Two facts about the answers, read straight off the equation.
Without solving, find the sum of the roots
2x² − 7x + 3 = 0
The sum of the roots is −ba = −−72 = 72. D misses that b is already negative, so −b comes out positive. A is the product, ca = 32.
For any ax² + bx + c = 0 with roots r₁ and r₂.
r₁ + r₂ = −ba
note the minus
r₁ · r₂ = ca
no minus here
Both sit over a, so a quadratic starting with a bare x² makes them even easier — the sum is just −b and the product is just c.
2x² − 7x + 3 = 0
a
2
b
−7
c
3
b is −7, carrying its sign along.
sum = −−72
minus a minus
sum = 3.5
or 72
product = 32
straight from c ÷ a
product = 1.5
done
and the equation was never solved
Writing sum = ba drops the minus: −7 ÷ 2 = −3.5, the right size with the wrong sign. Writing sum = −b and forgetting the a gives 7, twice too big. The roots are 0.5 and 3, so the sum must be positive and below 7 — only 3.5 fits.
Multiply a factored quadratic back out and watch.
(x − r₁)(x − r₂) = 0
roots r₁ and r₂
x² − (r₁ + r₂)x + r₁r₂ = 0
FOIL it out
is −(sum), so sum = −b
is the product, so product = c
Divide the whole line by a to handle a leading coefficient, and the two formulas appear.
Given two roots, you can write the quadratic in one line.
x² − (sum)x + (product) = 0
roots 5 and −2
the target
sum = 3, product = −10
5 + (−2), 5 · (−2)
x² − 3x − 10 = 0
the equation
Multiple-choice exams love this direction — it is far quicker than expanding (x − 5)(x + 2) by hand.
A fractional sum or product: multiply the whole line by the LCD. Sum 34 and product −12 give x² − (34)x − 12 = 0; times 4, that is 4x² − 3x − 2 = 0.
Rewrite each one using only r + s and rs.
| We want | Rewrite it as |
|---|---|
| r² + s² | (r + s)² − 2rs |
| 1r + 1s | (r + s) ÷ rs |
| (r − s)² | (r + s)² − 4rs |
| r²s + rs² | rs(r + s) |
r + s = 5, rs = 2
−ba and ca
r² + s² = 25 − 4 = 21
(r + s)² − 2rs
1r + 1s = 52
sum over product
(r − s)² = 25 − 8 = 17
(r + s)² − 4rs
Squaring the sum gives 25, which still contains the 2rs. Take it back out, or the answer is too big by twice the product.
Three common disguises of the same two shortcuts.
Terms out of order, negative a: 5 − 2x² = 3x
−2x² − 3x + 5 = 0
everything on one side, x² first
sum = −−3−2 = −32
a = −2, b = −3
product = 5−2 = −52
c = 5
A negative a flips both signs. The roots are 1 and −52, which do add to −32 and multiply to −52.
Building from surd roots: 3 + √2 and 3 − √2
sum = 6
the surds cancel
product = 3² − (√2)² = 9 − 2 = 7
(p + q)(p − q) = p² − q²
x² − 6x + 7 = 0
x² − (sum)x + (product)
Conjugate roots always give whole-number coefficients. Expanding (x − 3 − √2)(x − 3 + √2) term by term works too, but takes three times as long.
New roots from old: the roots of x² − 3x − 5 = 0, each doubled
r + s = 3, rs = −5
from the old equation
2r + 2s = 2(3) = 6
new sum
(2r)(2s) = 4rs = −20
new product
x² − 6x − 20 = 0
x² − (sum)x + (product)
The old roots are surds, 3 ± √292, and we never needed them.
Each root is doubled, so the product picks up 2 × 2. Writing x² − 6x − 10 = 0 doubles the product only once.
Name both roots with one letter, then use the sum.
roots r and 4r
the condition
r + 4r = 15, so r = 3
the sum is −ba = 15
roots 3 and 12
k = 3 × 12 = 36
k is the product
Other ties work the same way. Roots that differ by 3 in x² − 9x + k = 0 add to 9 and differ by 3, so they are 6 and 3 and k = 18. If one root is given, the other is the sum minus that root. When the condition fixes the product instead, use c: roots of x² − kx + 50 = 0 in the ratio 1 : 2 are t and 2t, so 2t² = 50, t = 5, the roots are 5 and 10, and k = 15 (taking k > 0).
These items reward not solving: an expression in r and s, a condition linking the roots, or an equation built from a given sum and product. Rewrite in r + s and rs first. If a value looks odd, test the method on a quadratic whose roots you know.
2x² − 7x + 3 factors as (2x − 1)(x − 3).
Find the product of the roots
3x² + 11x − 4 = 0
The product is ca = −43. B is the sum (−ba = −113). Both read straight off the coefficients with no solving at all — which is exactly why the exam likes this question.
Hint: c is negative, so the product comes out negative — meaning the two roots have opposite signs. That is a useful sanity check on its own.
