Question

find the roots of the following quadratic equations if they exist by the method of completing square. (i) x^2 - 6x - 27 = 0 (i) 4t^2 - 8t+ 3 = 0​

Answer 1

$\huge{\pink{\red{\sf \boxed{Answer}}}}$

(i) x²-6x-27=0

=> x²-9x+3x-27=0

=> x(x-9)+3(x-9)=0

=> (x+3)(x-9)=0

=> x+3=0 or x-9=0

x= -3 or x= 9

Ignore negative value

Hence, x=9

(ii) 4t²-8t+3=0

=> 4t²-6t-2t+3=0

=> 2t(2t-3)-1(2t+3)=0

=> (2t-1)(2t+3)=0

2t-1=0 or 2t+3=0

2t=1 or 2t=-3

t=1/2 or t=-3/2

Ignore negative value

Hence, t=1/2

$<font color = "red"><marquee>❤❤❤Thanks❤❤❤</marquee> </font>$

Answer 2

Answer:

(i) x²-6x-27=0

(i) x²-6x-27=0=> x²-9x+3x-27=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative value

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0=> 2t(2t-3)-1(2t+3)=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0=> 2t(2t-3)-1(2t+3)=0=> (2t-1)(2t+3)=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0=> 2t(2t-3)-1(2t+3)=0=> (2t-1)(2t+3)=02t-1=0 or 2t+3=0

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0=> 2t(2t-3)-1(2t+3)=0=> (2t-1)(2t+3)=02t-1=0 or 2t+3=02t=1 or 2t=-3

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0=> 2t(2t-3)-1(2t+3)=0=> (2t-1)(2t+3)=02t-1=0 or 2t+3=02t=1 or 2t=-3t=1/2 or t=-3/2

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0=> 2t(2t-3)-1(2t+3)=0=> (2t-1)(2t+3)=02t-1=0 or 2t+3=02t=1 or 2t=-3t=1/2 or t=-3/2Ignore negative value

(i) x²-6x-27=0=> x²-9x+3x-27=0=> x(x-9)+3(x-9)=0=> (x+3)(x-9)=0=> x+3=0 or x-9=0x= -3 or x= 9Ignore negative valueHence, x=9(ii) 4t²-8t+3=0=> 4t²-6t-2t+3=0=> 2t(2t-3)-1(2t+3)=0=> (2t-1)(2t+3)=02t-1=0 or 2t+3=02t=1 or 2t=-3t=1/2 or t=-3/2Ignore negative valueHence, t=1/2

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