x^2-(2b-1)x+(b^2-b-20)=0 find the roots of this quadratic equation by factorization method
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Answered by
134
x^2-(2b-1)x+(b^2-b-20)=0
=>x^2-2(b-1/2)x+(b^2-b+1/4)-20.25=0
=>{x-(b-1/2)}^2-(4.5)^2=0
=>(x-b+0.5)^2-(4.5)^2=0
=>(x-b+0.5-4.5)(x-b+0.5+4.5)=0
=>(x-b-4)(x-b+5)=0
=>x=b+4 and b-5
=>x^2-2(b-1/2)x+(b^2-b+1/4)-20.25=0
=>{x-(b-1/2)}^2-(4.5)^2=0
=>(x-b+0.5)^2-(4.5)^2=0
=>(x-b+0.5-4.5)(x-b+0.5+4.5)=0
=>(x-b-4)(x-b+5)=0
=>x=b+4 and b-5
Answered by
41
Hey mate!!!!
This sum can easily be solved by splitting the middle term:
x^2 - (b-5)x + (b+4)x + (b^2-b-20) =0
x [x-(b-5)] + [(b+4) x+ (b-5)]=0
x = b-5. OR
x = b+4
Hope u understand!
Mark me as the BRAINLIEST!
This sum can easily be solved by splitting the middle term:
x^2 - (b-5)x + (b+4)x + (b^2-b-20) =0
x [x-(b-5)] + [(b+4) x+ (b-5)]=0
x = b-5. OR
x = b+4
Hope u understand!
Mark me as the BRAINLIEST!
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