true

Take Class 10 Tuition from the Best Tutors

• Affordable fees
• 1-1 or Group class
• Flexible Timings
• Verified Tutors

Search in

# Learn UNIT II: Algebra with Free Lessons & Tips

Post a Lesson

All

All

Lessons

Discussion

To have infinitely many solutions, the two equations must represent the same line, meaning one equation is a multiple of the other. Given equations: read more

To have infinitely many solutions, the two equations must represent the same line, meaning one equation is a multiple of the other.

Given equations:$T1.&space;$$2x&space;-&space;3y&space;=&space;7$$&space;2.&space;$$(k+1)x&space;+&space;(1-2k)y&space;=&space;5k&space;-&space;4$$&space;To&space;check&space;if&space;one&space;equation&space;is&space;a&space;multiple&space;of&space;the&space;other,&space;we'll&space;compare&space;their&space;slopes&space;and&space;intercepts.&space;The&space;slope-intercept&space;form&space;of&space;the&space;first&space;equation&space;is&space;$$y&space;=&space;\frac{2}{3}x&space;-&space;\frac{7}{3}$$.&space;The&space;second&space;equation&space;can&space;be&space;rewritten&space;as:&space;$(k+1)x&space;+&space;(1-2k)y&space;=&space;5k&space;-&space;4$&space;$y&space;=&space;\frac{k+1}{1-2k}x&space;+&space;\frac{5k-4}{1-2k}$&space;Comparing&space;the&space;slopes:&space;$\frac{2}{3}&space;=&space;\frac{k+1}{1-2k}$&space;This&space;implies:&space;$2(1-2k)&space;=&space;3(k+1)$&space;$2&space;-&space;4k&space;=&space;3k&space;+&space;3$&space;$2&space;-&space;3&space;=&space;3k&space;+&space;4k$&space;$-1&space;=&space;7k$&space;$k&space;=&space;-\frac{1}{7}$&space;So,&space;for&space;$$k&space;=&space;-\frac{1}{7}$$,&space;the&space;given&space;equations&space;will&space;have&space;infinitely&space;many&space;solutions.$

Dislike Bookmark

Let's denote the numerator of the fraction as xx and the denominator as yy. According to the given conditions: The sum of the denominator and numerator of a fraction is 3 less than twice the denominator: read more

Let's denote the numerator of the fraction as xx and the denominator as yy.

According to the given conditions:

1. The sum of the denominator and numerator of a fraction is 3 less than twice the denominator:$L&space;$x&space;+&space;y&space;=&space;2y&space;-&space;3$&space;2.&space;If&space;each&space;of&space;the&space;numerator&space;and&space;denominator&space;is&space;decreased&space;by&space;1,&space;the&space;fraction&space;becomes&space;$$&space;\frac{1}{2}&space;$$:&space;$\frac{x-1}{y-1}&space;=&space;\frac{1}{2}$&space;We&space;have&space;a&space;system&space;of&space;equations.&space;Let's&space;solve&space;it:&space;From&space;the&space;first&space;equation,&space;we&space;can&space;express&space;$$x$$&space;in&space;terms&space;of&space;$$y$$:&space;$x&space;=&space;2y&space;-&space;3&space;-&space;y$&space;$x&space;=&space;y&space;-&space;3$&space;Substitute&space;this&space;value&space;of&space;$$x$$&space;into&space;the&space;second&space;equation:&space;$\frac{y-3-1}{y-1}&space;=&space;\frac{1}{2}$&space;$\frac{y-4}{y-1}&space;=&space;\frac{1}{2}$&space;Cross&space;multiply:&space;$2(y&space;-&space;4)&space;=&space;y&space;-&space;1$&space;$2y&space;-&space;8&space;=&space;y&space;-&space;1$&space;$2y&space;-&space;y&space;=&space;8&space;-&space;1$&space;$y&space;=&space;7$&space;Now&space;that&space;we&space;have&space;found&space;$$y$$,&space;let's&space;find&space;$$x$$:&space;$x&space;=&space;y&space;-&space;3$&space;$x&space;=&space;7&space;-&space;3$&space;$x&space;=&space;4$&space;So,&space;the&space;fraction&space;is&space;$$&space;\frac{4}{7}&space;$$.$

