One formula solves every quadratic equation, even those that refuse to factorise. Its key ingredient, the discriminant $D=b^2-4ac$, tells you before you solve whether there are two roots, one repeated root or no real root at all.
At the end of Section 4.2 we met equations that would not factorise. Try $x^2+4x-1=0$: we need two whole numbers with product $-1$ and sum $4$, and there are none. Yet the parabola $y=x^2+4x-1$ does cross the $x$-axis twice — once near $x=0.2$ and once near $x=-4.2$. The roots exist; they are simply not whole numbers or neat fractions, so splitting the middle term cannot find them.
In this section you learn the quadratic formula, which works for every quadratic equation, whether it factorises or not. You also meet the discriminant $D=b^2-4ac$, the number under the square root in the formula. Just by looking at its sign you can tell whether the equation has two different roots, one repeated root, or no real root — often without solving anything.
The quadratic formula
📐 Theorem 4.3 — The quadratic formula
Let $a$, $b$, $c$ be real numbers with $a\ne0$. If $b^2-4ac\ge0$, the roots of the quadratic equation $ax^2+bx+c=0$ are
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$$
that is, one root uses the $+$ sign and the other the $-$ sign.
If $b^2-4ac\lt0$, the equation has no real roots.
Why it works. Start from $ax^2+bx+c=0$ and multiply both sides by $4a$ (allowed, since $a\ne0$):
$$4a^2x^2+4abx+4ac=0 .$$
Add $b^2$ to both sides and move $4ac$ to the right: $\;4a^2x^2+4abx+b^2=b^2-4ac$. The left side is exactly $(2ax+b)^2$ — expand it and see. So the equation becomes
$$(2ax+b)^2=b^2-4ac .$$
If $b^2-4ac\ge0$, take square roots: $2ax+b=\pm\sqrt{b^2-4ac}$, so $2ax=-b\pm\sqrt{b^2-4ac}$, and dividing by $2a$ gives the formula. If $b^2-4ac\lt0$, the right side is negative, but the square of a real number is never negative — so no real $x$ works. This neat trick of making the left side a perfect square was used by the Indian mathematician Sridharacharya about a thousand years ago, and the formula is often named after him. You only need the trick to see why the formula is true; to solve equations, use the formula itself.
📘 Definition 4.3 — Discriminant
For the quadratic equation $ax^2+bx+c=0$ ($a\ne0$), the number
$$D=b^2-4ac$$
is called its discriminant. With this letter the quadratic formula reads $\;x=\dfrac{-b\pm\sqrt D}{2a}$ (when $D\ge0$).
🧭 Problem-solving strategy — using the quadratic formula
Bring the equation to standard form $ax^2+bx+c=0$. Clear fractions and remove common factors first — smaller numbers mean fewer slips.
Write down $a$, $b$ and $c$ with their signs. In $3x^2-4x-1=0$: $a=3$, $b=-4$, $c=-1$.
Work out $D=b^2-4ac$ on its own line. If $D\lt0$, stop: there are no real roots.
Substitute: $x=\dfrac{-b\pm\sqrt D}{2a}$. Simplify $\sqrt D$ (e.g. $\sqrt{20}=2\sqrt5$) and cancel a common factor from the whole numerator and the denominator.
Check with $\alpha+\beta=-\tfrac ba$ and $\alpha\beta=\tfrac ca$, or substitute a decimal value. In a word problem, reject roots that make no sense.
✏️ Example 1 — The formula on an equation that factorises
Solve $2x^2-7x+3=0$ using the quadratic formula, and compare with factorisation.
💡 Show solution▶
Coefficients $a=2$, $b=-7$, $c=3$.
Discriminant $D=b^2-4ac=(-7)^2-4(2)(3)=49-24=25$. Since $D\gt0$, there are two roots, and $\sqrt D=5$.
Formula $x=\dfrac{-(-7)\pm5}{2\times2}=\dfrac{7\pm5}{4}$, so $x=\dfrac{12}{4}=3$ or $x=\dfrac{2}{4}=\dfrac12$.
