Quadratic Equations

www.akankshaclasses.com
आ
CLASS X Mathematics ~5 marks/year Ch 4 of 14
Quadratic Equations

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • A quadratic equation is any equation that can be tidied to the standard form $ax^{2}+bx+c=0$ with $a\neq0$ — degree exactly $2$.
  • A root (or solution) is a value of $x$ that makes the equation true; a quadratic has at most two roots.
  • Three ways to find roots: factorisation (splitting the middle term), completing the square, and the quadratic formula $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
  • The discriminant $D=b^{2}-4ac$ decides the nature of roots: $D>0$ two distinct real, $D=0$ two equal real, $D<0$ no real roots.
  • Word problems (numbers, ages, areas, speed–distance–time) are modelled as quadratics, solved, then the meaningless (e.g. negative) root is rejected.
  • Board weightage: ~5 marks/year — typically one solve-by-method question (2–3 marks) and one application/word problem (3 marks).
Detailed notes

1. Where this chapter comes from

In Chapter 2 you met the quadratic polynomial $ax^{2}+bx+c$ (with $a\neq0$). When we set such a polynomial equal to zero, we get a quadratic equation. These pop up everywhere in real life.

Motivating example (NCERT prayer hall). A charity wants a prayer hall with carpet area $300\text{ m}^{2}$, where the length is one metre more than twice the breadth. Let breadth $=x$ m, so length $=(2x+1)$ m. Then

$$\text{area}=(2x+1)\,x=2x^{2}+x=300\ \Rightarrow\ 2x^{2}+x-300=0.$$

The breadth must satisfy this — a quadratic equation. Historically Babylonians, Euclid, Brahmagupta (598–665 CE) and Sridharacharya (c. 1025 CE, who gave the quadratic formula) all studied these.

2. Standard form and definition

A quadratic equation in the variable $x$ is any equation that can be written as

$$ax^{2}+bx+c=0,\qquad a\neq0,$$

where $a,b,c$ are real numbers. The condition $a\neq0$ is essential — if $a=0$ the equation is only linear. When the terms are arranged in descending powers of $x$ this is called the standard form. In fact, any equation of the form $p(x)=0$ where $p(x)$ is a degree-$2$ polynomial is a quadratic equation.

Examples: $2x^{2}+x-300=0$, $\ 2x^{2}-3x+1=0$, $\ 4x-3x^{2}+2=0$ and $1-x^{2}+300=0$ are all quadratic equations. The key test: simplify first, then check the highest power is exactly $2$.

NCERT Example 1 — model two situations as quadratics

(i) John and Jivanti together have $45$ marbles. Both lose $5$ each, and now the product of what they have is $124$. Let John start with $x$, so Jivanti has $45-x$. After losing $5$ each: John has $x-5$, Jivanti has $45-x-5=40-x$. Their product:

$(x-5)(40-x)=40x-x^{2}-200+5x=-x^{2}+45x-200$. Set $=124$: $-x^{2}+45x-200=124\Rightarrow -x^{2}+45x-324=0\Rightarrow x^{2}-45x+324=0$.

(ii) A cottage makes toys; cost per toy $=(55-\text{number of toys})$. On a day, total cost $=\rupee 750$. Let toys $=x$, cost each $=55-x$. Total $=x(55-x)=750\Rightarrow 55x-x^{2}=750\Rightarrow x^{2}-55x+750=0$.

NCERT Example 2 — which are quadratic equations?

(i) $(x-2)^{2}+1=2x-3$: LHS $=x^{2}-4x+4+1=x^{2}-4x+5$. So $x^{2}-4x+5=2x-3\Rightarrow x^{2}-6x+8=0$. Of the form $ax^{2}+bx+c=0$ — yes, quadratic.

(ii) $x(x+1)+8=(x+2)(x-2)$: LHS $=x^{2}+x+8$, RHS $=x^{2}-4$. So $x^{2}+x+8=x^{2}-4\Rightarrow x+12=0$ — only degree $1$. Not quadratic.

(iii) $x(2x+3)=x^{2}+1$: LHS $=2x^{2}+3x$, so $2x^{2}+3x=x^{2}+1\Rightarrow x^{2}+3x-1=0$ — yes, quadratic.