Dislike Bookmark

Let's denote the tens digit of the two-digit number as xx and the units digit as yy. The original number can then be represented as 10x+y. According to the given conditions: The sum of the digits of the two-digit number is 12: read more

Let's denote the tens digit of the two-digit number as xx and the units digit as yy. The original number can then be represented as 10x+y.$L1.&space;The&space;sum&space;of&space;the&space;digits&space;of&space;the&space;two-digit&space;number&space;is&space;12:&space;$x&space;+&space;y&space;=&space;12$&space;2.&space;The&space;number&space;obtained&space;by&space;interchanging&space;the&space;two&space;digits&space;exceeds&space;the&space;given&space;number&space;by&space;18:&space;$10y&space;+&space;x&space;=&space;10x&space;+&space;y&space;+&space;18$&space;We&space;have&space;a&space;system&space;of&space;equations.&space;Let's&space;solve&space;it:&space;From&space;the&space;first&space;equation,&space;we&space;can&space;express&space;$$x$$&space;in&space;terms&space;of&space;$$y$$:&space;$x&space;=&space;12&space;-&space;y$&space;Substitute&space;this&space;value&space;of&space;$$x$$&space;into&space;the&space;second&space;equation:&space;$10y&space;+&space;(12&space;-&space;y)&space;=&space;10(12&space;-&space;y)&space;+&space;y&space;+&space;18$&space;Let's&space;solve&space;for&space;$$y$$:&space;$10y&space;+&space;12&space;-&space;y&space;=&space;120&space;-&space;10y&space;+&space;y&space;+&space;18$&space;$9y&space;+&space;12&space;=&space;138&space;-&space;9y$&space;$9y&space;+&space;9y&space;=&space;138&space;-&space;12$&space;$18y&space;=&space;126$&space;$y&space;=&space;\frac{126}{18}$&space;$y&space;=&space;7$&space;Now&space;that&space;we&space;have&space;found&space;$$y$$,&space;let's&space;find&space;$$x$$&space;using&space;the&space;first&space;equation:&space;$x&space;+&space;7&space;=&space;12$&space;$x&space;=&space;12&space;-&space;7$&space;$x&space;=&space;5$&space;So,&space;the&space;original&space;number&space;is&space;$$10x&space;+&space;y&space;=&space;10(5)&space;+&space;7&space;=&space;50&space;+&space;7&space;=&space;57$$.$

According to the given conditions:

1. The sum of the digits of the two-digit number is 12:

Dislike Bookmark

Take Class 10 Tuition from the Best Tutors

• Affordable fees
• Flexible Timings
• Choose between 1-1 and Group class
• Verified Tutors

The given sequence is an arithmetic progression (AP) with a common difference. To find this common difference, let's subtract each term from the next one: read more

The given sequence is an arithmetic progression (AP) with a common difference. To find this common difference, let's subtract each term from the next one:
$T&space;$&space;\frac{3&space;-&space;a}{3a}&space;-&space;\frac{1}{a}&space;=&space;\frac{3&space;-&space;a}{3a}&space;-&space;\frac{3}{3a}&space;=&space;\frac{(3&space;-&space;a)&space;-&space;3}{3a}&space;=&space;\frac{3&space;-&space;a&space;-&space;3}{3a}&space;=&space;\frac{-a}{3a}&space;=&space;-\frac{1}{3}&space;$&space;So,&space;the&space;common&space;difference&space;of&space;the&space;given&space;arithmetic&space;progression&space;is&space;$$&space;-\frac{1}{3}&space;$$.$

