Solve the following quadratic equations by factorization method: 1a+b+x=1a+1b+1x,a+b≠0

Question

Solve the following quadratic equations by factorization method:  1a+b+x=1a+1b+1x,a+b≠0

Melissa Moore · Accepted Answer

Step 1  Consider the given equation:  1a+1b+1x=1a+b+x,a,b,x≠0  ⇒bx+ax+ababx=1a+b+x  ⇒(bx+ax+ab)(a+b+x)=abx  ⇒abx+b2x+bx2+a2x+abx+ax2+a2b+ab2+abx=abx  ⇒(a+b)x2+a2x+2abx+b2x+a2b+ab2=0  ⇒(a+b)x2+(a+b)2x+a2b+ab2=0  ⇒(a+b)x2+(a+b)2x+ab(a+b)=0  Divide the above equation by a+b, we have,  x2+(a+b)x+ab=0  Step 2  Solving the above quadratic, we have,  x=−(a+b)±(a+b)2−4ab2  ⇒x=−(a+b)±a2+2ab+b2−4ab2  ⇒x=(−a+b)±a2−2ab+b22  ⇒x=−(a+b)±(a−b)22  ⇒x=−(a+b)±(a−b)2  ⇒x=−a,x=−b

Jillian Edgerton · Accepted Answer

Step 1  In this question we have to solve the quadratic equation by factorization method.  Step 2  Factorization method is method of breaking the quadratic equation into the product of their factors.  let us take an example  x2+3x+4=0  x2+4x−x+4=0  x(x+4)-1(x+4)=0  (x-1)(x+4)=0  x=1, -4  Step 3  Now, we have   1a+b+x=1a+1b+1x  1a+b+x−1x=1a+1b  x−(a+b+x)x(a+b+x)=b+aab  x−a−b−xx(a+b+x)=b+aab  −1x(a+b+x)=b+aab  −ab=ax+bx+x2  x2+ax+bx+ab=0  x(x+a)+b(x+a)=0  (x+a)(x+b)=0  x=-a, -b

nick1337 · Accepted Answer

Step 1 1/(a+b+x)−1/x=(a+b)/ab(x−(a+b+x))/(x(a+b+x)=(a+b)/ab1/(x(a+b+x)=1/abx2+(a+b)x+ab=0x=−(a+b)+−((a+b)2−4ab)1/2/2x=(−(a+b)+−(a−b))/2x=−a,−b Step 2 Used quadractic equation.

Solve the following quadratic equations by factorization method: \frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}, a+b\ne 0

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