Compare Splitting the middle term ($ac=6$, sum $-7$: $-6$ and $-1$) gives $2x^2-6x-x+3=(2x-1)(x-3)$, so $x=\tfrac12$ or $x=3$ — the same roots. The formula never needs a lucky pair, but when a pair is easy to spot, factorising is quicker.
✏️ Example 2 — Roots that are not rational
Solve $x^2+4x-1=0$. Give the roots exactly and to two decimal places.
💡 Show solution▶
Coefficients $a=1$, $b=4$, $c=-1$.
Discriminant $D=4^2-4(1)(-1)=16+4=20$, and $\sqrt{20}=\sqrt{4\times5}=2\sqrt5$.
Formula $x=\dfrac{-4\pm2\sqrt5}{2}=-2\pm\sqrt5$ (divide both terms of the numerator by $2$).
Decimals With $\sqrt5\approx2.236$: $\;x\approx0.24$ or $x\approx-4.24$ — the two crossings we saw on the graph.
Check Sum $(-2+\sqrt5)+(-2-\sqrt5)=-4=-\tfrac ba$ ✓; product $(-2)^2-(\sqrt5)^2=4-5=-1=\tfrac ca$ ✓.
⚠️ Four slips to avoid with the formula
1. The sign of $b$. The formula starts with $-b$. If $b=-7$, then $-b=+7$. Write $-(-7)$ first and then simplify.
2. Standard form first. In $x^2=4x+1$ the coefficients are not $1$, $4$, $1$. Rewrite it as $x^2-4x-1=0$, so $b=-4$ and $c=-1$.
3. $2a$ divides the whole numerator. $x=\dfrac{-b\pm\sqrt D}{2a}$, not $-b\pm\dfrac{\sqrt D}{2a}$. Draw one long fraction bar.
4. Cancel from every term. $\dfrac{-4\pm2\sqrt5}{2}=-2\pm\sqrt5$, not $-4\pm\sqrt5$: the $2$ must divide both $-4$ and $2\sqrt5$.
Interactive 1 — Inside the Formula: Centre and Spread
Read the formula as two pieces: $x=-\dfrac{b}{2a}\pm\dfrac{\sqrt D}{2a}$. The first piece is the centre — the purple line, where the parabola turns. The second is the spread — how far you walk left and right from the centre to reach the roots. Press Play to watch the roots being built for several equations, or set $a$, $b$, $c$ yourself.
2−73
$D=b^2-4ac$—
centre $-\tfrac{b}{2a}$—
spread $\tfrac{\sqrt D}{2|a|}$—
roots—
The discriminant and the nature of roots
Look again at $x=\dfrac{-b\pm\sqrt D}{2a}$. Everything hangs on $\sqrt D$. If $D$ is positive, adding and subtracting $\sqrt D$ gives two different answers. If $D=0$, adding or subtracting $0$ changes nothing, so the two answers are the same. If $D$ is negative, there is no real square root at all. Deciding which case you are in is called finding the nature of the roots.
📐 Theorem 4.4 — Nature of roots
Let $ax^2+bx+c=0$ be a quadratic equation with real coefficients, $a\ne0$, and discriminant $D=b^2-4ac$.
If $D\gt0$, the equation has two distinct real roots, $\;\dfrac{-b+\sqrt D}{2a}$ and $\dfrac{-b-\sqrt D}{2a}$.
If $D=0$, the equation has two equal real roots, $\;x=-\dfrac{b}{2a},\ -\dfrac{b}{2a}$ — a repeated root.
If $D\lt0$, the equation has no real roots.
The picture. By Theorem 4.1 (Section 4.1) the real roots are where the parabola $y=ax^2+bx+c$ meets the $x$-axis. So $D\gt0$ means the parabola cuts the axis at two points, $D=0$ means it just touches the axis at its turning point $x=-\tfrac{b}{2a}$, and $D\lt0$ means it misses the axis completely, staying entirely above it (when $a\gt0$) or entirely below it (when $a\lt0$).
A useful extra. When $a$, $b$, $c$ are integers and $D$ is a perfect square ($0, 1, 4, 9, 16, 25,\dots$), $\sqrt D$ is a whole number and the roots are rational — the equation can be factorised by splitting the middle term. When $D\gt0$ is not a perfect square, the roots are irrational and the formula is the way to go.
