is a solution of 
The given equation is $x^{2}+6 x+9=0$ Putting $x=-3$ in the given equation, we get $L H S=(-3)^{2}+6 \times(-3)+9=9-18+9=0=R H S$ $\therefore x=-3$ is a solution of the given equation.
at a uniform speed. If the speed had been 
less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.Let the usual speed of the train be $x \mathrm{~km} / \mathrm{h}$. $\therefore$ Reduced speed of the train $=(x-8) \mathrm{km} / \mathrm{h}$ Total distance to be covered $=480 \mathrm{~km}$ Time...
Let there be $x$ rows. Then, the number of students in each row will also be $x$. $\therefore$ Total number of students $=\left(x^{2}+24\right)$ According to the question: $\begin{array}{l}...
Find the number.Let the required natural number be $x$. According to the given condition, $x+x^{2}=156$ $\Rightarrow x^{2}+x-156=0$ $\Rightarrow x^{2}+13 x-12 x-156=0$ $\Rightarrow x(x+13)-12(x+13)=0$...
.The given equation is $\sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0$. Comparing it with $a x^{2}+b x+c=0$, we get $a=\sqrt{3}, b=-2 \sqrt{2}$ and $c=-2 \sqrt{3}$ $\therefore$ Discriminant, $D=b^{2}-4 a...
Given: $x^{2}-6 x+4=0$ On comparing it with $a x^{2}+b x+c=0$, we get: $a=1, b=-6$ and $c=4$ Discriminant $D$ is given by: $ \begin{array}{l} D=\left(b^{2}-4 a c\right) \\ =(-6)^{2}-4 \times 1...
$\frac{2}{x^{2}}-\frac{5}{x}+2=0$ $\Rightarrow \frac{2-5 x+2 x^{2}}{x^{2}}=0$ $\Rightarrow 2 x^{2}-5 x+2=0$ $\Rightarrow 4 x^{2}-10 x+4=0 \quad$ (Multiplying both sides by 2) $\Rightarrow 4 x^{2}-10...
$\begin{array}{l} x^{2}-6 x+3=0 \\ \Rightarrow x^{2}-6 x=-3 \\ \Rightarrow x^{2}-2 \times x \times 3+3^{2}=-3+3^{2} \\ \Rightarrow(x-3)^{2}=-3+9=6 \\ \Rightarrow x-3=\pm \sqrt{6} \end{array}$...
We write, $6 x=(a+4) x-(a-2) x$ as $\begin{array}{l} x^{2} \times\left[-\left(a^{2}+2 a-8\right)\right]=-\left(a^{2}+2 a-8\right) x^{2}=(a+4) x \times[-(a-2) x] \\ \therefore x^{2}+6 x-\left(a^{2}+2...
$\begin{array}{l} x^{2}-(1+\sqrt{2}) x+\sqrt{2}=0 \\ \Rightarrow x^{2}-x-\sqrt{2} x+\sqrt{2}=0 \\ \Rightarrow x(x-1)-\sqrt{2}(x-1)=0 \\ \Rightarrow(x-\sqrt{2})(x-1)=0 \\ \Rightarrow x-\sqrt{2}=0...
$\begin{array}{l} 4 \sqrt{6} x^{2}-13 x-2 \sqrt{6}=0 \\ \Rightarrow 4 \sqrt{6} x^{2}-16 x+3 x-2 \sqrt{6}=0 \\ \Rightarrow 4 \sqrt{2} x(\sqrt{3} x-2 \sqrt{2})+\sqrt{3}(\sqrt{3} x-2 \sqrt{2})=0 \\...
Given: tan θ+ sec θ = l … eq. 1 Duplicating and isolating by (sec θ – tan θ) on numerator and denominator of L.H.S, Along these lines, sec θ – tan θ = 1 … eq.2 Adding eq. 1and eq. 2, we get \[\left(...
Considering that an upward banner staff of stature h is overcomed on an upward pinnacle of tallness H(say), with the end goal that FP = h and FO = H. The point of height of the base and top of the...
Considering that an upward banner staff of stature h is overcomed on an upward pinnacle of tallness H(say), with the end goal that FP = h and FO = H. The point of height of the base and top of the...
Let SQ = h be the pinnacle. ∠SPQ = 30° and ∠SRQ = 60° As per the inquiry, the length of shadow is 50 m long hen point of rise of the sun is 30° than when it was 60°. Thus, PR = 50 m and RQ = x m So...
Allow Zeba's to age = x As indicated by the inquiry, \[\left( x-5 \right){}^\text{2}=11+5x\] \[x{}^\text{2}+25-10x=11+5x\] \[x{}^\text{2}-15x+14=0\] \[x{}^\text{2}-14x-x+14=0\] \[x\left( x-14...
Let unique speed of train = x km/h We know, Time = distance/speed As indicated by the inquiry, we have, Time taken via train = 360/x hour What's more, Time taken via train its speed increment 5 km/h...
Let the normal number = 'x'. As per the inquiry, We get the condition, \[x{}^\text{2}\text{ }\text{ }84\text{ }=\text{ }3\left( x+8 \right)\] \[x{}^\text{2}\text{ }\text{ }84\text{ }=\text{...
(½)x2– √11x + 1 = 0 using the formulae: we get,
x2 + 2 √2x – 6 = 0x2 – 3 √5x + 10 = 0 using formulae: 5. we get, 6.
–3x2 + 5x + 12 = 0–x2 + 7x – 10 = 0 using the formulae: 3. we get, 4.
2 x2 – 3x – 5 = 05x2 + 13x + 8 = 0by using ths formulae below: 1. we get, 2.
No, a quadratic condition with essential coefficients might possibly have indispensable roots. Support Think about the accompanying condition, \[8x2\text{ }\text{ }2x\text{ }\text{ }1\text{ }=\text{...
