Quadratic Equations — CBSE Class 10 Maths Important Questions

Answer: (a) $-8$-8 — $D=b^2-4ac=(-4)^2-4(2)(3)=16-24=-8$D=b²-4ac=(-4)²-4(2)(3)=16-24=-8. Since $D<0$D<0, the equation has no real roots.

Answer: (a) $5,\ -2$5, -2 — $x^2-3x-10=(x-5)(x+2)=0$x²-3x-10=(x-5)(x+2)=0, so $x=5$x=5 or $x=-2$x=-2.

Answer: (a) $\pm2\sqrt6$±2√6 — For equal roots $D=0$D=0: $\ k^2-4(2)(3)=0\Rightarrow k^2=24\Rightarrow k=\pm2\sqrt6$k²-4(2)(3)=0⇒ k²=24⇒ k=±2√6.

Answer: (a) $7$7 — Substituting $x=-5$x=-5: $\ 2(25)+p(-5)-15=0\Rightarrow 50-5p-15=0\Rightarrow 5p=35\Rightarrow p=7$2(25)+p(-5)-15=0⇒ 50-5p-15=0⇒ 5p=35⇒ p=7.

Answer: (a) Both are true and R explains A. For $x^2+x+1=0$x²+x+1=0, $\ D=1^2-4(1)(1)=-3<0$D=1²-4(1)(1)=-3<0, so it has no real roots — exactly the condition in R.

Let Rohan's present age be $x$x years; his mother's is $(x+26)$(x+26) years. In $3$3 years: $(x+3)(x+29)=360$(x+3)(x+29)=360.
$x^2+32x+87=360\Rightarrow x^2+32x-273=0\Rightarrow (x+39)(x-7)=0$x²+32x+87=360⇒ x²+32x-273=0⇒ (x+39)(x-7)=0.
Age cannot be negative, so $x=7$x=7. Rohan is $7$7 years old (his mother is $33$33). (Check: $10\times36=360$10×36=360 ✓)

(i) $x^2+2x+1=2x-6\Rightarrow x^2+7=0$x²+2x+1=2x-6⇒ x²+7=0. This is of the form $ax^2+bx+c=0$ax²+bx+c=0 with $a=1\neq0$a=1≠0, so it is a quadratic equation.
(ii) $x^2-x-2=x^2+2x-3\Rightarrow -3x+1=0$x²-x-2=x²+2x-3⇒ -3x+1=0. The $x^2$x² terms cancel, leaving a linear equation, so it is not a quadratic equation.

Let the integers be $x$x and $x+1$x+1.

$x^2+(x+1)^2=365\Rightarrow 2x^2+2x+1=365\Rightarrow 2x^2+2x-364=0\Rightarrow x^2+x-182=0.$x²+(x+1)²=365⇒ 2x²+2x+1=365⇒ 2x²+2x-364=0⇒ x²+x-182=0.

$(x+14)(x-13)=0\Rightarrow x=13$(x+14)(x-13)=0⇒ x=13 (rejecting $x=-14$x=-14 since the integers are positive).

The integers are $13$13 and $14$14.

Let the larger number be $x$x and the smaller be $y$y. Given $x^2-y^2=180$x²-y²=180 and $y^2=8x$y²=8x.

Substituting: $x^2-8x=180\Rightarrow x^2-8x-180=0\Rightarrow (x-18)(x+10)=0\Rightarrow x=18$x²-8x=180⇒ x²-8x-180=0⇒ (x-18)(x+10)=0⇒ x=18 (taking $x>0$x>0 so that $y^2=8x>0$y²=8x>0).

Then $y^2=8(18)=144\Rightarrow y=\pm12$y²=8(18)=144⇒ y=±12.

The numbers are $18$18 and $12$12 (or $18$18 and $-12$-12).

The coefficients sum to zero: $(b-c)+(c-a)+(a-b)=0$(b-c)+(c-a)+(a-b)=0, so $x=1$x=1 is always a root. For equal roots both roots must equal $1$1, hence the product of the roots is $1$1:
$\frac{a-b}{b-c}=1\Rightarrow a-b=b-c\Rightarrow a+c=2b.$(a-b)/(b-c)=1⇒ a-b=b-c⇒ a+c=2b.
Hence proved.

(Equivalently, $D=(c-a)^2-4(b-c)(a-b)=0$D=(c-a)²-4(b-c)(a-b)=0 simplifies to $(a+c-2b)^2=0$(a+c-2b)²=0.)

Here $a=2,\ b=1,\ c=-4$a=2, b=1, c=-4, so $D=b^2-4ac=1+32=33$D=b²-4ac=1+32=33.
$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-1\pm\sqrt{33}}{4}$x=(-b± √D)/2a=(-1± √33)/4
Since $D=33>0$D=33>0, the roots are real and distinct, and since $33$33 is not a perfect square, they are irrational. This equation cannot be solved by simple factorisation, which is why the formula is needed.

Let the speed be $x$x km/h, so the time taken is $\dfrac{360}{x}$360/x hours.

$\dfrac{360}{x}-\dfrac{360}{x+5}=1\Rightarrow 360\left(\dfrac{(x+5)-x}{x(x+5)}\right)=1\Rightarrow \dfrac{1800}{x^2+5x}=1.$360/x-360/(x+5)=1⇒ 360(((x+5)-x)/(x(x+5)))=1⇒ 1800/(x²+5x)=1.

$x^2+5x-1800=0.$x²+5x-1800=0. Here $D=25+7200=7225=85^2$D=25+7200=7225=85², so $x=\dfrac{-5\pm85}{2}$x=-5±85/2.

Taking the positive value, $x=\dfrac{80}{2}=40$x=80/2=40.

The speed of the train is $40$40 km/h.

Let the smaller tap take $x$x hours; then the larger takes $(x-10)$(x-10) hours. Together they fill the tank in $\dfrac{75}{8}$75/8 hours, so
$\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}.$1/x+1/(x-10)=8/75.
$\dfrac{(x-10)+x}{x(x-10)}=\dfrac{8}{75}\Rightarrow 75(2x-10)=8(x^2-10x)\Rightarrow 8x^2-230x+750=0\Rightarrow 4x^2-115x+375=0.$((x-10)+x)/(x(x-10))=8/75⇒ 75(2x-10)=8(x²-10x)⇒ 8x²-230x+750=0⇒ 4x²-115x+375=0.

$D=115^2-4(4)(375)=13225-6000=7225=85^2$D=115²-4(4)(375)=13225-6000=7225=85², so $x=\dfrac{115\pm85}{8}$x=115±85/8, giving $x=25$x=25 or $x=3.75$x=3.75.

Since $x-10$x-10 must be positive, reject $x=3.75$x=3.75. Thus $x=25$x=25.

Smaller tap: $25$25 hours; larger tap: $15$15 hours.

(i) Length $=(2x+1)$=(2x+1) m, so $x(2x+1)=528\Rightarrow 2x^2+x-528=0$x(2x+1)=528⇒ 2x²+x-528=0.

(ii) $D=b^2-4ac=1^2-4(2)(-528)=1+4224=4225=65^2$D=b²-4ac=1²-4(2)(-528)=1+4224=4225=65².

(iii) $x=\dfrac{-1\pm65}{4}$x=-1±65/4. Taking the positive root, $x=\dfrac{64}{4}=16$x=64/4=16. So breadth $=16$=16 m and length $=2(16)+1=33$=2(16)+1=33 m. (Check: $16\times33=528$16×33=528 ✓.)