✏️ Example 3 — Deciding the nature of the roots
Find the nature of the roots of each equation. If real roots exist, find them.
(a) $2x^2-4x+3=0$ (b) $4x^2-12x+9=0$ (c) $3x^2-5x-2=0$
💡 Show solution▶
(a) $D=(-4)^2-4(2)(3)=16-24=-8\lt0$. No real roots. (The parabola $y=2x^2-4x+3$ stays above the $x$-axis.)
(b) $D=(-12)^2-4(4)(9)=144-144=0$. Two equal real roots, $x=-\dfrac{b}{2a}=\dfrac{12}{8}=\dfrac32$, i.e. $\tfrac32,\tfrac32$. Indeed $4x^2-12x+9=(2x-3)^2$.
(c) $D=(-5)^2-4(3)(-2)=25+24=49\gt0$. Two distinct real roots; $49=7^2$ is a perfect square, so they are rational: $x=\dfrac{5\pm7}{6}$, i.e. $x=2$ or $x=-\tfrac13$.
Check (c): sum $2-\tfrac13=\tfrac53=-\tfrac ba$ ✓, product $-\tfrac23=\tfrac ca$ ✓.
✏️ Example 4 — Finding an unknown coefficient
(a) Find the values of $k$ for which $2x^2+kx+8=0$ has two equal roots, and find the root in each case.
(b) Find the values of $k$ for which $kx^2+6x+1=0$ has real roots.
💡 Show solution▶
(a) Condition Equal roots $\iff D=0$: $\;k^2-4(2)(8)=0$, so $k^2=64$ and $k=8$ or $k=-8$.
(a) Roots The repeated root is $x=-\dfrac{b}{2a}=-\dfrac{k}{4}$. For $k=8$: $x=-2$ (and $2x^2+8x+8=2(x+2)^2$ ✓). For $k=-8$: $x=2$ (and $2x^2-8x+8=2(x-2)^2$ ✓).
(b) Condition Real roots $\iff D\ge0$ (two distinct or two equal): $\;36-4k\ge0$, so $k\le9$.
(b) Careful We also need $k\ne0$, or the equation is $6x+1=0$, which is not quadratic. Answer: $k\le9$ and $k\ne0$. At $k=9$ the roots are equal: $9x^2+6x+1=(3x+1)^2$, root $-\tfrac13$.
Interactive 2 — The Discriminant Dial: Two, One or No Roots
Changing the constant $c$ slides the parabola up and down without changing its shape. Watch the D-meter on the right: while $D$ is in the green zone the parabola cuts the axis twice, at $D=0$ (the amber mark) it just touches, and in the red zone it misses the axis. Press Play, or drag the $c$ slider.
1
equation—
$D=b^2-4ac$—
nature—
roots—
Word problems: solve, or show it is impossible
Real measurements rarely give equations that factorise, so in practice the formula is the everyday tool. And the discriminant answers a second kind of question: is the situation possible at all? If the model has $D\lt0$, no real number fits, so no such object can exist.
✏️ Example 5 — Framing a photograph
A photograph is $12$ cm by $8$ cm. It is placed in a frame of uniform width so that the outer area of the frame is three times the area of the photograph. Find the width of the frame, to two decimal places.
💡 Show solution▶
Model Let the width be $x$ cm. The outer size is $(12+2x)$ cm by $(8+2x)$ cm, and the photo's area is $96$ cm². So $(12+2x)(8+2x)=3\times96=288$.
Standard form $96+40x+4x^2=288$, so $4x^2+40x-192=0$; dividing by $4$, $\;x^2+10x-48=0$.
Formula $D=100+192=292$, which is not a perfect square — no factorisation with whole numbers. $\sqrt{292}=\sqrt{4\times73}=2\sqrt{73}$, so $x=\dfrac{-10\pm2\sqrt{73}}{2}=-5\pm\sqrt{73}$.
Reject $-5-\sqrt{73}$ is negative, so it cannot be a width. With $\sqrt{73}\approx8.544$: $\;x=-5+\sqrt{73}\approx3.54$.
Answer The frame is about $3.54$ cm wide. Check: $(12+7.088)(8+7.088)=19.088\times15.088\approx288.0$ cm² ✓.
