2 \]
Result
\[ \boxed{x_1 = 4 \qquad x_2 = -2} \]
Exercise 5 — level ★★☆☆☆
\[ x^2 + 4x + 4 = 0 \]
Answer
\[ x_1 = x_2 = -2 \quad \text{(double root)} \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = 4, \quad c = 4 \]
Discriminant
\[ \Delta = 4^2 - 4 \cdot 1 \cdot 4 = 16 - 16 = 0 \]
Since \(\Delta = 0\), the equation has a double root.
Complete formula
\[ x_{1,2} = \frac{-4 \pm \sqrt{0}}{2} = \frac{-4}{2} = -2 \]
Reduced formula
\(b = 4 = 2k\), so \(k = 2\): \[ k^2 - ac = 4 - 4 = 0 \]
\[ x_{1,2} = \frac{-2 \pm \sqrt{0}}{1} = -2 \]
Result
\[ \boxed{x_1 = x_2 = -2} \]
Exercise 6 — level ★★☆☆☆
\[ x^2 - 6x + 9 = 0 \]
Answer
\[ x_1 = x_2 = 3 \quad \text{(double root)} \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = -6, \quad c = 9 \]
Observation
Note that \(x^2 - 6x + 9 = (x-3)^2\): this is the square of a binomial. We therefore expect a double root.
Discriminant
\[ \Delta = (-6)^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0 \]
Complete formula
\[ x_{1,2} = \frac{6 \pm \sqrt{0}}{2} = \frac{6}{2} = 3 \]
Reduced formula
\(b = -6 = 2k\), so \(k = -3\): \[ k^2 - ac = 9 - 9 = 0 \]
\[ x_{1,2} = \frac{3 \pm \sqrt{0}}{1} = 3 \]
Result
\[ \boxed{x_1 = x_2 = 3} \]
Exercise 7 — level ★★☆☆☆
\[ 2x^2 - 6x + 4 = 0 \]
Answer
\[ x_1 = 2 \qquad x_2 = 1 \]
Solution
Standard form and coefficients
\[ a = 2, \quad b = -6, \quad c = 4 \]
Discriminant
\[ \Delta = (-6)^2 - 4 \cdot 2 \cdot 4 = 36 - 32 = 4 \]
Since \(\Delta > 0\), the equation has two distinct real solutions.
Complete formula
\[ x_{1,2} = \frac{6 \pm \sqrt{4}}{2 \cdot 2} = \frac{6 \pm 2}{4} \]
\[ x_1 = \frac{6 + 2}{4} = 2 \qquad x_2 = \frac{6 - 2}{4} = 1 \]
Reduced formula
\(b = -6 = 2k\), so \(k = -3\): \[ k^2 - ac = 9 - 2 \cdot 4 = 9 - 8 = 1 \]
\[ x_{1,2} = \frac{3 \pm \sqrt{1}}{2} = \frac{3 \pm 1}{2} \]
\[ x_1 = \frac{3 + 1}{2} = 2 \qquad x_2 = \frac{3 - 1}{2} = 1 \]
Result
\[ \boxed{x_1 = 2 \qquad x_2 = 1} \]
Exercise 8 — level ★★☆☆☆
\[ x^2 + 2x + 5 = 0 \]
Answer
The equation has no real solutions (\(\Delta < 0\)).
Solution
Standard form and coefficients
\[ a = 1, \quad b = 2, \quad c = 5 \]
Discriminant
\[ \Delta = 2^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16 \]
Since \(\Delta < 0\), the equation has no real solutions. The square root of a negative number is not defined in the set of real numbers.
Complete formula
\[ x_{1,2} = \frac{-2 \pm \sqrt{-16}}{2} \]
The expression \(\sqrt{-16}\) is not real: this confirms the absence of solutions.
Reduced formula
\(b = 2 = 2k\), so \(k = 1\): \[ k^2 - ac = 1 - 5 = -4 \]
Even with the reduced formula, the value under the radical is negative, confirming the absence of real solutions.
Result
\[ \boxed{\Delta = -16 < 0 \implies \text{no real solutions}} \]
Exercise 9 — level ★★★☆☆
\[ 3x^2 - 12x + 9 = 0 \]
Answer
\[ x_1 = 3 \qquad x_2 = 1 \]
Solution
Standard form and coefficients
\[ a = 3, \quad b = -12, \quad c = 9 \]
Discriminant
\[ \Delta = (-12)^2 - 4 \cdot 3 \cdot 9 = 144 - 108 = 36 \]
Since \(\Delta > 0\), the equation has two distinct real solutions.
