Quadratic equations in Class 10: how to choose the right method in ten seconds

A quadratic equation can be solved by factorising, by completing the square or by the formula. Most students pick one by habit, and lose time when the habit does not fit the equation. This guide gives a quick test that tells you which method to use before you write the first line, with every example worked in full. The examples use fresh numbers, so you are practising the choice of method, not recognising a question you have seen before.

Step 1: read the discriminant

• $D > 0$: two different real roots.
• $D = 0$: two equal real roots, so one repeated value.
• $D < 0$: no real roots. Stop. There is nothing to find.

When $a, b, c$ are integers and $D$ is a perfect square (such as 1, 4, 9, 25 or 529), the roots are rational and the equation factorises. When $D$ is positive but not a perfect square (such as 20), the roots contain a square root and factorising by inspection will not work. Go straight to the formula.

Step 2: if D is a perfect square, factorise

Example 1. Solve $x^2 + 13x + 36 = 0$.

Here $D = 13^2 - 4(1)(36) = 169 - 144 = 25$, a perfect square, so factorising will work. Find two numbers with product 36 and sum 13: they are 4 and 9.

Example 2, when the first coefficient is not 1. Solve $6x^2 - 5x - 21 = 0$.

$D = (-5)^2 - 4(6)(-21) = 25 + 504 = 529 = 23^2$, a perfect square. Now multiply $a$ and $c$: $6 \times (-21) = -126$. Find two numbers with product $-126$ and sum $-5$: they are 9 and $-14$. Split the middle term with them.

Notice the discriminant did something useful here: it told you the roots were rational before you started, so a fraction like $\frac{7}{3}$ was expected, not a sign of a mistake.

Step 3: if D is not a perfect square, use the formula

Example 3. Solve $x^2 + 6x + 4 = 0$.

$D = 6^2 - 4(1)(4) = 20$. It is positive, so there are two real roots, but 20 is not a perfect square, so no pair of whole numbers will factorise this. Use $x = \dfrac{-b \pm \sqrt{D}}{2a}$.

Completing the square: the method behind the formula

Example 4. Solve $x^2 + 10x - 11 = 0$ by completing the square. Move the constant across, then add the square of half the coefficient of $x$, which is $5^2 = 25$, to both sides.

This is the same working that, done once with letters instead of numbers, produces the quadratic formula. That is why the formula always works, even when factorising does not.

Step 4: when D is negative, say so

Example 5. Does $3x^2 - 5x + 4 = 0$ have real roots?

Questions that ask for the value of k

Some questions give a quadratic containing $k$ and describe the roots in words. Translate the words into a condition on $D$, then solve for $k$.

• “Equal roots” means $D = 0$.
• “Two distinct real roots” means $D > 0$.
• “No real roots” means $D < 0$.

Example 6. Find the values of $k$ for which $x^2 + kx + 16 = 0$ has equal roots.

Check both of them. With $k = 8$ the equation is $x^2 + 8x + 16 = (x + 4)^2 = 0$, which has the repeated root $x = -4$. With $k = -8$ it is $(x - 4)^2 = 0$, repeated root $x = 4$.

Example 7. For which values of $k$ does $x^2 - 5x + k = 0$ have two distinct real roots?

Test it: $k = 6$ gives $D = 1 > 0$, so two roots (they are 2 and 3). $k = 7$ gives $D = -3 < 0$, so none.

The routine in four lines

1. Write $a, b, c$ with their signs, then work out $D$.
2. If $D < 0$, write “no real roots” and stop.
3. If $D$ is a perfect square, factorise by splitting the middle term.
4. Otherwise use the formula, and divide the whole numerator by $2a$.