(iv) $(x+2)^{3}=x^{3}-4$: LHS $=x^{3}+6x^{2}+12x+8$, so $x^{3}+6x^{2}+12x+8=x^{3}-4\Rightarrow 6x^{2}+12x+12=0\Rightarrow x^{2}+2x+2=0$ — the cubic terms cancel, so it is yes, quadratic. Lesson: always simplify before deciding.

3. Roots of a quadratic equation

A real number $\alpha$ is a root of $ax^{2}+bx+c=0$ if $a\alpha^{2}+b\alpha+c=0$ — that is, substituting $x=\alpha$ makes the LHS equal $0$. We also say $\alpha$ is a solution, or that $\alpha$ satisfies the equation.

For instance, put $x=1$ in $2x^{2}-3x+1$: $(2\times1)-(3\times1)+1=0$. So $1$ is a root. The roots of the equation $ax^{2}+bx+c=0$ are exactly the zeroes of the polynomial $ax^{2}+bx+c$ — same numbers. Since a quadratic polynomial has at most two zeroes, a quadratic equation has at most two roots.

4. Method 1 — Solving by factorisation

If we can write $ax^{2}+bx+c$ as a product of two linear factors, then setting the product to zero means each factor can be zero (zero-product rule). The trick is splitting the middle term: write $b$ as a sum of two numbers whose product equals $a\times c$.

NCERT Example 3 — solve $2x^{2}-5x+3=0$

Here $a\times c=2\times3=6$. Split $-5x$ as $-2x-3x$ (since $(-2)(-3)=6$). Then $2x^{2}-2x-3x+3=2x(x-1)-3(x-1)=(2x-3)(x-1)$.

So $(2x-3)(x-1)=0\Rightarrow 2x-3=0$ or $x-1=0\Rightarrow x=\dfrac{3}{2}$ or $x=1$. Roots: $\dfrac{3}{2}$ and $1$.

NCERT Example 4 — solve $6x^{2}-x-2=0$

$a\times c=6\times(-2)=-12$; split $-x$ as $+3x-4x$. Then $6x^{2}+3x-4x-2=3x(2x+1)-2(2x+1)=(3x-2)(2x+1)$.

So $3x-2=0$ or $2x+1=0\Rightarrow x=\dfrac{2}{3}$ or $x=-\dfrac{1}{2}$. Roots: $\dfrac{2}{3}$ and $-\dfrac{1}{2}$.

NCERT Example 5 — solve $3x^{2}-2\sqrt{6}\,x+2=0$

Split the middle term using $\sqrt6=\sqrt3\cdot\sqrt2$: $3x^{2}-\sqrt6\,x-\sqrt6\,x+2=\sqrt3\,x(\sqrt3\,x-\sqrt2)-\sqrt2(\sqrt3\,x-\sqrt2)=(\sqrt3\,x-\sqrt2)(\sqrt3\,x-\sqrt2)$.

So $\sqrt3\,x-\sqrt2=0\Rightarrow x=\dfrac{\sqrt2}{\sqrt3}=\sqrt{\dfrac{2}{3}}$. The factor repeats, so the root is repeated: both roots are $\sqrt{\dfrac{2}{3}}$.

NCERT Example 6 — the prayer hall dimensions

Breadth $x$ satisfies $2x^{2}+x-300=0$. Split: $2x^{2}-24x+25x-300=2x(x-12)+25(x-12)=(x-12)(2x+25)=0$.

So $x=12$ or $x=-12.5$. Breadth can't be negative, so $x=12$ m and length $=2x+1=25$ m.

5. Method 2 — Completing the square

Not every quadratic factorises with nice integers. Completing the square always works: rearrange $ax^{2}+bx+c=0$ into a perfect square $\big(x+\tfrac{b}{2a}\big)^{2}=$ (a number), then take square roots.

