Solving quadratic equations

Where it fits. This topic opens the Quadratic Equations sub-unit of Algebra (Section A, ISC Class 11 Mathematics), building on the factorisation and formula methods already met for real-rooted quadratics in ICSE Class X. It introduces the Fundamental Theorem of Algebra and prepares the ground for later topics on the nature and relations of roots, both of which depend on being fluent with the discriminant.

The formula itself. For real coefficients $a,b,c$ with $a\neq0$, the equation

has roots given by

where: $a,b,c$ are the real coefficients, $x$ is the unknown, and the quantity $D=b^2-4ac$ under the root is the discriminant.

What the discriminant decides. If $D\ge0$, $\sqrt D$ is real and the formula gives two real roots (equal roots when $D=0$). If $D<0$, $\sqrt D$ is imaginary, and the two roots form a complex-conjugate pair

A negative discriminant never means "no solution" — the Fundamental Theorem of Algebra guarantees a degree-2 equation has exactly two roots; when $D<0$ those roots simply are not real.

Equations reducible to quadratic form. Many equations that are not literally quadratic become quadratic after a substitution. For example, $x^4-13x^2+36=0$ is quadratic in $y=x^2$. Spotting the repeated block — here $x^2$ — and substituting $y$ for it turns an unfamiliar-looking equation into an ordinary quadratic, which is solved as usual and then translated back to the original variable. The same trick works whenever an expression and its square (or an expression and its reciprocal squared, as in $x+\dfrac1x$ type problems) both appear: rename the repeated block, solve the resulting quadratic in the new variable, then undo the substitution for every valid value found. Two things make this step-family error-prone in practice — students solve for the substituted variable and stop, forgetting to translate back, or they take only the positive square root when recovering $x$ from $y=x^2$, silently discarding half the answers.

Why the sign of the discriminant behaves this way. Geometrically, the graph of $y=ax^2+bx+c$ is a parabola, and the roots of $ax^2+bx+c=0$ are exactly where that parabola crosses the $x$-axis. When $D>0$ the parabola crosses the axis at two distinct points, giving two real roots. When $D=0$ the parabola is tangent to the axis at its vertex, giving one repeated real root. When $D<0$ the entire parabola lies above or below the axis without touching it — there is no real crossing at all, which is exactly the case where the algebra hands back complex roots instead. The completing-the-square derivation below makes this precise without needing a picture: the sign of $D$ is literally the sign of the quantity being square-rooted, so it directly controls whether that square root is real or imaginary.

Solve $2x^2+3x+5=0$.

Step 1. Identify the coefficients: $a=2$, $b=3$, $c=5$.

Step 2. Compute the discriminant: $D=b^2-4ac=3^2-4(2)(5)=9-40=-31$. Since $D<0$, the roots will be a complex-conjugate pair, not real numbers.

Step 3. Apply the quadratic formula: $x=\dfrac{-3\pm\sqrt{-31}}{4}$. Writing $\sqrt{-31}=i\sqrt{31}$ gives $x=\dfrac{-3\pm i\sqrt{31}}{4}$.

Step 4. Check using sum and product of roots: sum should equal $-b/a=-3/2$, and indeed $\dfrac{-3+i\sqrt{31}}{4}+\dfrac{-3-i\sqrt{31}}{4}=\dfrac{-6}{4}=-\dfrac32$. The roots are $x=\dfrac{-3+i\sqrt{31}}{4}$ and $x=\dfrac{-3-i\sqrt{31}}{4}$.

Competitive stretch

Solve $x^4-13x^2+36=0$.

Step 1. This is quadratic in $y=x^2$: substituting gives $y^2-13y+36=0$.

Step 2. Factorise: two numbers multiplying to $36$ and adding to $-13$ are $-4$ and $-9$, so $(y-4)(y-9)=0$, giving $y=4$ or $y=9$.

Step 3. Substitute back $y=x^2$: from $y=4$, $x=\pm2$; from $y=9$, $x=\pm3$. The four roots of the quartic are $x=2,-2,3,-3$, consistent with the quartic factorising fully as $(x-2)(x+2)(x-3)(x+3)$.

• Concluding a quadratic "has no solution" when $D<0$ — it always has exactly two solutions over the complex numbers; only real solutions fail to exist.
• Losing roots when reducing to quadratic form — each valid value of $y$ can give up to two values of $x$, so both $\pm$ square roots must be taken, not just the positive one.
• Sign errors inside the formula when $b$ is negative — always substitute the actual signed coefficient so $-b$ flips correctly.
• Writing $\sqrt{-31}$ carelessly instead of pulling out $i$ explicitly as $i\sqrt{31}$.
• Confusing "reducible to quadratic form" with an actual quadratic — after finding $y$, the problem is not finished until $y$ has been translated back into values of the original variable $x$.
• Assuming a quadratic must factorise neatly — many quadratics, like $2x^2+3x+5=0$ above, have no rational (or even real) factorisation at all, and the formula is the only reliable route.

Ask the tutor: If a discriminant comes out negative and the complex-root notation feels unfamiliar, ask the AI tutor to walk through the completing-the-square derivation once more using your own coefficients.