equations
A equation has a highest of two and can be written as ax² + bx + c = 0, where a is not zero. Its solutions are values that make the whole expression zero. A can have two distinct real solutions, one repeated real solution, or no real solution. Finding one value and stopping may therefore miss part of the answer.
Factorisation rewrites a as a product. For x² − 5x + 6, look for two numbers whose product is 6 and whose sum is −5. They are −2 and −3, so the expression becomes (x − 2)(x − 3). The equation equals zero when either factor is zero; hence x = 2 or x = 3. Expanding the brackets verifies the factorisation.
If the of x² is not one, keep that in the factors. For example, 2x² + 7x + 3 factors as (2x + 1)(x + 3). Check all three terms on expansion. The roots are −1/2 and −3. A frequent error is to read a root straight from a factor without solving the small linear equation represented by that factor.
Completing the square uses the identity. Starting with x² + 6x + 5 = 0, add and subtract 9 to get (x + 3)² − 4 = 0. Therefore. Both signs matter, giving x = −1 or x = −5. This method also explains how the general formula is obtained.
The formula is ± √(b² − 4ac)]/(2a). Substitute signed values of a, b and c inside brackets. The discriminant, b² − 4ac, indicates the number of real roots: positive gives two, zero gives a repeated root, and negative gives no real roots. Round only at the end when an exact surd answer is not required.
A model must still be interpreted. A rectangle with width x and length x + 3 has area x(x + 3). If its area is 28 square units, x² + 3x − 28 = 0 gives x = 4 or x = −7. Only the positive width fits the geometry. State the dimensions with units and explain why the unsuitable root is rejected.
Recognise the structure before choosing a method
A equation can be written ax² + bx + c = 0 with a ≠ 0. Always collect every term on one side before identifying a, b and c. In, the standard form is 2x² − 7x − 4 = 0, so b is −7, not 7. The signs belong to the terms. equations can have two distinct real solutions, one repeated real solution, or no real solution. A sketch and the discriminant help explain these possibilities.
Factorisation is efficient when suitable integer factors are visible. For x² − x − 12 = 0, find two numbers with product −12 and sum −1: −4 and 3. Therefore (x − 4)(x + 3) = 0. A product of two real numbers is zero only when at least one factor is zero, so x = 4 or x = −3. This zero-product rule applies after the equation has zero on the other side. You cannot conclude x = 4 or x = −3 directly from (x − 4)(x + 3) = 10.
Factor a with a before x²
For 2x² + 7x + 3, split the middle term using numbers whose product is 2 × 3 = 6 and sum is 7. Write 2x² + 6x + x + 3, then group as 2x(x + 3) + 1(x + 3). The expression factors to (2x + 1)(x + 3). If it equals zero, or x = −3. Expand the factors to check them. A factorisation that produces the correct constant but the wrong middle is not valid.
Completing the square exposes the turning point of a . Since, the equation x² + 6x + 5 = 0 becomes (x + 3)² = 4. Taking square roots gives, so x = −1 or x = −5. The symbol ± matters: both 2 and −2 have square 4. For, the minimum point is (−3, −4), because a square cannot be negative.
Interpret the formula
For ax² + bx + c = 0, ± √(b² − 4ac)]/(2a). The entire numerator is divided by 2a. Let. If D > 0, the graph crosses the x-axis at two distinct points; if D = 0, it touches at one repeated root; if D < 0, it has no real x-intercepts. For x² + 2x + 5 = 0, D = 4 − 20 = −16, so no real solutions exist. This is a mathematical conclusion, not an arithmetic failure.
A can model an area or a height, but the context can restrict which root is usable. A rectangle has width w metres and length w + 3 metres, with area 40 m². Then w(w + 3) = 40, so w² + 3w − 40 = 0 and (w + 8)(w − 5) = 0. The algebraic roots are −8 and 5. Ordinary width must be positive, so w = 5 m and the length is 8 m. State why the negative root is rejected rather than silently omitting it.
Connect roots, graphs and simultaneous equations
For roots r and s, the monic is x² − (r + s)x + rs. More generally the roots of ax² + bx + c = 0 satisfy r + s = −b/a and. This provides a quick check, not a substitute for deriving uncertain roots. Intersecting a line with a also produces a equation. If and. Factoring x² − x − 2 gives x = 2 or −1, then the points are (2, 4) and (−1, 1).
Self-check: solve x² − 9 = 0 and decide how many real roots 3x² − 6x + 3 = 0 has. The first roots are 3 and −3, using a difference of squares. The second has D = 36 − 36 = 0 and repeated root 1. Avoid dividing x² = 3x by x without considering x = 0: factor x(x − 3) = 0 to preserve both roots. When using a calculator, enter negative coefficients in brackets and retain unrounded intermediate values until the required final accuracy.
Do not confuse solving with evaluating
Solving x² − 5x + 6 = 0 gives inputs 2 and 3. Evaluating the same expression at x = 4 gives output 2. A root is an input where the output is zero; it is not every output produced by the expression. State roots individually or as a solution set, and preserve exact fractions or surds when an approximation is not requested.