The recipe: divide through by $a$, move the constant to the right, add $\big(\tfrac{b}{2a}\big)^{2}$ to both sides to complete the square, then solve.

$$ax^{2}+bx+c=0\ \Longrightarrow\ \left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}-4ac}{4a^{2}}.$$
Worked illustration — solve $x^{2}-4x-5=0$ by completing the square

Move constant: $x^{2}-4x=5$. Half of $-4$ is $-2$; add $(-2)^{2}=4$ to both sides: $x^{2}-4x+4=5+4\Rightarrow(x-2)^{2}=9$.

Take roots: $x-2=\pm3\Rightarrow x=2+3=5$ or $x=2-3=-1$. Roots: $5$ and $-1$.

Worked illustration — solve $2x^{2}-5x+3=0$ by completing the square

Divide by $2$: $x^{2}-\dfrac{5}{2}x+\dfrac{3}{2}=0\Rightarrow x^{2}-\dfrac{5}{2}x=-\dfrac{3}{2}$. Half of $-\tfrac52$ is $-\tfrac54$; add $\tfrac{25}{16}$: $\left(x-\dfrac54\right)^{2}=-\dfrac32+\dfrac{25}{16}=\dfrac{1}{16}$.

So $x-\dfrac54=\pm\dfrac14\Rightarrow x=\dfrac{6}{4}=\dfrac32$ or $x=\dfrac{4}{4}=1$ — agreeing with Example 3.

Carrying this method out on the general equation $ax^{2}+bx+c=0$ is exactly how the quadratic formula is derived (next section).

6. Method 3 — The quadratic formula

Completing the square on $ax^{2}+bx+c=0$ gives $\left(x+\dfrac{b}{2a}\right)^{2}=\dfrac{b^{2}-4ac}{4a^{2}}$. Taking square roots and isolating $x$:

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a},\qquad\text{provided } b^{2}-4ac\ge0.$$

This is Sridharacharya's formula. Plug in $a,b,c$ from the standard form and read off both roots from the $\pm$. It works for every quadratic that has real roots, even when factorisation is hard.

Worked illustration — solve $x^{2}+7x-60=0$ by the formula

$a=1,\ b=7,\ c=-60$. Discriminant $b^{2}-4ac=49+240=289=17^{2}$. So $x=\dfrac{-7\pm\sqrt{289}}{2}=\dfrac{-7\pm17}{2}$, giving $x=5$ or $x=-12$. (This is the equation behind Example 8.)

7. The discriminant and nature of roots

Under the square root in the formula sits $b^{2}-4ac$. This quantity decides whether the roots are real, and is called the discriminant, written $D$:

$$D=b^{2}-4ac.$$

Because $\sqrt{D}$ only gives a real number when $D\ge0$, the discriminant tells us the nature of the roots without even solving:

  • $D>0$: two distinct real roots, $-\dfrac{b}{2a}+\dfrac{\sqrt{D}}{2a}$ and $-\dfrac{b}{2a}-\dfrac{\sqrt{D}}{2a}$.
  • $D=0$: two equal real roots (coincident), each $=-\dfrac{b}{2a}$.
  • $D<0$: no real roots (the square root of a negative number is not real).
NCERT Example 7 — discriminant of $2x^{2}-4x+3=0$

$a=2,\ b=-4,\ c=3$. $D=(-4)^{2}-4(2)(3)=16-24=-8<0$. Since $D<0$, the equation has no real roots.

NCERT Example 8 — the pole on a circular park

A pole is to stand on the boundary of a circular park of diameter $13$ m so that the difference of its distances from two diametrically opposite gates $A,B$ is $7$ m. Let $BP=x$, then $AP=x+7$. Since $AB=13$ is a diameter, $\angle APB=90^\circ$, so by Pythagoras $AP^{2}+BP^{2}=AB^{2}$:

$(x+7)^{2}+x^{2}=13^{2}\Rightarrow x^{2}+14x+49+x^{2}=169\Rightarrow 2x^{2}+14x-120=0\Rightarrow x^{2}+7x-60=0$.

$D=7^{2}-4(1)(-60)=49+240=289>0$, so real roots exist — the pole can be placed. By the formula $x=\dfrac{-7\pm\sqrt{289}}{2}=\dfrac{-7\pm17}{2}=5$ or $-12$. Distance must be positive, so $x=5$. The pole is $5$ m from gate $B$ and $12$ m from gate $A$.

