3) Solve the following quadratic equation and give your answer correct

1.	3) Solve the following quadratic equation and give your answer correct to two significant figures: x+1/2x+5=x+3/3x+4
Answer» $\large\underline{\sf{Solution-}}$ 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

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

Answer»

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

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.