If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.If the coefficient of x2 and the constant term have the same...
Every quadratic equation has at least two roots.Every quadratic equations has at most two roots. (iii) False. For instance, a quadratic condition \[x2\text{ }\text{ }4x\text{ }+\text{ }4\text{...
Every quadratic equation has exactly one root.Every quadratic equation has at least one real root. (I) False. For instance, a quadratic condition \[x2\text{ }\text{ }9\text{ }=\text{ }0\] has two...
(x – 1) (x + 2) + 2 = 0(x + 1) (x – 2) + x = 0 (ix) The condition (x – 1) (x + 2) + 2 = 0 has two genuine and unmistakable roots. Working on the above condition, \[x2\text{ }\text{ }x\text{ }+\text{...
√2 x2 –(3/√2)x + 1/√2 = 0x (1 – x) – 2 = 0 (vii) The condition √2x2 – 3x/√2 + ½ = 0 has two genuine and unmistakable roots. \[\begin{array}{*{35}{l}} D\text{ }=\text{ }b2\text{ }\text{...
(x + 4)2 – 8x = 0(x – √2)2 – 2(x + 1) = 0 (v) The condition (x + 4)2 – 8x = 0 has no genuine roots. Working on the above condition, \[x2\text{ }+\text{ }8x\text{ }+\text{...
2x2 – 6x + 9/2 = 03x2 – 4x + 1 = 0 (iii) The condition 2x2 – 6x + (9/2) = 0 has genuine and equivalent roots. \[D\text{ }=\text{ }b2\text{ }\text{ }4ac\] \[=\text{ }\left( -\text{ }6 \right)2\text{...
x2 – 3x + 4 = 02x2 + x – 1 = 0 (I) The condition x2 – 3x + 4 = 0 has no genuine roots. \[D\text{ }=\text{ }b2\text{ }\text{ }4ac\] \[=\text{ }\left( -\text{ }3 \right)2\text{ }\text{...
(A) 2x2 – 3x + 6 = 0 (B) –x2 + 3x – 3 = 0 (C) √2x2 – 3/√2x+1=0 (D) 3x2 – 3x + 3 = 0 (B) – \[x2\text{ }+\text{ }3x\text{ }\text{ }3\text{ }=\text{ }0\] The amount of the foundations of a...
(A) x2 – 4x + 5 = 0 (B) x2 + 3x – 12 = 0 (C) 2x2 – 7x + 6 = 0 (D) 3x2 – 6x – 2 = 0 (C) \[2x2\text{ }\text{ }7x\text{ }+\text{ }6\text{ }=\text{ }0\] Assuming 2 is a...
(A) 2(x – 1)2 = 4x2 – 2x + 1 (B) 2x – x2 = x2 + 5 (C) (√2x + √3)2 + x2 = 3x2 − 5x (D) (x2 + 2x)2 = x4 + 3 + 4x3 (D) \[\left( x2\text{ }+\text{ }2x...
Solution: Let’s consider the original number of people as ‘a’. Given that, $Rs.9000$ were divided equally among a certain number of persons. Had there been $20$ more persons, each would have got...
Solution: Let the original price of the toy be ‘x’. Given that, when the list price of a toy is reduced by $Rs.2$, the person can buy $2$ toys more for $Rs.360$. The number of toys he can buy at the...
Solution: Let the cost price be assumed as Rs x. Given, the dealer sells an article for $Rs.24$ and gains as much percent as the cost price of the article. It’s given that he gains as much as the...
Solution: Let the number of students who planned the picnic be ‘x’. And given, budget for the food was $Rs.480$ So, cost of food for each member $= 480/x$ Also given, eight of these failed to go and...
Solution: Let’s assume the length of the cloth to be ‘a’ meters. Given, piece of cloth costs $Rs.35$ and if the piece were $4m$ longer and each metre costs $Rs.1$ less, the cost remains unchanged....
(A) x2 + 2x + 1 = (4 – x)2 + 3 (B) –2x2 = (5 – x)(2x-(2/5)) (C) (k + 1) x2 + (3/2) x = 7, where k = –1 (D) x3 – x2 = (x – 1)3 (D) \[x3\text{ }\text{ }x2\text{ }=\text{ }\left( x\text{ }\text{ }1...
hour less for a journey of 
if its speed is increased by 
from its usual speed of the plane. Find its usual speed.Solution: ...
km was 
more than the time taken in the return journey. If he returned at the speed of 
more than the speed of going, what was the speed per hour in each direction?Solution: Let the ongoing speed of person be x km/hr, Then, the returning speed of the person is $= (x + 10) km/hr$ (from the question) Using, speed = distance/ time Time taken by the person in...
km if its speed is increased 
from its usual speed. Find the usual speed of the train.Solution: Let’s assume the usual speed of train as x km/hr Then, the increased speed of the train $= (x + 5) km/hr$ Using, speed = distance/ time Time taken by the train under usual speed to...
. If the speed of the slow train is less than that of the fast train, find the speed of the two trainsSolution: Let’s consider the speed of the fast train as x km/hr Then, the speed of the slow train will be $= (x -10) km/hr$ Using, speed = distance/ time Time taken by the fast train to cover $200...
km, would have taken 
minutes less to travel the same distance if its speed were 
km/hr more. Find the original speed of the train.Solution: Let the original speed of train be x km/hr When increased by $5$, speed of the train $= (x + 5) km/hr$ Using, speed = distance/...
km/hr. It can go 
km upstream and 
km downstream in hours. Find the speed of the stream.Solution: Let the speed of stream be x km/hr Given, speed of boat in still water is $8km/hr$. So, speed of downstream $= (8 + x) km/hr$ And, speed of upstream $= (8 – x) km/hr$ Using, speed =...