NCERT Example 9 — discriminant of $3x^{2}-2x+\tfrac13=0$

$a=3,\ b=-2,\ c=\dfrac13$. $D=(-2)^{2}-4(3)\!\left(\dfrac13\right)=4-4=0$. Since $D=0$, two equal real roots, each $=-\dfrac{b}{2a}=\dfrac{2}{6}=\dfrac13$. Roots: $\dfrac13,\dfrac13$.

8. Word problems — the modelling routine

Application questions are worth the most marks. Follow the same routine every time:

  • Name the unknown with a variable (let speed $=x$, age $=x$, etc.).
  • Translate the conditions into an equation; simplify to standard form $ax^{2}+bx+c=0$.
  • Solve by factorisation or formula.
  • Reject impossible roots — a length, age, speed or count cannot be negative (or non-integer when counting objects).
  • State the answer in words with units.

Common types: two numbers with a given sum & product, consecutive integers, ages, rectangle area/dimensions, and speed–distance–time ($\text{time}=\dfrac{\text{distance}}{\text{speed}}$). All the Exercise 4.2 and 4.3 word problems below use this routine.

9. NCERT Exercise 4.1 — fully solved

Q1. Check whether these are quadratic equations.

  • (i) $(x+1)^{2}=2(x-3)$: $x^{2}+2x+1=2x-6\Rightarrow x^{2}+7=0$ — yes.
  • (ii) $x^{2}-2x=(-2)(3-x)$: $x^{2}-2x=-6+2x\Rightarrow x^{2}-4x+6=0$ — yes.
  • (iii) $(x-2)(x+1)=(x-1)(x+3)$: $x^{2}-x-2=x^{2}+2x-3\Rightarrow -3x+1=0$ — no (linear).
  • (iv) $(x-3)(2x+1)=x(x+5)$: $2x^{2}-5x-3=x^{2}+5x\Rightarrow x^{2}-10x-3=0$ — yes.
  • (v) $(2x-1)(x-3)=(x+5)(x-1)$: $2x^{2}-7x+3=x^{2}+4x-5\Rightarrow x^{2}-11x+8=0$ — yes.
  • (vi) $x^{2}+3x+1=(x-2)^{2}$: $x^{2}+3x+1=x^{2}-4x+4\Rightarrow 7x-3=0$ — no (linear).
  • (vii) $(x+2)^{3}=2x(x^{2}-1)$: $x^{3}+6x^{2}+12x+8=2x^{3}-2x\Rightarrow x^{3}-6x^{2}-14x-8=0$ — no (cubic).
  • (viii) $x^{3}-4x^{2}-x+1=(x-2)^{3}$: RHS $=x^{3}-6x^{2}+12x-8$, so $-4x^{2}-x+1=-6x^{2}+12x-8\Rightarrow 2x^{2}-13x+9=0$ — yes.

Q2. Represent as quadratic equations.

  • (i) Rectangular plot, area $528\text{ m}^{2}$, length is one more than twice breadth. Let breadth $=x$, length $=2x+1$: $x(2x+1)=528\Rightarrow 2x^{2}+x-528=0$.
  • (ii) Product of two consecutive positive integers is $306$. Let them be $x$ and $x+1$: $x(x+1)=306\Rightarrow x^{2}+x-306=0$.
  • (iii) Rohan's mother is $26$ years older; product of their ages $3$ years from now is $360$. Let Rohan $=x$: $(x+3)(x+29)=360\Rightarrow x^{2}+32x-273=0$.
  • (iv) Train travels $480$ km; if speed were $8$ km/h less it would take $3$ h more. Let speed $=x$: $\dfrac{480}{x-8}-\dfrac{480}{x}=3\Rightarrow x^{2}-8x-1280=0$.

10. NCERT Exercise 4.2 — fully solved

Q1. Find the roots by factorisation.