Dislike Bookmark

To find which term of the arithmetic progression read more

To find which term of the arithmetic progression$T$$&space;-7,&space;-12,&space;-17,&space;-22,&space;\ldots&space;$$&space;will&space;be&space;$$&space;-82&space;$$,&space;we&space;can&space;use&space;the&space;formula&space;for&space;the&space;$$&space;n&space;$$th&space;term&space;of&space;an&space;arithmetic&space;progression:&space;$&space;a_n&space;=&space;a_1&space;+&space;(n&space;-&space;1)&space;\cdot&space;d&space;$&space;where:&space;-&space;$$&space;a_n&space;$$&space;is&space;the&space;$$&space;n&space;$$th&space;term,&space;-&space;$$&space;a_1&space;$$&space;is&space;the&space;first&space;term,&space;-&space;$$&space;d&space;$$&space;is&space;the&space;common&space;difference,&space;and&space;-&space;$$&space;n&space;$$&space;is&space;the&space;term&space;number.&space;For&space;the&space;given&space;sequence,&space;$$&space;a_1&space;=&space;-7&space;$$&space;and&space;$$&space;d&space;=&space;-5&space;$$&space;(because&space;each&space;term&space;decreases&space;by&space;5).&space;Let's&space;plug&space;these&space;values&space;into&space;the&space;formula&space;and&space;solve&space;for&space;$$&space;n&space;$$&space;when&space;$$&space;a_n&space;=&space;-82&space;$$:&space;$&space;-7&space;+&space;(n&space;-&space;1)&space;\cdot&space;(-5)&space;=&space;-82&space;$&space;$&space;-7&space;-&space;5n&space;+&space;5&space;=&space;-82&space;$&space;$&space;-2&space;-&space;5n&space;=&space;-82&space;$&space;$&space;-5n&space;=&space;-82&space;+&space;2&space;$&space;$&space;-5n&space;=&space;-80&space;$&space;$&space;n&space;=&space;\frac{-80}{-5}&space;$&space;$&space;n&space;=&space;16&space;$&space;So,&space;the&space;16th&space;term&space;of&space;the&space;arithmetic&space;progression&space;is&space;$$&space;-82&space;$$.$

$To&space;determine&space;if&space;$$&space;-100&space;$$&space;is&space;any&space;term&space;of&space;the&space;arithmetic&space;progression,&space;we&space;can&space;check&space;if&space;it&space;can&space;be&space;obtained&space;using&space;the&space;formula&space;for&space;the&space;$$&space;n&space;$$th&space;term:&space;$&space;a_n&space;=&space;-7&space;+&space;(n&space;-&space;1)&space;\cdot&space;(-5)&space;$&space;Let's&space;solve&space;for&space;$$&space;n&space;$$&space;when&space;$$&space;a_n&space;=&space;-100&space;$$:&space;$&space;-7&space;+&space;(n&space;-&space;1)&space;\cdot&space;(-5)&space;=&space;-100&space;$&space;$&space;-7&space;-&space;5n&space;+&space;5&space;=&space;-100&space;$&space;$&space;-2&space;-&space;5n&space;=&space;-100&space;$&space;$&space;-5n&space;=&space;-100&space;+&space;2&space;$&space;$&space;-5n&space;=&space;-98&space;$&space;$&space;n&space;=&space;\frac{-98}{-5}&space;$&space;$&space;n&space;=&space;19.6&space;$$

Dislike Bookmark

To find the value of p for which read more

To find the value of p for which$To&space;find&space;the&space;value&space;of&space;$$&space;p&space;$$&space;for&space;which&space;$$&space;-4&space;$$&space;is&space;a&space;zero&space;of&space;the&space;polynomial&space;$$&space;x^2&space;-&space;2x&space;-&space;(7p&space;+&space;3)&space;$$,&space;we&space;substitute&space;$$&space;x&space;=&space;-4&space;$$&space;into&space;the&space;polynomial&space;and&space;solve&space;for&space;$$&space;p&space;$$.&space;Substituting&space;$$&space;x&space;=&space;-4&space;$$&space;into&space;the&space;polynomial&space;equation:&space;$&space;(-4)^2&space;-&space;2(-4)&space;-&space;(7p&space;+&space;3)&space;=&space;0&space;$&space;$&space;16&space;+&space;8&space;-&space;7p&space;-&space;3&space;=&space;0&space;$&space;$&space;24&space;-&space;7p&space;-&space;3&space;=&space;0&space;$&space;$&space;21&space;-&space;7p&space;=&space;0&space;$&space;Now,&space;let's&space;solve&space;for&space;$$&space;p&space;$$:&space;$&space;21&space;-&space;7p&space;=&space;0&space;$&space;$&space;7p&space;=&space;21&space;$&space;$&space;p&space;=&space;\frac{21}{7}&space;$&space;$&space;p&space;=&space;3&space;$&space;So,&space;for&space;$$&space;p&space;=&space;3&space;$$,&space;$$&space;-4&space;$$&space;is&space;a&space;zero&space;of&space;the&space;polynomial&space;$$&space;x^2&space;-&space;2x&space;-&space;(7p&space;+&space;3)&space;$$.$