✏️ Example 6 — Is such a garden possible?
Riya has $40$ m of fencing for a rectangular vegetable patch. Is it possible to fence a patch of area (a) $120$ m² (b) $100$ m²? If so, find its length and breadth.
💡 Show solution▶
Model The perimeter is $40$ m, so length $+$ breadth $=20$ m. If the length is $x$ m, the breadth is $(20-x)$ m and the area is $x(20-x)$ m².
(a) $x(20-x)=120$ gives $x^2-20x+120=0$. $D=400-480=-80\lt0$: no real roots. Not possible — no rectangle with perimeter $40$ m has area $120$ m².
(b) $x(20-x)=100$ gives $x^2-20x+100=0$. $D=400-400=0$: equal roots, $x=\tfrac{20}{2}=10$. Possible: length $10$ m and breadth $20-10=10$ m — the patch is a square.
Insight For a fixed perimeter of $40$ m, $D=400-4A\ge0$ only when $A\le100$. So $100$ m² is the largest area possible, and it is reached by the square.
Exercises
I
Solve using the quadratic formula. Give irrational roots exactly (simplest surd form) and to two decimal places.
1 $x^2-5x+6=0$
2 $x^2+2x-4=0$
3 $2x^2+5x-3=0$
4 $3x^2-4x-1=0$
5 $x^2=6x-4$
6 $4x^2+4\sqrt3\,x+3=0$
7 $2x^2-2\sqrt2\,x+1=0$
8 $5x^2+x-1=0$
9 $x^2+x+1=0$
10 $9x^2-6x-2=0$
11 $\dfrac1x-\dfrac1{x-2}=3$ ($x\ne0$, $x\ne2$)
Check: type whole-number coefficients and compare $D$, the nature and the roots with yours
$D$—
nature—
roots—
💡 Show solutions▶
1 $D=25-24=1$; $x=\dfrac{5\pm1}{2}$, so $x=3$ or $x=2$.
11 Left side $=\dfrac{(x-2)-x}{x(x-2)}=\dfrac{-2}{x^2-2x}$. So $-2=3x^2-6x$, i.e. $3x^2-6x+2=0$. $D=36-24=12$; $x=\dfrac{6\pm2\sqrt3}{6}=1\pm\dfrac{\sqrt3}{3}\approx1.58,\ 0.42$ (neither is $0$ or $2$, so both are allowed).
II
Find the nature of the roots. If they are real, find them.
1 $2x^2-3x+5=0$
2 $3x^2-4\sqrt3\,x+4=0$
3 $2x^2-6x+3=0$
4 $6x^2-x-2=0$
5 $x^2+5=0$
6 $x^2-(\sqrt2+1)x+\sqrt2=0$
7 $16x^2-24x+9=0$
8 $0.2x^2-x+1.5=0$
💡 Show solutions▶
1 $D=9-40=-31\lt0$: no real roots.
2 $D=48-48=0$: two equal real roots, $x=\dfrac{4\sqrt3}{6}=\dfrac{2\sqrt3}{3}$ (that is, $\tfrac{2}{\sqrt3}$).
3 $D=36-24=12\gt0$, not a perfect square: two distinct irrational roots, $x=\dfrac{6\pm2\sqrt3}{4}=\dfrac{3\pm\sqrt3}{2}$.
4 $D=1+48=49=7^2$: two distinct rational roots, $x=\dfrac{1\pm7}{12}$, i.e. $\tfrac23$ and $-\tfrac12$.
5 $a=1$, $b=0$, $c=5$: $D=0-20=-20\lt0$, no real roots (a square plus $5$ is never $0$).
6 $D=(\sqrt2+1)^2-4\sqrt2=3+2\sqrt2-4\sqrt2=3-2\sqrt2=(\sqrt2-1)^2\gt0$: two distinct real roots. $\sqrt D=\sqrt2-1$, so $x=\dfrac{(\sqrt2+1)\pm(\sqrt2-1)}{2}$, i.e. $\sqrt2$ and $1$.
7 $D=576-576=0$: two equal roots, $x=\dfrac{24}{32}=\dfrac34$.
8 Multiply by $10$: $2x^2-10x+15=0$. $D=100-120=-20\lt0$: no real roots.
III
Unknown coefficients.