  • (i) $x^{2}-3x-10=0\Rightarrow(x-5)(x+2)=0\Rightarrow x=5$ or $x=-2$.
  • (ii) $2x^{2}+x-6=0\Rightarrow 2x^{2}+4x-3x-6=(x+2)(2x-3)=0\Rightarrow x=-2$ or $x=\dfrac32$.
  • (iii) $\sqrt2\,x^{2}+7x+5\sqrt2=0\Rightarrow \sqrt2\,x^{2}+2x+5x+5\sqrt2=\sqrt2\,x(x+\sqrt2)+5(x+\sqrt2)=(x+\sqrt2)(\sqrt2\,x+5)=0\Rightarrow x=-\sqrt2$ or $x=-\dfrac{5}{\sqrt2}$.
  • (iv) $2x^{2}-x+\dfrac18=0\Rightarrow 16x^{2}-8x+1=0\Rightarrow(4x-1)^{2}=0\Rightarrow x=\dfrac14,\dfrac14$.
  • (v) $100x^{2}-20x+1=0\Rightarrow(10x-1)^{2}=0\Rightarrow x=\dfrac{1}{10},\dfrac{1}{10}$.

Q2. Solve the problems in Example 1.

  • Marbles: $x^{2}-45x+324=0\Rightarrow(x-36)(x-9)=0\Rightarrow x=36$ or $x=9$. So John had $36$ and Jivanti $9$ (or vice versa).
  • Toys: $x^{2}-55x+750=0\Rightarrow(x-25)(x-30)=0\Rightarrow x=25$ or $x=30$.

Q3. Two numbers, sum $27$, product $182$. Let one be $x$, other $27-x$: $x(27-x)=182\Rightarrow x^{2}-27x+182=0\Rightarrow(x-13)(x-14)=0\Rightarrow x=13$ or $14$. The numbers are $\mathbf{13}$ and $\mathbf{14}$.

Q4. Two consecutive positive integers, sum of squares $365$. Let them be $x,x+1$: $x^{2}+(x+1)^{2}=365\Rightarrow 2x^{2}+2x-364=0\Rightarrow x^{2}+x-182=0\Rightarrow(x-13)(x+14)=0$. Take $x=13$ (positive). The integers are $\mathbf{13}$ and $\mathbf{14}$.

Q5. Right triangle: altitude is $7$ cm less than base, hypotenuse $13$ cm. Let base $=x$, altitude $=x-7$: $x^{2}+(x-7)^{2}=13^{2}\Rightarrow 2x^{2}-14x+49=169\Rightarrow x^{2}-7x-60=0\Rightarrow(x-12)(x+5)=0\Rightarrow x=12$. So base $=\mathbf{12}$ cm, altitude $=\mathbf{5}$ cm.

Q6. Cottage pottery: cost per article $=(2\times\text{number}+3)$, total cost $\rupee90$. Let number $=x$: $x(2x+3)=90\Rightarrow 2x^{2}+3x-90=0\Rightarrow(2x+15)(x-6)=0\Rightarrow x=6$ (reject $-7.5$). So $\mathbf{6}$ articles, cost each $=2(6)+3=\rupee\mathbf{15}$.

11. NCERT Exercise 4.3 — fully solved

Q1. Find the nature of roots; if real, find them.

  • (i) $2x^{2}-3x+5=0$: $D=(-3)^{2}-4(2)(5)=9-40=-31<0$ — no real roots.
  • (ii) $3x^{2}-4\sqrt3\,x+4=0$: $D=(-4\sqrt3)^{2}-4(3)(4)=48-48=0$ — two equal real roots, each $=-\dfrac{b}{2a}=\dfrac{4\sqrt3}{6}=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}$.
  • (iii) $2x^{2}-6x+3=0$: $D=(-6)^{2}-4(2)(3)=36-24=12>0$ — two distinct real roots. $x=\dfrac{6\pm\sqrt{12}}{4}=\dfrac{6\pm2\sqrt3}{4}=\dfrac{3\pm\sqrt3}{2}$.

Q2. Find $k$ so that the equation has two equal roots (need $D=0$).