Dislike Bookmark

Take Class 10 Tuition from the Best Tutors

• Affordable fees
• Flexible Timings
• Choose between 1-1 and Group class
• Verified Tutors

If 11 is a zero of the polynomial p(x)=ax2−3(a−1)x−1 read more

If 11 is a zero of the polynomial p(x)=ax2−3(a−1)x−1$Substituting&space;$$x&space;=&space;1$$&space;into&space;the&space;polynomial&space;equation:&space;$&space;p(1)&space;=&space;a(1)^2&space;-&space;3(a&space;-&space;1)(1)&space;-&space;1&space;=&space;0&space;$&space;$&space;a&space;-&space;3(a&space;-&space;1)&space;-&space;1&space;=&space;0&space;$&space;Now,&space;let's&space;solve&space;for&space;$$a$$:&space;$&space;a&space;-&space;3a&space;+&space;3&space;-&space;1&space;=&space;0&space;$&space;$&space;-2a&space;+&space;2&space;=&space;0&space;$&space;$&space;-2a&space;=&space;-2&space;$&space;$&space;a&space;=&space;\frac{-2}{-2}&space;$&space;$&space;a&space;=&space;1&space;$&space;So,&space;the&space;value&space;of&space;$$a$$&space;for&space;which&space;$$1$$&space;is&space;a&space;zero&space;of&space;the&space;polynomial&space;$$p(x)&space;=&space;ax^2&space;-&space;3(a&space;-&space;1)x&space;-&space;1$$&space;is&space;$$a&space;=&space;1$$.$

Dislike Bookmark

To find aa if (x+a) is a factor of read more

To find aa if (x+a) is a factor of$$$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10$$,&space;we'll&space;perform&space;polynomial&space;division.&space;Given&space;that&space;$$(x&space;+&space;a)$$&space;is&space;a&space;factor,&space;we&space;have:&space;$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10&space;=&space;(x&space;+&space;a)(2x&space;+&space;b)$&space;Expanding&space;$$(x&space;+&space;a)(2x&space;+&space;b)$$,&space;we&space;get:&space;$2x^2&space;+&space;(2a&space;+&space;b)x&space;+&space;ab$&space;Comparing&space;coefficients&space;of&space;corresponding&space;terms&space;in&space;$$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10$$&space;and&space;$$2x^2&space;+&space;(2a&space;+&space;b)x&space;+&space;ab$$,&space;we&space;have:&space;For&space;$$x$$&space;terms:&space;$2ax&space;+&space;5x&space;=&space;(2a&space;+&space;b)x$&space;So,&space;$$2a&space;+&space;b&space;=&space;5$$&space;For&space;constant&space;terms:&space;$10&space;=&space;ab$&space;Given&space;$$2a&space;+&space;b&space;=&space;5$$,&space;we&space;can&space;express&space;$$b$$&space;in&space;terms&space;of&space;$$a$$:&space;$b&space;=&space;5&space;-&space;2a$&space;Substituting&space;this&space;expression&space;for&space;$$b$$&space;into&space;$$10&space;=&space;ab$$,&space;we&space;get:&space;$10&space;=&space;a(5&space;-&space;2a)$&space;Expanding&space;and&space;rearranging,&space;we&space;have:&space;$10&space;=&space;5a&space;-&space;2a^2$&space;$2a^2&space;-&space;5a&space;+&space;10&space;=&space;0$&space;This&space;is&space;a&space;quadratic&space;equation.&space;We&space;can&space;solve&space;it&space;using&space;the&space;quadratic&space;formula:$