Complete formula
\[ x_{1,2} = \frac{12 \pm \sqrt{36}}{2 \cdot 3} = \frac{12 \pm 6}{6} \]
\[ x_1 = \frac{12 + 6}{6} = 3 \qquad x_2 = \frac{12 - 6}{6} = 1 \]
Reduced formula
\(b = -12 = 2k\), so \(k = -6\): \[ k^2 - ac = 36 - 3 \cdot 9 = 36 - 27 = 9 \]
\[ x_{1,2} = \frac{6 \pm \sqrt{9}}{3} = \frac{6 \pm 3}{3} \]
\[ x_1 = \frac{6 + 3}{3} = 3 \qquad x_2 = \frac{6 - 3}{3} = 1 \]
Result
\[ \boxed{x_1 = 3 \qquad x_2 = 1} \]
Exercise 10 — level ★★★☆☆
\[ x^2 - 8x + 12 = 0 \]
Answer
\[ x_1 = 6 \qquad x_2 = 2 \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = -8, \quad c = 12 \]
Discriminant
\[ \Delta = (-8)^2 - 4 \cdot 1 \cdot 12 = 64 - 48 = 16 \]
Complete formula
\[ x_{1,2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2} \]
\[ x_1 = \frac{8 + 4}{2} = 6 \qquad x_2 = \frac{8 - 4}{2} = 2 \]
Reduced formula
\(b = -8 = 2k\), so \(k = -4\): \[ k^2 - ac = 16 - 12 = 4 \]
\[ x_{1,2} = \frac{4 \pm \sqrt{4}}{1} = 4 \pm 2 \]
\[ x_1 = 6 \qquad x_2 = 2 \]
Result
\[ \boxed{x_1 = 6 \qquad x_2 = 2} \]
Exercise 11 — level ★★★☆☆
\[ 2x^2 + 8x + 6 = 0 \]
Answer
\[ x_1 = -1 \qquad x_2 = -3 \]
Solution
Standard form and coefficients
\[ a = 2, \quad b = 8, \quad c = 6 \]
Discriminant
\[ \Delta = 8^2 - 4 \cdot 2 \cdot 6 = 64 - 48 = 16 \]
Complete formula
\[ x_{1,2} = \frac{-8 \pm \sqrt{16}}{2 \cdot 2} = \frac{-8 \pm 4}{4} \]
\[ x_1 = \frac{-8 + 4}{4} = -1 \qquad x_2 = \frac{-8 - 4}{4} = -3 \]
Reduced formula
\(b = 8 = 2k\), so \(k = 4\): \[ k^2 - ac = 16 - 2 \cdot 6 = 16 - 12 = 4 \]
\[ x_{1,2} = \frac{-4 \pm \sqrt{4}}{2} = \frac{-4 \pm 2}{2} \]
\[ x_1 = \frac{-4 + 2}{2} = -1 \qquad x_2 = \frac{-4 - 2}{2} = -3 \]
Result
\[ \boxed{x_1 = -1 \qquad x_2 = -3} \]
Exercise 12 — level ★★★☆☆
\[ x^2 - 10x + 16 = 0 \]
Answer
\[ x_1 = 8 \qquad x_2 = 2 \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = -10, \quad c = 16 \]
Discriminant
\[ \Delta = (-10)^2 - 4 \cdot 1 \cdot 16 = 100 - 64 = 36 \]
Complete formula
\[ x_{1,2} = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2} \]
\[ x_1 = \frac{10 + 6}{2} = 8 \qquad x_2 = \frac{10 - 6}{2} = 2 \]
Reduced formula
\(b = -10 = 2k\), so \(k = -5\): \[ k^2 - ac = 25 - 16 = 9 \]
\[ x_{1,2} = \frac{5 \pm \sqrt{9}}{1} = 5 \pm 3 \]
\[ x_1 = 8 \qquad x_2 = 2 \]
Result
\[ \boxed{x_1 = 8 \qquad x_2 = 2} \]
Exercise 13 — level ★★★☆☆
\[ 3x^2 - 6x - 9 = 0 \]
Answer
\[ x_1 = 3 \qquad x_2 = -1 \]
Solution
Standard form and coefficients
\[ a = 3, \quad b = -6, \quad c = -9 \]
Discriminant
\[ \Delta = (-6)^2 - 4 \cdot 3 \cdot (-9) = 36 + 108 = 144 \]
Complete formula
\[ x_{1,2} = \frac{6 \pm \sqrt{144}}{2 \cdot 3} = \frac{6 \pm 12}{6} \]
\[ x_1 = \frac{6 + 12}{6} = 3 \qquad x_2 = \frac{6 - 12}{6} = -1 \]
