Ravi Kant Kumar

Solve the Following Quadratic Equation by Factorization Question: \[ a(x^2+1)-x(a^2+1)=0 \] Solution Given, \[ a(x^2+1)-x(a^2+1)=0 \] Expanding, \[ ax^2+a-a^2x-x=0 \] \[ ax^2-(a^2+1)x+a=0 \] Splitting the middle term: \[ ax^2-a^2x-x+a=0 \] Taking common factors: \[ ax(x-a)-1(x-a)=0 \] \[ (x-a)(ax-1)=0 \] Therefore, \[ x-a=0 \] or \[ ax-1=0 \] \[ x=a \] or \[ x=\frac{1}{a} \] Final […]

Solve the Following Quadratic Equation by Factorization Question: \[ (a+b)^2x^2-4abx-(a-b)^2=0 \] Solution Given, \[ (a+b)^2x^2-4abx-(a-b)^2=0 \] Since \[ (a-b)^2=(a+b)^2-4ab \] Substituting, we get \[ (a+b)^2x^2-4abx-(a+b)^2+4ab=0 \] \[ (a+b)^2(x^2-1)-4ab(x-1)=0 \] \[ (x-1)\Big[(a+b)^2(x+1)-4ab\Big]=0 \] Now, \[ (a+b)^2(x+1)-4ab \] \[ =(a+b)^2x+(a+b)^2-4ab \] \[ =(a+b)^2x+(a-b)^2 \] Therefore, \[ (x-1)\Big[(a+b)^2x+(a-b)^2\Big]=0 \] Hence, \[ x-1=0 \] or \[ (a+b)^2x+(a-b)^2=0 \] \[

Solve the Following Quadratic Equation by Factorization Question: \[ x^2+2ab=(2a+b)x \] Solution Bringing all terms to one side: \[ x^2-(2a+b)x+2ab=0 \] We need two terms whose sum is \((2a+b)\) and product is \(2ab\). \[ 2a+b=(2a)+b \] \[ (2a)\times b=2ab \] Splitting the middle term: \[ x^2-2ax-bx+2ab=0 \] Taking common factors: \[ x(x-2a)-b(x-2a)=0 \] \[ (x-2a)(x-b)=0

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c} \] Solution Given, \[ \frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c} \] Multiplying both sides by \((x-a)(x-b)(x-c)\), we get \[ a(x-b)(x-c)+b(x-a)(x-c) = 2c(x-a)(x-b) \] Expanding: \[ a(x^2-bx-cx+bc) + b(x^2-ax-cx+ac) = 2c(x^2-ax-bx+ab) \] \[ (a+b)x^2 – 2abx – c(a+b)x + 2abc = 2cx^2 – 2c(a+b)x + 2abc \] \[ (a+b-2c)x^2 + (a+b-2c)cx

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x+1}{x-1}+\frac{x-2}{x+2} = 4-\frac{2x+3}{x-2}, \qquad x\ne 1,-2,2 \] Solution Given: \[ \frac{x+1}{x-1}+\frac{x-2}{x+2} = 4-\frac{2x+3}{x-2} \] Multiplying both sides by \((x-1)(x+2)(x-2)\): \[ (x+1)(x+2)(x-2) +(x-2)^2(x-1) = 4(x-1)(x+2)(x-2) -(2x+3)(x-1)(x+2) \] Simplifying and expanding: \[ 2x^3-3x^2-6x = 2x^3-5x^2-14x+20 \] \[ 2x^2+8x-20=0 \] \[ x^2+4x-10=0 \] Using factorization through surds: \[ x^2+4x-10

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{a}{x-b}+\frac{b}{x-a}=2, \qquad x\ne a,b \] Solution Given: \[ \frac{a}{x-b}+\frac{b}{x-a}=2 \] Taking LCM on the left side: \[ \frac{a(x-a)+b(x-b)} {(x-a)(x-b)} =2 \] \[ \frac{(a+b)x-(a^2+b^2)} {(x-a)(x-b)} =2 \] Cross-multiplying: \[ (a+b)x-(a^2+b^2) = 2(x-a)(x-b) \] \[ (a+b)x-(a^2+b^2) = 2x^2-2(a+b)x+2ab \] \[ 2x^2-3(a+b)x+(a^2+2ab+b^2)=0 \] \[ 2x^2-3(a+b)x+(a+b)^2=0 \] Factorizing: \[ 2x^2-2(a+b)x-(a+b)x+(a+b)^2=0

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac16 \] Solution Using \[ \frac{1}{(x-a)(x-b)} = \frac{1}{a-b} \left( \frac{1}{x-a} – \frac{1}{x-b} \right) \] we get \[ \frac{1}{(x-1)(x-2)} = \frac{1}{x-2} – \frac{1}{x-1} \] \[ \frac{1}{(x-2)(x-3)} = \frac{1}{x-3} – \frac{1}{x-2} \] \[ \frac{1}{(x-3)(x-4)} = \frac{1}{x-4} – \frac{1}{x-3} \] Adding, the middle

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x-a}{x-b}+\frac{x-b}{x-a} = \frac{a}{b}+\frac{b}{a} \] Solution Given: \[ \frac{x-a}{x-b}+\frac{x-b}{x-a} = \frac{a}{b}+\frac{b}{a} \] Taking LCM on both sides: \[ \frac{(x-a)^2+(x-b)^2}{(x-a)(x-b)} = \frac{a^2+b^2}{ab} \] Cross-multiplying: \[ ab\left[(x-a)^2+(x-b)^2\right] = (a^2+b^2)(x-a)(x-b) \] Expanding and simplifying: \[ 2abx^2-2ab(a+b)x+ab(a^2+b^2) = (a^2+b^2)\left[x^2-(a+b)x+ab\right] \] \[ (a-b)^2x^2-(a-b)^2(a+b)x=0 \] \[ (a-b)^2x\,[x-(a+b)]=0 \] Therefore, \[ x=0 \] or

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{m}{n}x^2+\frac{n}{m}=1-2x \] Solution Given: \[ \frac{m}{n}x^2+\frac{n}{m}=1-2x \] Multiplying both sides by \(mn\): \[ m^2x^2+n^2=mn-2mnx \] \[ m^2x^2+2mnx+n^2-mn=0 \] Observe that: \[ m^2x^2+2mnx+n^2=(mx+n)^2 \] Therefore, \[ (mx+n)^2-mn=0 \] Using the identity \(a^2-b^2=(a-b)(a+b)\): \[ (mx+n-\sqrt{mn})(mx+n+\sqrt{mn})=0 \] Hence, \[ mx+n-\sqrt{mn}=0 \] or \[ mx+n+\sqrt{mn}=0 \] Therefore, \[ x=\frac{\sqrt{mn}-n}{m} \]

Solve √2x² + 7x + 5√2 = 0 by Factorization Question: \[ \sqrt2x^2+7x+5\sqrt2=0 \] Solution Given: \[ \sqrt2x^2+7x+5\sqrt2=0 \] Product of the coefficient of \(x^2\) and the constant term: \[ (\sqrt2)(5\sqrt2)=10 \] We split the middle term \(7x\) as \(5x+2x\): \[ \sqrt2x^2+5x+2x+5\sqrt2=0 \] Taking common factors: \[ x(\sqrt2x+5)+\sqrt2(\sqrt2x+5)=0 \] \[ (\sqrt2x+5)(x+\sqrt2)=0 \] Therefore, \[ \sqrt2x+5=0