Ravi Kant Kumar

Quadratic Equation – Exercise 4.4 Solutions (Step-by-Step Guide)     Find the roots of the following quadratic equations (if they exist) by the method of completing the square. x² − 4√2x + 6 = 0 Watch Solution 2x² − 7x + 3 = 0 Watch Solution 3x² + 11x + 10 = 0 Watch Solution 2x² […]

Solve the Following Quadratic Equation by Factorization Question: \[ (x-5)(x-6)=\frac{25}{(24)^2} \] Solution Given, \[ (x-5)(x-6)=\frac{25}{576} \] Write the left side as a difference of squares: \[ (x-5)(x-6) = \left(x-\frac{11}{2}\right)^2 -\left(\frac12\right)^2 \] Therefore, \[ \left(x-\frac{11}{2}\right)^2-\frac14 = \frac{25}{576} \] \[ \left(x-\frac{11}{2}\right)^2 = \frac14+\frac{25}{576} \] \[ = \frac{144+25}{576} = \frac{169}{576} = \left(\frac{13}{24}\right)^2 \] \[ \left(x-\frac{11}{2}\right)^2 = \left(\frac{13}{24}\right)^2 \]

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{3(7x+1)}{5x-3} – \frac{4(5x-3)}{7x+1} = 11, \qquad x\ne\frac35,-\frac17 \] Solution Given, \[ \frac{3(7x+1)}{5x-3} – \frac{4(5x-3)}{7x+1} = 11 \] Multiplying both sides by \((5x-3)(7x+1)\), we get \[ 3(7x+1)^2 – 4(5x-3)^2 = 11(5x-3)(7x+1) \] \[ 3(49x^2+14x+1) – 4(25x^2-30x+9) = 11(35x^2-16x-3) \] \[ 147x^2+42x+3 – 100x^2+120x-36 = 385x^2-176x-33 \] \[

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{3(3x-1)}{2x+3} – \frac{2(2x+3)}{3x-1} =5, \qquad x\ne\frac13,-\frac32 \] Solution Given, \[ \frac{3(3x-1)}{2x+3} – \frac{2(2x+3)}{3x-1} =5 \] Multiplying both sides by \((2x+3)(3x-1)\), we get \[ 3(3x-1)^2-2(2x+3)^2 = 5(2x+3)(3x-1) \] \[ 3(9x^2-6x+1)-2(4x^2+12x+9) = 5(6x^2+7x-3) \] \[ 27x^2-18x+3-8x^2-24x-18 = 30x^2+35x-15 \] \[ 19x^2-42x-15 = 30x^2+35x-15 \] \[ 11x^2+77x=0 \] \[

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{1}{2a+b+2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \] Solution Given, \[ \frac{1}{2a+b+2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \] Taking LCM on the right-hand side: \[ \frac{bx+2ax+ab} {2abx} = \frac{1} {2a+b+2x} \] Cross-multiplying: \[ 2abx = (2a+b+2x)(bx+2ax+ab) \] Since \[ bx+2ax+ab=a(2x+b)+bx \] \[ =(a+x)(2a+b) \] Therefore,

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}, \qquad x\ne2,4 \] Solution Given, \[ \frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3} \] Multiplying both sides by \(3(x-2)(x-4)\), we get \[ 3(x-1)(x-4)+3(x-3)(x-2) = 10(x-2)(x-4) \] \[ 3(x^2-5x+4)+3(x^2-5x+6) = 10(x^2-6x+8) \] \[ 6x^2-30x+30 = 10x^2-60x+80 \] \[ 4x^2-30x+50=0 \] \[ 2x^2-15x+25=0 \] Factorizing: \[ 2x^2-10x-5x+25=0 \] \[ 2x(x-5)-5(x-5)=0 \] \[ (x-5)(2x-5)=0

Solve the Following Quadratic Equation by Factorization Question: \[ a^2b^2x^2+b^2x-a^2x-1=0 \] Solution Given, \[ a^2b^2x^2+b^2x-a^2x-1=0 \] Grouping the terms: \[ (a^2b^2x^2+b^2x)-(a^2x+1)=0 \] Taking common factors: \[ b^2x(a^2x+1)-1(a^2x+1)=0 \] \[ (a^2x+1)(b^2x-1)=0 \] Therefore, \[ a^2x+1=0 \] or \[ b^2x-1=0 \] \[ x=-\frac{1}{a^2} \] or \[ x=\frac{1}{b^2} \] Final Answer \[ \boxed{x=-\frac{1}{a^2} \quad \text{or} \quad x=\frac{1}{b^2}} \]

Solve the Following Quadratic Equation by Factorization Question: \[ abx^2+(b^2-ac)x-bc=0 \] Solution Given, \[ abx^2+(b^2-ac)x-bc=0 \] Split the middle term: \[ abx^2+b^2x-acx-bc=0 \] Taking common factors: \[ bx(ax+b)-c(ax+b)=0 \] \[ (ax+b)(bx-c)=0 \] Therefore, \[ ax+b=0 \] or \[ bx-c=0 \] \[ x=-\frac{b}{a} \] or \[ x=\frac{c}{b} \] Final Answer \[ \boxed{x=-\frac{b}{a} \quad \text{or} \quad x=\frac{c}{b}}

Solve the Following Quadratic Equation by Factorization Question: \[ x^2+\left(a+\frac1a\right)x+1=0 \] Solution Given, \[ x^2+\left(a+\frac1a\right)x+1=0 \] Splitting the middle term: \[ x^2+ax+\frac{x}{a}+1=0 \] Taking common factors: \[ x(x+a)+\frac1a(x+a)=0 \] \[ (x+a)\left(x+\frac1a\right)=0 \] Therefore, \[ x+a=0 \] or \[ x+\frac1a=0 \] \[ x=-a \] or \[ x=-\frac1a \] Final Answer \[ \boxed{x=-a \quad \text{or} \quad x=-\frac1a}

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