Reduced formula
\(b = -6 = 2k\), so \(k = -3\): \[ k^2 - ac = 9 - 3 \cdot (-9) = 9 + 27 = 36 \]
\[ x_{1,2} = \frac{3 \pm \sqrt{36}}{3} = \frac{3 \pm 6}{3} \]
\[ x_1 = \frac{3 + 6}{3} = 3 \qquad x_2 = \frac{3 - 6}{3} = -1 \]
Result
\[ \boxed{x_1 = 3 \qquad x_2 = -1} \]
Exercise 14 — level ★★★☆☆
\[ 2x^2 - 4x - 6 = 0 \]
Answer
\[ x_1 = 3 \qquad x_2 = -1 \]
Solution
Standard form and coefficients
\[ a = 2, \quad b = -4, \quad c = -6 \]
Discriminant
\[ \Delta = (-4)^2 - 4 \cdot 2 \cdot (-6) = 16 + 48 = 64 \]
Complete formula
\[ x_{1,2} = \frac{4 \pm \sqrt{64}}{2 \cdot 2} = \frac{4 \pm 8}{4} \]
\[ x_1 = \frac{4 + 8}{4} = 3 \qquad x_2 = \frac{4 - 8}{4} = -1 \]
Reduced formula
\(b = -4 = 2k\), so \(k = -2\): \[ k^2 - ac = 4 - 2 \cdot (-6) = 4 + 12 = 16 \]
\[ x_{1,2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \]
\[ x_1 = \frac{2 + 4}{2} = 3 \qquad x_2 = \frac{2 - 4}{2} = -1 \]
Result
\[ \boxed{x_1 = 3 \qquad x_2 = -1} \]
Exercise 15 — level ★★★★☆
\[ x^2 + 6x + 2 = 0 \]
Answer
\[ x_1 = -3 + \sqrt{7} \qquad x_2 = -3 - \sqrt{7} \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = 6, \quad c = 2 \]
Discriminant
\[ \Delta = 6^2 - 4 \cdot 1 \cdot 2 = 36 - 8 = 28 = 4 \cdot 7 \]
We simplify \(\sqrt{28} = 2\sqrt{7}\).
Complete formula
\[ x_{1,2} = \frac{-6 \pm 2\sqrt{7}}{2} = -3 \pm \sqrt{7} \]
Reduced formula
\(b = 6 = 2k\), so \(k = 3\): \[ k^2 - ac = 9 - 2 = 7 \]
\[ x_{1,2} = \frac{-3 \pm \sqrt{7}}{1} = -3 \pm \sqrt{7} \]
The reduced formula leads directly to the result without having to simplify the radical.
Result
\[ \boxed{x_1 = -3 + \sqrt{7} \qquad x_2 = -3 - \sqrt{7}} \]
Exercise 16 — level ★★★★☆
\[ 2x^2 - 4x - 1 = 0 \]
Answer
\[ x_1 = \frac{2 + \sqrt{6}}{2} \qquad x_2 = \frac{2 - \sqrt{6}}{2} \]
Solution
Standard form and coefficients
\[ a = 2, \quad b = -4, \quad c = -1 \]
Discriminant
\[ \Delta = (-4)^2 - 4 \cdot 2 \cdot (-1) = 16 + 8 = 24 = 4 \cdot 6 \]
We simplify \(\sqrt{24} = 2\sqrt{6}\).
Complete formula
\[ x_{1,2} = \frac{4 \pm 2\sqrt{6}}{4} = \frac{2 \pm \sqrt{6}}{2} \]
Reduced formula
\(b = -4 = 2k\), so \(k = -2\): \[ k^2 - ac = 4 - 2 \cdot (-1) = 4 + 2 = 6 \]
\[ x_{1,2} = \frac{2 \pm \sqrt{6}}{2} \]
The reduced formula avoids the simplification of \(\sqrt{24}\) and provides the result directly.
Result
\[ \boxed{x_1 = \frac{2 + \sqrt{6}}{2} \qquad x_2 = \frac{2 - \sqrt{6}}{2}} \]
Exercise 17 — level ★★★★☆
\[ x^2 - 2x - 5 = 0 \]
Answer
\[ x_1 = 1 + \sqrt{6} \qquad x_2 = 1 - \sqrt{6} \]
Solution
Standard form and coefficients
\[ a = 1, \quad b = -2, \quad c = -5 \]
Discriminant
\[ \Delta = (-2)^2 - 4 \cdot 1 \cdot (-5) = 4 + 20 = 24 = 4 \cdot 6 \]
We simplify \(\sqrt{24} = 2\sqrt{6}\).
