Math, asked by narendersingh3925535, 9 months ago

find the root of x minus one upon X + 2 + x minus 3 upon x minus 2 is equal to 11 upon 8 by using quadratic formula​

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Answered by tirlotkarswayam06
14

Answer:

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Answered by hukam0685
4

Step-by-step explanation:

Given that:find the root of x minus one upon X + 2 + x minus 3 upon x minus 2 is equal to 11 upon 8 by using quadratic formula

Expression :

 \bold{\frac{x - 1}{x + 2}  +  \frac{x - 3}{x - 2}  =  \frac{11}{8}}  \\

Solution:

To find the roots of given expression by quadratic equation,first we have to simply the expression and convert it into quadratic equation

Steps for simplification

1.Take LCM

 \frac{(x - 1)(x - 2) + (x - 3)(x + 2)}{ {x}^{2}  - 4}  =  \frac{11}{8}  \\  \\

2.Multiply terms in bracket and put common terms together

 \frac{ {x}^{2} - 2x - x + 2 +  {x}^{2}  + 2x - 3x - 6 }{ {x}^{2}  - 4}  =  \frac{11}{8}  \\  \\  \frac{2 {x}^{2} - 4x - 4 }{ {x}^{2} - 4 }  =  \frac{11}{8}  \\  \\

3. Cross multiply

8(2 {x}^{2}  - 4x - 4) = 11( {x}^{2} - 4) \\  \\ 16 {x}^{2}   - 32x - 32 = 11 {x}^{2}  - 44 \\  \\ 5 {x}^{2}  - 32x + 12 = 0 \\  \\

Now apply quadratic formula

a {x}^{2}  + bx + c = 0 \\  \\\bold{ x_{1,2} =  \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} } \\  \\

On compare with standard equation

here

a = 5 \\ b =  - 32 \\ c = 12 \\  \\ x_{1,2} =  \frac{ - ( - 32) ±  \sqrt{ {( - 32)}^{2}  - 4 \times 5 \times 12}}{5 \times 2}  \\  \\x_{1,2} =  \frac{32 ±  \sqrt{1024 - 240} }{10}  \\  \\ x_{1,2} =  \frac{32 ±  \sqrt{784} }{10}  \\  \\ x_{1,2} =  \frac{32± 28}{10}  \\

So,to find both the roots,put negative and positive value of 28

x_1 =  \frac{32 + 28}{10}  \\  \\ x_1 =  \frac{60}{10}  \\  \\ \bold{x_1 = 6 }\\  \\ x _2=  \frac{32 - 28}{10}  \\  \\ x_2 =  \frac{4}{10 }  \\  \\ \bold{x_2=  \frac{2}{5}}  \\  \\

Thus,roots of given expression are 6 and 2/5.

Hope it helps you.

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