Math, asked by sayanikadutta28, 2 months ago

3) Solve the following quadratic equation and give your answer correct to two significant figures: x+1/2x+5=x+3/3x+4​

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Answers

Answered by kiran12355
1

Step-by-step explanation:

(x+1)/(2x+5)=(x+3)/(3x+4)

(x+1)(3x+4)=(x+3)(2x+5)

3x^2+4x+3x+4=2x^2+5x+6x+15

3x^2+7x+4=2x^2+11x+15

3x^2-2x^2+7x-11x=15-4

x^2-4x=11

x^2-4x-11=0

a=1 ,b= -4 ,c=-11

roots= [-b(+-)√(b^2-4ac)]/2a

=[4(+-)√16-(4*-11)]/2

=[4(+-)√(16+44)]/2

=[4(+-)√60]/2

=[4+-√15*4]/2

=[4(+-)2√15]/2

=(4+2√15)/2. or. (4-2√15)/2

=2+√15. or. 2-√15

Answered by mathdude500
10

\large\underline{\sf{Solution-}}

Given equation is

\rm :\longmapsto\:\dfrac{x + 1}{2x + 5}  = \dfrac{x + 3}{3x + 4}

\rm :\longmapsto\:(x + 1)(3x + 4) = (x + 3)(2x + 5)

\rm :\longmapsto\:x(3x + 4) + 1(3x + 4) = x(2x + 3) + 5(x + 3)

\rm :\longmapsto\: {3x}^{2} + 4x + 3x + 4 = 2 {x}^{2} + 6x + 5x + 15

\rm :\longmapsto\: {3x}^{2} + 7x + 4 = 2 {x}^{2} + 11x + 15

\rm :\longmapsto\: {3x}^{2} + 7x + 4  -  2 {x}^{2}  -  11x  -  15 = 0

\rm :\longmapsto\: {x}^{2} - 4x - 11 = 0

Its a quadratic in x and to solve this quadratic equation, we use Quadratic Formula, which is given by

 \red{\boxed{ \rm{ x = \dfrac{ - b \:  \pm \:  \sqrt{ {b}^{2}  - 4ac} }{2a} }}}

Here,

\red{\rm :\longmapsto\:a = 1}

\red{\rm :\longmapsto\:b =  - 4}

\red{\rm :\longmapsto\:c =  - 11}

So, on substituting the values, we get

\rm :\longmapsto\:x = \dfrac{ - ( - 4) \:  \pm \:  \sqrt{ {( - 4)}^{2}  - 4(1)( - 11)} }{2 \times 1}

\rm :\longmapsto\:x = \dfrac{4 \:  \pm \:  \sqrt{16 +   44} }{2 }

\rm :\longmapsto\:x = \dfrac{4 \:  \pm \:  \sqrt{60} }{2 }

\rm :\longmapsto\:x = \dfrac{4 \:  \pm \:  \sqrt{2 \times 2 \times 15} }{2 }

\rm :\longmapsto\:x = \dfrac{4 \:  \pm \:  2\sqrt{ 15} }{2 }

\rm :\longmapsto\:x = \dfrac{2(1 \:  \pm \:  \sqrt{ 15} )}{2 }

\rm :\longmapsto\:x = 2 \:  \pm \:  \sqrt{15}

Let evaluate now

\red{\rm :\longmapsto\: \sqrt{15}}

 \red{\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:3.872 \:\:}}}\\ {\underline{\sf{3}}}& {\sf{\:\:15.000000 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \: \: 9 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{68}}}& {\sf{\:\: \: \: \: \: 600 \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: 544 \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{767}}}& {\sf{\:\:5600  \:\:}} \\{\sf{}}& \underline{\sf{\:\:5369\:\:}} \\ {\underline{\sf{7742}}}& {\sf{\:\: \:  \: \:  \:  \:  23100\:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \:    \: 15484\:\:}} \\ {\underline{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  7616\:\:}}{\sf{}}&{\sf{\:\:\:\:}}\end{array}\end{gathered}}

\bf\implies \: \sqrt{15} = 3.872

So, on substituting the value, we get

\rm :\longmapsto\:x = 2 \:  \pm \:  3.87

\rm :\longmapsto\:x = 2 \:   +  \:  3.87 \:  \:  \: or \:  \:  \: 2 \:  -  \: 3.87

\bf :\longmapsto\:x =  \:  5.87 \:  \:  \: or \:  \:  \:  -  \: 1.87

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

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