Complete formula
\[ x_{1,2} = \frac{2 \pm 2\sqrt{6}}{2} = 1 \pm \sqrt{6} \]
Reduced formula
\(b = -2 = 2k\), so \(k = -1\): \[ k^2 - ac = 1 - 1 \cdot (-5) = 1 + 5 = 6 \]
\[ x_{1,2} = \frac{1 \pm \sqrt{6}}{1} = 1 \pm \sqrt{6} \]
Result
\[ \boxed{x_1 = 1 + \sqrt{6} \qquad x_2 = 1 - \sqrt{6}} \]
Exercise 18 — level ★★★★★
\[ 3x^2 + 6x - 1 = 0 \]
Answer
\[ x_1 = \frac{-3 + 2\sqrt{3}}{3} \qquad x_2 = \frac{-3 - 2\sqrt{3}}{3} \]
Solution
Standard form and coefficients
\[ a = 3, \quad b = 6, \quad c = -1 \]
Discriminant
\[ \Delta = 6^2 - 4 \cdot 3 \cdot (-1) = 36 + 12 = 48 = 16 \cdot 3 \]
We simplify \(\sqrt{48} = 4\sqrt{3}\).
Complete formula
\[ x_{1,2} = \frac{-6 \pm 4\sqrt{3}}{6} = \frac{-3 \pm 2\sqrt{3}}{3} \]
Reduced formula
\(b = 6 = 2k\), so \(k = 3\): \[ k^2 - ac = 9 - 3 \cdot (-1) = 9 + 3 = 12 = 4 \cdot 3 \]
We simplify \(\sqrt{12} = 2\sqrt{3}\): \[ x_{1,2} = \frac{-3 \pm 2\sqrt{3}}{3} \]
The reduced formula halves the discriminant calculation: instead of \(\Delta = 48\) we work with \(k^2 - ac = 12\), which is more manageable.
Result
\[ \boxed{x_1 = \frac{-3 + 2\sqrt{3}}{3} \qquad x_2 = \frac{-3 - 2\sqrt{3}}{3}} \]
Exercise 19 — level ★★★★★
\[ 2x^2 - 8x + 3 = 0 \]
Answer
\[ x_1 = \frac{4 + \sqrt{10}}{2} \qquad x_2 = \frac{4 - \sqrt{10}}{2} \]
Solution
Standard form and coefficients
\[ a = 2, \quad b = -8, \quad c = 3 \]
Discriminant
\[ \Delta = (-8)^2 - 4 \cdot 2 \cdot 3 = 64 - 24 = 40 = 4 \cdot 10 \]
We simplify \(\sqrt{40} = 2\sqrt{10}\).
Complete formula
\[ x_{1,2} = \frac{8 \pm 2\sqrt{10}}{4} = \frac{4 \pm \sqrt{10}}{2} \]
Reduced formula
\(b = -8 = 2k\), so \(k = -4\): \[ k^2 - ac = 16 - 2 \cdot 3 = 16 - 6 = 10 \]
\[ x_{1,2} = \frac{4 \pm \sqrt{10}}{2} \]
Here too the reduced formula is more efficient: the value under the radical is \(10\) instead of \(40\), making the simplification immediate.
Result
\[ \boxed{x_1 = \frac{4 + \sqrt{10}}{2} \qquad x_2 = \frac{4 - \sqrt{10}}{2}} \]
Exercise 20 — level ★★★★★
\[ 5x^2 - 10x + 3 = 0 \]
Answer
\[ x_1 = \frac{5 + \sqrt{10}}{5} \qquad x_2 = \frac{5 - \sqrt{10}}{5} \]
Solution
Standard form and coefficients
\[ a = 5, \quad b = -10, \quad c = 3 \]
Discriminant
\[ \Delta = (-10)^2 - 4 \cdot 5 \cdot 3 = 100 - 60 = 40 = 4 \cdot 10 \]
We simplify \(\sqrt{40} = 2\sqrt{10}\).
Complete formula
\[ x_{1,2} = \frac{10 \pm 2\sqrt{10}}{10} = \frac{5 \pm \sqrt{10}}{5} \]
Reduced formula
\(b = -10 = 2k\), so \(k = -5\): \[ k^2 - ac = 25 - 5 \cdot 3 = 25 - 15 = 10 \]
\[ x_{1,2} = \frac{5 \pm \sqrt{10}}{5} \]
With the reduced formula the reduced discriminant is \(10\) instead of \(40\): the result can be read directly without further simplifications.
Result
\[ \boxed{x_1 = \frac{5 + \sqrt{10}}{5} \qquad x_2 = \frac{5 - \sqrt{10}}{5}} \]