Three methods, one decision — and the decision takes about five seconds.
What is the positive solution to the given equation?
2x2 + 7x = 15
Move the 15 over: 2x² + 7x − 15 = 0, which factors as (2x − 3)(x + 5) = 0, so x = 32 or x = −5. The positive one is 32.C reads the 3 straight out of (2x − 3) and forgets to divide by the 2 — the single most common slip when a is not 1. D reads 5 out of (x + 5) and ignores the sign; that factor gives x = −5. B comes from the wrong split, (2x − 5)(x + 3), which multiplies out to 2x² + x − 15.
1
Zero on one side, then factor — because only zero forces a factor to vanish.
If A · B = 0 then A = 0 or B = 0. That is the zero-product property, and it is false for every other number.
x2 + 5x = 24
not zero yet — nothing works
x2 + 5x − 24 = 0
− 24 both sides
(x + 8)(x − 3) = 0
8 · (−3) = −24 and 8 + (−3) = 5
x = −8 or x = 3
set each bracket to zero
x = −8 or x = 3
When a = 1, factoring is a search for two numbers: they multiply to c and add to b.
Signs are the whole game — a negative c means the two numbers have opposite signs.
Never divide both sides by x
From x² = 7x it is tempting to cancel an x and write x = 7. That throws away
x = 0, which really is a solution. Move everything over instead:
x² − 7x = 0, so x(x − 7) = 0 and x = 0 or x = 7. Dividing by a variable
destroys a root every time, and the SAT stocks that missing root in the choices.
2
When a is not 1, split the middle term.
Multiply a by c, find two numbers with that product and the right sum, then group in pairs.
2x² + 7x − 15ac2(−15) = −30need sum+7so+10 and −3
2x2 + 10x − 3x − 15 = 0
7x split into 10x − 3x
2x(x + 5) − 3(x + 5) = 0
factor each pair
(x + 5)(2x − 3) = 0
the bracket is the common factor
x = −5 or 2x = 3
each bracket to zero
x = −5 or x = 32
divide that one by 2
x = −5 or x = 32
The favorite distractor is 3
A bracket like (2x − 3) does not give x = 3. It gives 2x = 3, so x = 32.
Whenever the leading coefficient is not 1 at least one root is a fraction, and
the whole-number version of it is always on the answer list.
3
No linear term? Take square roots and keep both signs.
Whenever the x appears only inside a square, the fastest route is one square root — never the formula.
3x² − 75 = 0
x² = 25
x = ±5
two solutions, one line
(2x + 1)² = 36
2x + 1 = ±6
2x = 5 or 2x = −7
x = 52 or x = −72
The square root need not come out whole. x² = 20 gives x = ±√20 = ±2√5, and the
SAT is perfectly happy to put 2√5 in the choices.
± or you lose half the answer
√36 by itself means 6, the non-negative root. But solving y² = 36 is a
different question, and its answers are 6 and −6. Items that ask for "the
positive solution" or "which of the following is a solution" exist because
one of the two is so often dropped.
4
The quadratic formula never fails — it just costs more.
Write ax² + bx + c = 0 first, then substitute a, b and c with their signs.
x =
−b ± √(b2 − 4ac)2a
a = 2b = −5c = −4
2x2 − 5x − 4 = 0
already equals zero
b2 − 4ac = (−5)2 − 4(2)(−4)
inside the root first
= 25 + 32 = 57
two minuses make a plus
x = (5 ± √57) ÷ 4
−b = −(−5) = +5, and 2a = 4
x = 5 + √574 ≈ 3.14 or x = 5 − √574 ≈ −0.64
SAT answers often arrive in the form "6 + √k2", so the last move is usually
simplifying the radical and cancelling.
2x2 − 8x + 2 = 0 → x = (8 ± √48) ÷ 4
64 − 16 = 48
√48 = √16 · 3 = 4√3
pull out the largest perfect square
x = (8 ± 4√3) ÷ 4 = 2 ± √3
the 4 divides both terms
Cancel the whole numerator or none of it
8 ± 4√34 is 2 ± √3, not 8 ± √3 and not 2 ± 4√3. The 4 underneath has to
divide the 8 and the 4√3. If the numerator were (8 ± 2√3) nothing would cancel
cleanly, and the answer would stay as 4 ± √32.
5
Choose the method in five seconds, not five lines.
Glance at the equation, then at the answer choices. Both tell you which tool to reach for.
What you see
Use
Because
no x term, or ( … )² = number
square roots
one line, two answers
small whole-number coefficients
factor
fastest when it works
choices contain a √
the formula
it was never going to factor
a constant such as k in the equation
the formula or the discriminant
there is nothing to factor yet
Still unsure? Compute b² − 4ac first. If it is a perfect square the quadratic
factors over the integers, so hunting for the factors is worth it. If it is not,
stop hunting and write the formula.
Sometimes the fastest method is typing it into Desmos.
The Digital SAT gives you a graphing calculator for the whole section. Enter y = 2x² − 5x − 4 and click the x-intercepts.
graph it
choices are decimals or integers
"which is a solution"
"how many solutions"
coefficients are ugly
do it by hand
choices are exact radicals
"can be written as a + √k2"
a letter stands in for a number
the sum or product of the roots
The graph still helps when the answer is exact
Desmos gives −0.637 and 3.137 for this equation. That will not match a choice
written as 5 + √574 — but you can type that choice into Desmos too and see
3.137 come back. Reading a decimal off the graph and then matching it against
the radical choices is often quicker than simplifying by hand, and it is
immune to sign slips.
✓
Check: every root goes back into the original equation.
A quadratic has at most two solutions, and both are supposed to work. Test them.
x = 322(2.25) + 7(1.5) = 4.5 + 10.5 = 15the opening problem ✓
x = −52(25) + 7(−5) = 50 − 35 = 15the other root works too ✓
x = 32(9) + 21 = 39, not 15the distractor fails ✗
x = −8 and 364 − 40 = 24 and 9 + 15 = 24x² + 5x = 24 ✓
Zero on one side · square roots, or factor, or the formula · both signs
Check b² − 4ac before hunting for factors — it tells you whether factors exist.
Now you try
Practice problem
The solutions to the given equation can be written as −3 ± √k, where k is a constant. What is the value of k?
x2 + 6x − 4 = 0
b² − 4ac = 36 − 4(1)(−4) = 36 + 16 = 52, so x = −6 ± √522. Since √52 = 2√13, that is −6 ± 2√132 = −3 ± √13, and k = 13. D stops at the discriminant and never simplifies; −6 ± √522 is not in the required form. B halves 52 — you cannot divide inside a square root by 2, you factor a perfect square out of it. C makes 36 − 4(1)(−4) into 36 + 4 = 40, dropping the 4 in front of ac. Check: (−3 + √13)² + 6(−3 + √13) − 4 = (22 − 6√13) + (−18 + 6√13) − 4 = 0 ✓
Hint: the choices contain no radical sign, but the required form
does — so this is a quadratic formula item, not a factoring one. Run the formula,
then simplify the radical and cancel the 2 from both terms of the
numerator before you read off k.
