Nature of roots (discriminant)

For the general quadratic $ax^2+bx+c=0$ with $a\neq0$, the quadratic formula gives the roots as

where: $a$, $b$, $c$ are the real coefficients of the quadratic, and $x$ stands for either root, depending on the choice of sign in $\pm$. Everything interesting happens under the square root. Define the discriminant

where: $D$ is a single real number computed directly from the coefficients — no square root, no division, just arithmetic. Because $D$ sits inside $\sqrt{D}$ in the quadratic formula, its sign decides what kind of number $\sqrt{D}$ is, and that in turn decides the nature of the two roots.

• $D > 0$: $\sqrt{D}$ is a positive real number, so "$+\sqrt{D}$" and "$-\sqrt{D}$" are genuinely different — the formula produces two distinct real roots.
• $D = 0$: $\sqrt{D}=0$, so both signs collapse to the same value. The equation has two equal real roots (a repeated root), located at $x = -\dfrac{b}{2a}$.
• $D < 0$: there is no real square root of a negative number, so the formula produces no real roots. (Complex roots exist here, but they are outside the Class 10 syllabus — the correct answer to write is simply "no real roots.")

This is why the discriminant is so useful in "find the unknown" problems: if a question asks for the value of $k$ that makes an equation have equal roots, you never need to solve the equation at all — you just set $D=0$ and solve that simpler equation for $k$. The same test appears constantly in Class 10 word problems drawn from daily life, geometry, and mensuration — for instance, a problem about splitting a length into two parts satisfying some product condition, or finding a speed that makes a travel-time equation work out, often reduces to a quadratic in the unknown. Before grinding through the quadratic formula, checking the sign of $D$ tells you immediately whether the situation described even has a real, physically meaningful answer. If $D<0$, no real value fits the story — the problem as stated has no valid solution, which is itself a legitimate and often expected answer in exam questions phrased as "show that..." or "for what values of $k$ does a real solution exist." Recognising this saves time and avoids the common mistake of forcing an answer where none exists.

Find the nature of the roots of $4x^2-4x+1=0$ without solving the equation, and confirm using the value of $k$ that makes a related equation have equal roots.

Step 1. Identify the coefficients: $a=4,\ b=-4,\ c=1$.

Step 2. Compute the discriminant: $D = b^2-4ac = (-4)^2-4(4)(1) = 16-16 = 0$.

Step 3. Interpret the sign: since $D=0$, the equation has two equal real roots. The repeated root is $x = -\dfrac{b}{2a} = -\dfrac{-4}{2(4)} = \dfrac{1}{2}$. Indeed $4x^2-4x+1=(2x-1)^2$, confirming the double root $x=\dfrac12$.

Step 4. Now extend to the unknown-constant style of question that most often appears alongside this one: for what value of $k$ does $kx^2+6x+1=0$ ($k\neq0$) have equal roots? Set $D=0$: $(6)^2-4(k)(1)=0 \Rightarrow 36-4k=0 \Rightarrow k=9$. Check: with $k=9$, $D=36-4(9)(1)=0$, and indeed $9x^2+6x+1=(3x+1)^2$.

The final answer: $4x^2-4x+1=0$ has two equal real roots at $x=\dfrac12$, and the related equation $kx^2+6x+1=0$ has equal roots when $\mathbf{k=9}$.

Competitive stretch

If instead $k=10$ in $kx^2+6x+1=0$, what is the nature of the roots? Compute $D = 36-4(10)(1) = 36-40 = -4 < 0$. Since $D<0$, this equation has no real roots — a small change in $k$ flips the answer entirely, which is exactly why the discriminant test, not guesswork, must be used in these problems.

• Sign slip in $-4ac$, especially when $c$ is negative — that turns $-4ac$ positive and can flip the whole conclusion. Compute $4ac$ first, then subtract, tracking signs carefully.
• Reading $D<0$ as "the problem has no solution" rather than "no real number solves it." Never write down complex roots in this syllabus; simply state there are no real roots.
• Treating $D=0$ as "one root" instead of "two equal roots." The equation still has two roots conceptually; they just happen to coincide at $x=-\dfrac{b}{2a}$.
• Forgetting $a\neq0$ (or $k\neq0$ when the leading coefficient contains the unknown) — if the leading term can vanish, the equation stops being quadratic and that case needs separate handling.

Ask the tutor: Stuck on setting up the discriminant condition for a word problem, or want to see why $D<0$ rules out real roots — ask the AI tutor to walk through it with you.