1 Find $k$ if $kx^2+4x+1=0$ has two equal roots, and find the root.
2 Find $k$ if $2x^2+kx+3=0$ has two equal roots.
3 Find the values of $k$ for which $x^2-4x+k=0$ has real roots.
4 Find the values of $k$ for which $kx^2-6x+2=0$ has no real roots.
5 Find $k$ if $kx(x-3)+9=0$ has two equal roots, and find the root.
6 Find $k$ if $(k+1)x^2-2(k-1)x+1=0$ has two equal roots, and find the root for each value.
7 Show that $x^2+px-1=0$ has two distinct real roots for every real number $p$.
💡 Show solutions▶
1 $D=16-4k=0$, so $k=4$. Root $x=-\dfrac{b}{2a}=-\dfrac{4}{8}=-\dfrac12$ (and $4x^2+4x+1=(2x+1)^2$ ✓).
2 $D=k^2-24=0$, so $k=\pm\sqrt{24}=\pm2\sqrt6$.
3 $D=16-4k\ge0$, so $k\le4$.
4 $D=36-8k\lt0$, so $8k\gt36$, i.e. $k\gt\tfrac92$. (Such $k$ is never $0$, so the equation stays quadratic.)
5 Standard form $kx^2-3kx+9=0$. $D=9k^2-36k=9k(k-4)=0$, so $k=0$ or $k=4$. $k=0$ is rejected: the equation becomes $9=0$, not a quadratic. So $k=4$: $4x^2-12x+9=(2x-3)^2$, root $\tfrac32$.
6 $D=4(k-1)^2-4(k+1)=4(k^2-3k)=4k(k-3)=0$, so $k=0$ or $k=3$ (both keep $k+1\ne0$). $k=0$: $x^2+2x+1=(x+1)^2$, root $-1$. $k=3$: $4x^2-4x+1=(2x-1)^2$, root $\tfrac12$.
7 $D=p^2-4(1)(-1)=p^2+4$. Since $p^2\ge0$, $D\ge4\gt0$ for every $p$, so there are always two distinct real roots.
IV
Word problems. Form a quadratic equation, then solve it with the formula — or use the discriminant to show the situation is impossible.
1 A number exceeds its reciprocal by $1$. Find the number(s), exactly and to two decimal places.
2 A rectangular classroom notice board has area $30$ m², and its length is $4$ m more than its breadth. Find its dimensions to two decimal places.
3 A Diwali rocket's height is $h=30t-5t^2$ metres after $t$ seconds. (a) At what times is it $30$ m high? (b) Use the discriminant to show that it never reaches $50$ m.
4 A carpenter has $6$ m of edging strip for the border of a rectangular table-top. Can the table-top have area $2.5$ m²? Can it have area $2$ m²? Give the dimensions where possible.
5 The two shorter sides of a right triangle differ by $1$ cm and the hypotenuse is $6$ cm. Find the shorter sides to two decimal places.
6 A cyclist rides $24$ km at a uniform speed. Had she ridden $2$ km/h faster, she would have taken $1$ hour less. Find her speed.
7 The sum of the ages of two friends is $20$ years. Four years ago the product of their ages was $48$. Is this possible?
💡 Show solutions▶
1 $x-\dfrac1x=1$ ($x\ne0$) gives $x^2-x-1=0$. $D=1+4=5$; $x=\dfrac{1\pm\sqrt5}{2}\approx1.62$ or $-0.62$. Both work (e.g. $1.618-0.618=1$ ✓).
2 Breadth $x$ m: $x(x+4)=30$, $\;x^2+4x-30=0$. $D=16+120=136=4\times34$; $x=-2\pm\sqrt{34}$. Reject the negative root: breadth $-2+\sqrt{34}\approx3.83$ m, length $\approx7.83$ m.
3 (a) $30t-5t^2=30$ gives $t^2-6t+6=0$. $D=36-24=12$; $t=\dfrac{6\pm2\sqrt3}{2}=3\pm\sqrt3$, i.e. $t\approx1.27$ s (going up) and $t\approx4.73$ s (coming down). (b) $30t-5t^2=50$ gives $t^2-6t+10=0$, $D=36-40=-4\lt0$: no real time, so it never reaches $50$ m.