$$a&space;=&space;\frac{-b&space;\pm&space;\sqrt{b^2&space;-&space;4ac}}{2a}$&space;For&space;$$2a^2&space;-&space;5a&space;+&space;10&space;=&space;0$$,&space;we&space;have&space;$$a&space;=&space;2$$,&space;$$b&space;=&space;-5$$,&space;and&space;$$c&space;=&space;10$$.&space;Substituting&space;these&space;values&space;into&space;the&space;quadratic&space;formula:&space;$a&space;=&space;\frac{-(-5)&space;\pm&space;\sqrt{(-5)^2&space;-&space;4&space;\cdot&space;2&space;\cdot&space;10}}{2&space;\cdot&space;2}$&space;$a&space;=&space;\frac{5&space;\pm&space;\sqrt{25&space;-&space;80}}{4}$&space;$a&space;=&space;\frac{5&space;\pm&space;\sqrt{-55}}{4}$&space;Since&space;$$\sqrt{-55}$$&space;is&space;imaginary,&space;the&space;solutions&space;for&space;$$a$$&space;are&space;complex.&space;Thus,&space;there's&space;no&space;real&space;value&space;of&space;$$a$$&space;for&space;which&space;$$(x&space;+&space;a)$$&space;is&space;a&space;factor&space;of&space;$$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10$$.$

Dislike Bookmark

Let's denote the usual speed of the plane as S km/hr. read more

Let's denote the usual speed of the plane as S km/hr.$Since&space;the&space;plane&space;left&space;30&space;minutes&space;late,&space;it&space;had&space;less&space;time&space;to&space;reach&space;its&space;destination.&space;To&space;compensate&space;for&space;the&space;delay&space;and&space;still&space;reach&space;the&space;destination&space;on&space;time,&space;the&space;plane&space;had&space;to&space;increase&space;its&space;speed.&space;Let's&space;first&space;find&space;the&space;time&space;it&space;takes&space;for&space;the&space;plane&space;to&space;cover&space;1500&space;km&space;at&space;its&space;usual&space;speed&space;$$&space;S&space;$$.&space;The&space;time&space;taken&space;can&space;be&space;calculated&space;using&space;the&space;formula:&space;$&space;\text{Time}&space;=&space;\frac{\text{Distance}}{\text{Speed}}&space;$&space;At&space;the&space;usual&space;speed&space;$$&space;S&space;$$,&space;the&space;time&space;taken&space;to&space;cover&space;1500&space;km&space;is:&space;$&space;\text{Time&space;taken}&space;=&space;\frac{1500}{S}&space;$$

$Now,&space;since&space;the&space;plane&space;left&space;30&space;minutes&space;late,&space;it&space;has&space;$$&space;\frac{1}{2}&space;$$&space;hour&space;less&space;to&space;reach&space;the&space;destination.&space;So,&space;the&space;time&space;available&space;for&space;the&space;plane&space;to&space;reach&space;the&space;destination&space;on&space;time&space;is:&space;$&space;\text{Time&space;available}&space;=&space;\frac{1500}{S}&space;-&space;\frac{1}{2}&space;$&space;When&space;the&space;plane&space;increases&space;its&space;speed&space;by&space;100&space;km/hr,&space;its&space;new&space;speed&space;becomes&space;$$&space;S&space;+&space;100&space;$$&space;km/hr.&space;Using&space;this&space;new&space;speed,&space;the&space;time&space;taken&space;to&space;cover&space;1500&space;km&space;becomes:&space;$&space;\text{New&space;time&space;taken}&space;=&space;\frac{1500}{S&space;+&space;100}&space;$&space;Given&space;that&space;the&space;plane&space;needs&space;to&space;cover&space;the&space;same&space;distance&space;in&space;the&space;time&space;available,&space;we&space;can&space;equate&space;the&space;new&space;time&space;taken&space;with&space;the&space;time&space;available:&space;$&space;\frac{1500}{S&space;+&space;100}&space;=&space;\frac{1500}{S}&space;-&space;\frac{1}{2}&space;$&space;Now,&space;let's&space;solve&space;this&space;equation&space;to&space;find&space;the&space;value&space;of&space;$$&space;S&space;$$,&space;the&space;usual&space;speed&space;of&space;the&space;plane.&space;$&space;\frac{1500}{S&space;+&space;100}&space;=&space;\frac{1500}{S}&space;-&space;\frac{1}{2}&space;$$