  • (i) $2x^{2}+kx+3=0$: $D=k^{2}-4(2)(3)=k^{2}-24=0\Rightarrow k=\pm\sqrt{24}=\pm2\sqrt6$.
  • (ii) $kx(x-2)+6=0\Rightarrow kx^{2}-2kx+6=0$: $D=(-2k)^{2}-4(k)(6)=4k^{2}-24k=0\Rightarrow 4k(k-6)=0\Rightarrow k=6$ (reject $k=0$, else not quadratic).

Q3. Rectangular mango grove, length twice breadth, area $800\text{ m}^{2}$? Let breadth $=x$, length $=2x$: $2x^{2}=800\Rightarrow x^{2}=400\Rightarrow x=20$ (reject $-20$). $D=0+4(2)(800)>0$, so yes possible: breadth $=\mathbf{20}$ m, length $=\mathbf{40}$ m.

Q4. Sum of ages of two friends is $20$; four years ago product was $48$. Let ages $x$ and $20-x$: $(x-4)(20-x-4)=48\Rightarrow(x-4)(16-x)=48\Rightarrow -x^{2}+20x-64=48\Rightarrow x^{2}-20x+112=0$. Here $D=400-448=-48<0$ — no real roots, so this situation is not possible.

Q5. Rectangular park, perimeter $80$ m, area $400\text{ m}^{2}$? Then length $+$ breadth $=40$. Let breadth $=x$, length $=40-x$: $x(40-x)=400\Rightarrow x^{2}-40x+400=0\Rightarrow(x-20)^{2}=0\Rightarrow x=20$. $D=1600-1600=0$, so it is possible with equal sides: length $=$ breadth $=\mathbf{20}$ m (a square).

12. Common mistakes to avoid

  • Forgetting the $a\neq0$ condition, or not simplifying before declaring an equation quadratic (Example 2(iv) looked cubic but was quadratic).
  • When splitting the middle term, finding two numbers with the right sum but wrong product $a\times c$ (or vice versa).
  • In the formula, sign slips: $-b$ for negative $b$ means $-(-4)=+4$; and $b^{2}$ is always positive.
  • Stopping at one root from $\pm$ — a quadratic has up to two; always report both.
  • Keeping a negative/impossible root in a word problem (length, age, speed, count must be valid).
  • Confusing the discriminant cases: $D>0$ distinct, $D=0$ equal, $D<0$ none.
  • For "equal roots" questions, forgetting to reject the value of $k$ that makes $a=0$.