4 Length $+$ breadth $=3$ m, so the area is $x(3-x)$. For $2.5$ m²: $x^2-3x+2.5=0$, $D=9-10=-1\lt0$ — not possible. For $2$ m²: $x^2-3x+2=0$, $D=1$, $x=\dfrac{3\pm1}{2}=2$ or $1$ — possible, $2$ m by $1$ m.
5 $x^2+(x+1)^2=36$, $\;2x^2+2x-35=0$. $D=4+280=284=4\times71$; $x=\dfrac{-2\pm2\sqrt{71}}{4}=\dfrac{-1\pm\sqrt{71}}{2}$. Positive root $\approx3.71$ cm; the sides are about $3.71$ cm and $4.71$ cm (check $13.79+22.21\approx36$ ✓).
6 Speed $x$ km/h: $\dfrac{24}{x}-\dfrac{24}{x+2}=1$. Multiply by $x(x+2)$: $48=x^2+2x$, $\;x^2+2x-48=0$. $D=4+192=196$, $\sqrt D=14$; $x=\dfrac{-2\pm14}{2}=6$ or $-8$. Speed $6$ km/h ($4$ h at $6$ km/h, $3$ h at $8$ km/h ✓).
7 Ages $x$ and $20-x$ years. Four years ago: $(x-4)(16-x)=48$, so $-x^2+20x-64=48$, i.e. $x^2-20x+112=0$. $D=400-448=-48\lt0$: no real solution, so the situation is not possible.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main
How this is examined: board papers ask for the nature of roots (1 mark), "find $k$ for equal roots" (2 marks) and "solve using the quadratic formula" (3 marks — write $D$ on its own line, it carries a mark). Case studies end with a "can it happen?" part answered by the sign of $D$. JEE Main turns the discriminant condition into an inequality for $k$.
E1.Board · 1 mark Which of these equations has two distinct real roots? (A) $x^2+x+1=0$ (B) $2x^2-4x+2=0$ (C) $x^2-3x-1=0$ (D) $3x^2+2x+1=0$
Trap (B) does have real roots, but they are equal ($2(x-1)^2=0$), not distinct.
Answer (C), the only one with $D\gt0$. A shortcut: when $a$ and $c$ have opposite signs, $-4ac\gt0$, so $D\gt0$ automatically.
E2.Board · 1 markAssertion (A): The equation $x^2+4x+5=0$ has no real roots. Reason (R): A quadratic equation $ax^2+bx+c=0$ has no real roots when $b^2-4ac\lt0$. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
💡 Show solution▶
Check R This is Theorem 4.4 (iii). True.
Check A $D=16-20=-4\lt0$, so by R there are no real roots. True, and R is exactly the reason.
Answer (A). Option (B) would need A to be true for some other reason; here the discriminant is the whole argument.
E3.Board · 2 marks Find the value of $k$ for which $kx^2-2\sqrt5\,x+4=0$ has two equal roots. Hence find the root.
💡 Show solution▶
Condition Equal roots $\iff D=0$: $\;(-2\sqrt5)^2-4(k)(4)=0$, so $20-16k=0$ and $k=\tfrac54$. (1 mark)
Formula $x=\dfrac{2\sqrt2\pm4\sqrt2}{2\sqrt3}$, so $x=\dfrac{6\sqrt2}{2\sqrt3}=\dfrac{3\sqrt2}{\sqrt3}=\sqrt6$ or $x=\dfrac{-2\sqrt2}{2\sqrt3}=-\dfrac{\sqrt2}{\sqrt3}=-\dfrac{\sqrt6}{3}$. (1 mark)
E5.Board · Case study A garden fountain in a park in Pune throws a jet of water. With the nozzle at ground level, the height of the water above the ground is $h=6x-x^2$ metres when it is $x$ metres away horizontally from the nozzle. (i) At what horizontal distances is the water $5$ m high? Where does it land? (2 marks) (ii) Use the discriminant to show that the jet never reaches a height of $10$ m. (1 mark) (iii) For which height $H$ does $6x-x^2=H$ have two equal roots? What does this height mean for the fountain? (1 mark)
💡 Show solution▶
(i) $6x-x^2=5$ gives $x^2-6x+5=0$; $D=36-20=16$, $x=\dfrac{6\pm4}{2}=1$ or $5$. The water is $5$ m high at $1$ m and at $5$ m from the nozzle. It lands where $h=0$: $x(6-x)=0$, so $x=6$ m (the other root, $x=0$, is the nozzle itself).
