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Question

Solve each of the following quadratic equations: 
x1x2+x3x4=313,x2,4
 


Solution

As given,

x1x2+x3x4=313

x1x2+x3x4=103

L.C.M. of (x2) and (x4)=(x2)(x4)

Then,

=> [(x1)(x4)+(x3)(x2)](x2)(x4)=103

=> 3(x25x+4+x25x+6)=10(x2)(x4)

=> 3(2x210x+10)=10(x26x+8)

=> 6x230x+30=10x260x+80

Taking 2 as a common from both sides & cancelling it, we have,

=> 3x215x+15=5x230x+40

=>5x23x230x+15x+4015=0

=> 2x215x+25=0

Now, by the middle term splitting method,we have,

=> 2x210x5x+25=0

=> 2x(x5)5(x5)=0

=> (2x5)(x5)=0

=> 2x5=0

So,x=52

Or, x5=0

So, x=5

Hence, x=5 or 52


Mathematics
Secondary School Mathematics X
Standard X

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