13. Quick revision checklist

  • Standard form $ax^{2}+bx+c=0$, $a\neq0$; root = value that satisfies it; at most two roots.
  • Factorisation: split the middle term so the two parts multiply to $a\times c$.
  • Completing the square: make $\left(x+\tfrac{b}{2a}\right)^{2}=$ number, then take roots.
  • Quadratic formula $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (needs $D\ge0$).
  • Discriminant $D=b^{2}-4ac$: $>0$ two distinct, $=0$ two equal, $<0$ none.
  • Word problems: define variable → form equation → solve → reject impossible root → answer with units.
Practice MCQs
1. Which of the following is a quadratic equation?
  1. $x^{2}+\dfrac1x=2$
  2. $(x+1)^{2}=x^{2}+5$
  3. $x^{2}-3x+2=0$
  4. $x^{3}-x=0$
Answer: (C) — degree exactly $2$ with $a\neq0$. (A) is not polynomial, (B) simplifies to linear, (D) is cubic.
2. The roots of $x^{2}-3x-10=0$ are:
  1. $5,\,-2$
  2. $-5,\,2$
  3. $5,\,2$
  4. $-5,\,-2$
Answer: (A) $(x-5)(x+2)=0\Rightarrow x=5,-2.$
3. The discriminant of $2x^{2}-4x+3=0$ is:
  1. $8$
  2. $-8$
  3. $40$
  4. $-40$
Answer: (B) $D=(-4)^{2}-4(2)(3)=16-24=-8.$
4. A quadratic equation has two equal real roots when:
  1. $b^{2}-4ac>0$
  2. $b^{2}-4ac<0$
  3. $b^{2}-4ac=0$
  4. $b^{2}-4ac\neq0$
Answer: (C) $D=0$ gives equal (coincident) roots.
5. For what value of $k$ does $2x^{2}+kx+3=0$ have equal roots?
  1. $\pm2\sqrt6$
  2. $\pm\sqrt6$
  3. $\pm12$
  4. $\pm24$
Answer: (A) $D=k^{2}-24=0\Rightarrow k=\pm2\sqrt6.$
6. The sum of a number and its reciprocal is $\dfrac{10}{3}$. The number is:
  1. $2$ or $\dfrac12$
  2. $3$ or $\dfrac13$
  3. $4$ or $\dfrac14$
  4. $5$ or $\dfrac15$
Answer: (B) $x+\dfrac1x=\dfrac{10}{3}\Rightarrow 3x^{2}-10x+3=0\Rightarrow(3x-1)(x-3)=0\Rightarrow x=3$ or $\dfrac13.$
7. The product of two consecutive positive integers is $306$. The equation is:
  1. $x^{2}+x-306=0$
  2. $x^{2}-x-306=0$
  3. $x^{2}+306=0$
  4. $x^{2}-306x=0$
Answer: (A) $x(x+1)=306\Rightarrow x^{2}+x-306=0.$
8. If one root of $x^{2}+px+12=0$ is $4$, then $p=$
  1. $-7$
  2. $7$
  3. $-4$
  4. $3$
Answer: (A) Put $x=4$: $16+4p+12=0\Rightarrow 4p=-28\Rightarrow p=-7.$
9. The roots of $100x^{2}-20x+1=0$ are:
  1. $\dfrac1{10},\dfrac1{10}$
  2. $10,10$
  3. $\dfrac15,\dfrac15$
  4. $-\dfrac1{10},\dfrac1{10}$
Answer: (A) $(10x-1)^{2}=0\Rightarrow x=\dfrac1{10}$ (repeated).
10. For $3x^{2}-4\sqrt3\,x+4=0$, the nature of roots is:
  1. two distinct real
  2. two equal real
  3. no real roots
  4. cannot be determined
Answer: (B) $D=48-48=0$, so two equal real roots.
Assertion–Reason
A: The equation $2x^{2}-3x+5=0$ has no real roots.   R: A quadratic equation has no real roots when its discriminant $b^{2}-4ac<0$.
Answer: Both A and R are true, and R correctly explains A — here $D=9-40=-31<0$.
A: $x^{2}+x+1=0$ has two equal real roots.   R: The roots of $ax^{2}+bx+c=0$ are equal when $b^{2}-4ac=0$.
Answer: A is false (here $D=1-4=-3<0$, so no real roots), R is true. The reason does not explain a false assertion.
Previous-year questions
Q1. Solve $\dfrac{480}{x-8}-\dfrac{480}{x}=3$ to find the speed of a train covering $480$ km. (CBSE, 3 marks)
Answer: Simplify to $x^{2}-8x-1280=0\Rightarrow(x-40)(x+32)=0\Rightarrow x=40$ (reject $-32$). Speed $=40$ km/h.
Q2. Find the roots of $2x^{2}-6x+3=0$ using the quadratic formula. (CBSE, 2 marks)
Answer: $D=36-24=12>0$; $x=\dfrac{6\pm\sqrt{12}}{4}=\dfrac{3\pm\sqrt3}{2}.$
Q3. Is it possible to design a rectangular park of perimeter $80$ m and area $400\text{ m}^{2}$? If so, find its dimensions. (CBSE, 3 marks)
Answer: $x(40-x)=400\Rightarrow x^{2}-40x+400=0\Rightarrow(x-20)^{2}=0$; $D=0$ so possible — a square of side $20$ m.
Q4. The sum of the squares of two consecutive positive integers is $365$. Find them. (CBSE, 3 marks)
Answer: $x^{2}+(x+1)^{2}=365\Rightarrow x^{2}+x-182=0\Rightarrow(x-13)(x+14)=0\Rightarrow x=13$. Integers are $13$ and $14$.
Want personal coaching in Dwarka?
Book a free demo class
More Class 10 Mathematics chapters
Real Numbers · Polynomials · Pair of Linear Equations in Two Variables · Arithmetic Progressions · Triangles · Coordinate Geometry · Introduction to Trigonometry · Some Applications of Trigonometry