$Cross&space;multiply&space;to&space;get&space;rid&space;of&space;the&space;denominators:&space;$&space;1500S&space;=&space;1500(S&space;+&space;100)&space;-&space;\frac{1}{2}(S)(S&space;+&space;100)&space;$&space;$&space;1500S&space;=&space;1500S&space;+&space;150000&space;-&space;\frac{1}{2}(S^2&space;+&space;100S)&space;$&space;$&space;1500S&space;=&space;1500S&space;+&space;150000&space;-&space;\frac{1}{2}S^2&space;-&space;50S&space;$&space;$&space;0&space;=&space;150000&space;-&space;\frac{1}{2}S^2&space;-&space;50S&space;$&space;$&space;\frac{1}{2}S^2&space;+&space;50S&space;-&space;150000&space;=&space;0&space;$&space;Now,&space;we&space;have&space;a&space;quadratic&space;equation&space;in&space;terms&space;of&space;$$&space;S&space;$$.&space;Let's&space;solve&space;it&space;to&space;find&space;the&space;usual&space;speed&space;of&space;the&space;plane.&space;To&space;solve&space;the&space;quadratic&space;equation&space;$$&space;\frac{1}{2}S^2&space;+&space;50S&space;-&space;150000&space;=&space;0&space;$$,&space;we&space;can&space;multiply&space;both&space;sides&space;of&space;the&space;equation&space;by&space;2&space;to&space;get&space;rid&space;of&space;the&space;fraction:&space;$&space;S^2&space;+&space;100S&space;-&space;300000&space;=&space;0&space;$&space;Now,&space;we&space;can&space;use&space;the&space;quadratic&space;formula&space;to&space;find&space;the&space;values&space;of&space;$$&space;S&space;$$:&space;$&space;S&space;=&space;\frac{{-b&space;\pm&space;\sqrt{{b^2&space;-&space;4ac}}}}{{2a}}&space;$&space;For&space;the&space;quadratic&space;equation&space;$$&space;S^2&space;+&space;100S&space;-&space;300000&space;=&space;0&space;$$,&space;we&space;have&space;$$&space;a&space;=&space;1&space;$$,&space;$$&space;b&space;=&space;100&space;$$,&space;and&space;$$&space;c&space;=&space;-300000&space;$$.$

$Substituting&space;these&space;values&space;into&space;the&space;quadratic&space;formula:&space;$&space;S&space;=&space;\frac{{-100&space;\pm&space;\sqrt{{(100)^2&space;-&space;4&space;\cdot&space;1&space;\cdot&space;(-300000)}}}}{{2&space;\cdot&space;1}}&space;$&space;$&space;S&space;=&space;\frac{{-100&space;\pm&space;\sqrt{{10000&space;+&space;1200000}}}}{2}&space;$&space;$&space;S&space;=&space;\frac{{-100&space;\pm&space;\sqrt{{1210000}}}}{2}&space;$&space;$&space;S&space;=&space;\frac{{-100&space;\pm&space;1100}}{{2}}&space;$&space;Now,&space;we&space;have&space;two&space;possible&space;values&space;for&space;$$&space;S&space;$$:&space;1.&space;$$&space;S&space;=&space;\frac{{-100&space;+&space;1100}}{{2}}&space;=&space;\frac{{1000}}{{2}}&space;=&space;500&space;$$&space;2.&space;$$&space;S&space;=&space;\frac{{-100&space;-&space;1100}}{{2}}&space;=&space;\frac{{-1200}}{{2}}&space;=&space;-600&space;$$&space;Since&space;speed&space;cannot&space;be&space;negative,&space;we&space;discard&space;the&space;negative&space;value&space;of&space;$$&space;S&space;$$.&space;Therefore,&space;the&space;usual&space;speed&space;of&space;the&space;plane&space;is&space;$$&space;oxed{500&space;\text{&space;km/hr}}&space;$$.$

Dislike Bookmark

Take Class 10 Tuition from the Best Tutors

• Affordable fees
• Flexible Timings
• Choose between 1-1 and Group class
• Verified Tutors