(ii) $6x-x^2=10$ gives $x^2-6x+10=0$; $D=36-40=-4\lt0$. No real $x$, so the water is never $10$ m high.
(iii) $x^2-6x+H=0$ has equal roots when $D=36-4H=0$, i.e. $H=9$; the root is $x=3$. So $9$ m is the greatest height of the jet, reached $3$ m from the nozzle: any lower height is reached twice, $9$ m exactly once, and no higher height at all.
E6.JEE Main The greatest integer $k$ for which the equation $x^2-2kx+k^2+k-5=0$ has real roots is ______. (numerical answer)
💡 Show solution▶
Condition $D=(2k)^2-4(k^2+k-5)=4k^2-4k^2-4k+20=20-4k$. Real roots $\iff D\ge0\iff k\le5$.
Check the edge At $k=5$: $x^2-10x+25=(x-5)^2=0$, equal real roots — still real. At $k=6$: $D=-4\lt0$.
Answer $5$. The tempting answer $4$ comes from using $D\gt0$; "real roots" includes equal roots, so $D=0$ is allowed.
🔬 Research–Teaching NexusMini project · optional, about 45 minutes
Research question. Roll three dice and use the numbers as $a$, $b$ and $c$ in $ax^2+bx+c=0$. How likely is it that the equation you get has real roots?
Why it matters. Engineers and scientists often meet equations whose coefficients come from measurements, which vary a little each time. Whether the model still has a real solution — a beam that balances, an orbit that closes, a dose that reaches a target level — is a question about the sign of a discriminant. Here you explore it with dice, and you meet a first taste of probability (Chapter 14).
What to do.
Make a guess first: out of $100$ rolls, how many equations will have real roots?
With a partner, roll three dice $50$ times each. For every roll record $a$, $b$, $c$ and $D=b^2-4ac$, and mark it "real" ($D\ge0$) or "not real" ($D\lt0$). Mark the cases with $D=0$ specially.
Pool the class results and find the fraction of rolls that gave real roots.
Now count exactly. There are $6\times6\times6=216$ equally likely rolls. In a spreadsheet, list them all (or reason by hand: for each $b$, which products $ac\le\tfrac{b^2}{4}$ are possible?) and count how many have $D\ge0$.
Compare the class fraction, your guess and the exact count. Which values of $b$ produce most of the real-root equations, and why?
Tools: Three dice (or a dice app), a recording sheet, a spreadsheet
Hand in: Your table of rolls, the pooled class fraction, your exact count with the working (a table of $b$ against the number of $(a,c)$ pairs with $4ac\le b^2$), and two or three sentences comparing them.
Going further. Repeat with $a$, $b$, $c$ allowed to be $-6,\dots,-1,1,\dots,6$ (toss a coin for the sign). Why does the fraction with real roots jump up? (Hint: think about the sign of $ac$.)
Teacher/parent note: with $a,b,c\in\{1,\dots,6\}$ exactly $43$ of the $216$ rolls give $D\ge0$ (about $19.9\%$), and $5$ of them give $D=0$: $(a,b,c)$ $=(1,2,1)$, $(1,4,4)$, $(2,4,2)$, $(4,4,1)$, $(3,6,3)$. By value of $b$ the counts are $0,1,3,8,14,17$ for $b=1,\dots,6$. When signs are random, $ac\lt0$ half the time, and then $D\gt0$ for sure.
🧪 Self-check quiz
0/5 correct
Q1. The discriminant of $2x^2-5x+3=0$ is
Q2. The equation $x^2+6x+9=0$ has
Q3. The roots of $x^2-2x-1=0$ are
Q4. The equation $x^2+4x+k=0$ has no real roots exactly when
Q5. If $ax^2+bx+c=0$ ($a\ne0$) has $D=0$, its repeated root is