Chapter 4: Quadratic Equations (Algebra)

Expand and bring everything to one side. It is quadratic if the highest power left is exactly 2.

1. $x^2+2x+1=2x-6\Rightarrow x^2+7=0$. Yes.
2. $x^2-2x=-6+2x\Rightarrow x^2-4x+6=0$. Yes.
3. $x^2-x-2=x^2+2x-3\Rightarrow -3x+1=0$. Degree 1, No.
4. $2x^2-5x-3=x^2+5x\Rightarrow x^2-10x-3=0$. Yes.
5. $2x^2-7x+3=x^2+4x-5\Rightarrow x^2-11x+8=0$. Yes.
6. $x^2+3x+1=x^2-4x+4\Rightarrow 7x-3=0$. Degree 1, No.
7. $x^3+6x^2+12x+8=2x^3-2x\Rightarrow x^3-6x^2-14x-8=0$. Degree 3, No.
8. $x^3-4x^2-x+1=x^3-6x^2+12x-8\Rightarrow 2x^2-13x+9=0$. Yes.

Quadratic: (i), (ii), (iv), (v), (viii). Not quadratic: (iii), (vi), (vii).

1. Let the breadth be $x$ m; the length is $2x+1$. Then $x(2x+1)=528$, i.e. $2{\mathbf{x}}^2+\mathbf{x}-528=0$.
2. Let the integers be $x$ and $x+1$: $x(x+1)=306$, i.e. ${\mathbf{x}}^2+\mathbf{x}-306=0$.
3. Let Rohan be $x$ years old; his mother is $x+26$. Then $(x+3)(x+29)=360$, i.e. ${\mathbf{x}}^2+32x-273=0$.
4. Let the speed be $u$ km/h: $\frac{480}{u-8}-\frac{480}{u}=3\Rightarrow 480\times 8=3u(u-8)$, i.e. ${\mathbf{u}}^2-8u-1280=0$.

(i) 2x² + x – 528 = 0 (ii) x² + x – 306 = 0 (iii) x² + 32x – 273 = 0 (iv) u² – 8u – 1280 = 0

1. $x^2-5x+2x-10=(x-5)(x+2)=0$, so $x=5,-2$.
2. $2x^2+4x-3x-6=(2x-3)(x+2)=0$, so $x=\frac{3}{2},-2$.
3. Split $7x=5x+2x$ (since $\sqrt{2}\times 5\sqrt{2}=10=5\times 2$): $\sqrt{2}{x}^2+5x+2x+5\sqrt{2}=x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=(x+\sqrt{2})(\sqrt{2}x+5)=0$, so $x=-\sqrt{2},-\frac{5}{\sqrt{2}}$.
4. Multiply by 8: $16x^2-8x+1=(4x-1{)}^2=0$, so $x=\frac{1}{4},\frac{1}{4}$.
5. $(10x-1{)}^2=0$, so $x=\frac{1}{10},\frac{1}{10}$.

(i) 5, –2 (ii) 3/2, –2 (iii) –√2, –5/√2 (iv) 1/4, 1/4 (v) 1/10, 1/10

(i) Let John have $x$; Jivanti has $45-x$. Then $(x-5)(40-x)=124\Rightarrow -x^2+45x-200=124\Rightarrow x^2-45x+324=0\Rightarrow (x-36)(x-9)=0$.

So John had 36 and Jivanti 9, or John 9 and Jivanti 36.

(ii) Let $x$ toys be made: $x(55-x)=750\Rightarrow x^2-55x+750=0\Rightarrow (x-25)(x-30)=0$.

(i) They had 36 and 9 marbles. (ii) 25 or 30 toys were made that day.

Let the numbers be $x$ and $27-x$: $x(27-x)=182\Rightarrow x^2-27x+182=0\Rightarrow (x-13)(x-14)=0$.

13 and 14

$x^2+(x+1{)}^2=365\Rightarrow 2x^2+2x-364=0\Rightarrow x^2+x-182=0\Rightarrow (x+14)(x-13)=0$.

$x$ is positive, so $x=13$. Check: $169+196=365$ ✓

13 and 14

Let the base be $x$ cm; the altitude is $x-7$. By Pythagoras: $x^2+(x-7{)}^2=169\Rightarrow 2x^2-14x-120=0\Rightarrow x^2-7x-60=0\Rightarrow (x-12)(x+5)=0$.

A length cannot be negative, so $x=12$ and the altitude is 5.

Base 12 cm and altitude 5 cm.

Let $x$ articles be made; each costs $2x+3$. Then $x(2x+3)=90\Rightarrow 2x^2+3x-90=0\Rightarrow 2x^2+15x-12x-90=(2x+15)(x-6)=0$.

$x$ must be a positive whole number, so $x=6$, and each costs $2(6)+3=15$.

6 articles, each costing ₹15.

1. $D=9-40=-31<0$: no real roots.
2. $D=48-48=0$: two equal real roots, $x=\frac{4\sqrt{3}}{6}=\frac{2}{\sqrt{3}}$ each.
3. $D=36-24=12>0$: two distinct real roots, $x=\frac{6\pm \sqrt{12}}{4}=\frac{3\pm \sqrt{3}}{2}$.

(i) No real roots (ii) Equal roots, 2/√3 and 2/√3 (iii) Distinct roots, (3 + √3)/2 and (3 – √3)/2

For equal roots, $D=0$.

1. $k^2-4(2)(3)=0\Rightarrow k^2=24\Rightarrow k=\pm 2\sqrt{6}$.
2. $k{x}^2-2kx+6=0$: $4k^2-24k=0\Rightarrow 4k(k-6)=0$. Since $k=0$ would not give a quadratic, $k=6$.

(i) k = ±2√6 (ii) k = 6

Let the breadth be $x$ m: $x(2x)=800\Rightarrow x^2=400\Rightarrow x=20$ (positive). $D=0-4(2)(-800)>0$, so real roots exist.

Yes: breadth 20 m and length 40 m.

Let one friend be $x$ years; the other is $20-x$. Four years ago: $(x-4)(16-x)=48\Rightarrow -x^2+20x-64=48\Rightarrow x^2-20x+112=0$.

$D=400-448=-48<0$, so there are no real roots.

Not possible: the equation x² – 20x + 112 = 0 has no real roots.

Length + breadth $=40$. Let the length be $x$: $x(40-x)=400\Rightarrow x^2-40x+400=0\Rightarrow (x-20{)}^2=0$.

$D=0$, so $x=20$ and the breadth is also 20.

Yes: it is a square park, 20 m by 20 m.