To prove that the equation read more

To prove that the equation $$$&space;(a^2&space;+&space;b^2)x^2&space;+&space;2(ac&space;+&space;bd)x&space;+&space;(c^2&space;+&space;d^2)&space;=&space;0&space;$$&space;has&space;no&space;real&space;roots&space;when&space;$$&space;ad&space; eq&space;bc&space;$$,&space;we&space;will&space;use&space;the&space;discriminant.&space;The&space;discriminant&space;of&space;a&space;quadratic&space;equation&space;$$&space;ax^2&space;+&space;bx&space;+&space;c&space;=&space;0&space;$$&space;is&space;given&space;by&space;$$&space;\Delta&space;=&space;b^2&space;-&space;4ac&space;$$.&space;If&space;the&space;discriminant&space;is&space;negative,&space;then&space;the&space;quadratic&space;equation&space;has&space;no&space;real&space;roots.&space;In&space;our&space;given&space;equation&space;$$&space;(a^2&space;+&space;b^2)x^2&space;+&space;2(ac&space;+&space;bd)x&space;+&space;(c^2&space;+&space;d^2)&space;=&space;0&space;$$,&space;the&space;coefficients&space;are&space;$$&space;a^2&space;+&space;b^2&space;$$,&space;$$&space;2(ac&space;+&space;bd)&space;$$,&space;and&space;$$&space;c^2&space;+&space;d^2&space;$$.$

$Let's&space;calculate&space;the&space;discriminant&space;$$&space;\Delta&space;$$&space;for&space;this&space;quadratic&space;equation:&space;$&space;\Delta&space;=&space;(2(ac&space;+&space;bd))^2&space;-&space;4(a^2&space;+&space;b^2)(c^2&space;+&space;d^2)&space;$&space;$&space;\Delta&space;=&space;4(a^2c^2&space;+&space;2abcd&space;+&space;b^2d^2)&space;-&space;4(a^2c^2&space;+&space;a^2d^2&space;+&space;b^2c^2&space;+&space;b^2d^2)&space;$&space;$&space;\Delta&space;=&space;4a^2c^2&space;+&space;8abcd&space;+&space;4b^2d^2&space;-&space;4a^2c^2&space;-&space;4a^2d^2&space;-&space;4b^2c^2&space;-&space;4b^2d^2&space;$&space;$&space;\Delta&space;=&space;8abcd&space;-&space;4a^2d^2&space;-&space;4b^2c^2&space;$&space;$&space;\Delta&space;=&space;4(2abcd&space;-&space;a^2d^2&space;-&space;b^2c^2)&space;$$

$Now,&space;since&space;$$&space;ad&space; eq&space;bc&space;$$,&space;the&space;term&space;inside&space;the&space;parentheses&space;is&space;not&space;equal&space;to&space;zero.&space;Thus,&space;$$&space;\Delta&space;$$&space;will&space;be&space;positive&space;if&space;$$&space;2abcd&space;-&space;a^2d^2&space;-&space;b^2c^2&space;>&space;0&space;$$&space;and&space;negative&space;if&space;$$&space;2abcd&space;-&space;a^2d^2&space;-&space;b^2c^2&space;<&space;0&space;$$.&space;Since&space;the&space;discriminant&space;$$&space;\Delta&space;$$&space;is&space;proportional&space;to&space;this&space;expression,&space;if&space;the&space;expression&space;is&space;positive,&space;then&space;$$&space;\Delta&space;$$$

$is&space;positive,&space;and&space;the&space;quadratic&space;equation&space;has&space;real&space;roots.&space;If&space;the&space;expression&space;is&space;negative,&space;then&space;$$&space;\Delta&space;$$&space;is&space;negative,&space;and&space;the&space;quadratic&space;equation&space;has&space;no&space;real&space;roots.&space;Therefore,&space;when&space;$$&space;ad&space; eq&space;bc&space;$$,&space;the&space;equation&space;$$&space;(a^2&space;+&space;b^2)x^2&space;+&space;2(ac&space;+&space;bd)x&space;+&space;(c^2&space;+&space;d^2)&space;=&space;0&space;$$&space;has&space;no&space;real&space;roots.$

Dislike Bookmark

UrbanPro.com helps you to connect with the best Class 10 Tuition in India. Post Your Requirement today and get connected.

Overview

Questions 123

Lessons 18

Total Shares

124,685 Followers

## Top Contributors

Connect with Expert Tutors & Institutes for UNIT II: Algebra

## Class 10 Tuition in:

x

X

### Looking for Class 10 Tuition Classes?

The best tutors for Class 10 Tuition Classes are on UrbanPro

• Select the best Tutor
• Book & Attend a Free Demo
• Pay and start Learning

### Take Class 10 Tuition with the Best Tutors

The best Tutors for Class 10 Tuition Classes are on UrbanPro