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Solve x² − (√2 + 1)x + √2 = 0 by Factorization Question: \[ x^2-(\sqrt2+1)x+\sqrt2=0 \] Solution Given: \[ x^2-(\sqrt2+1)x+\sqrt2=0 \] We need two numbers whose sum is \((\sqrt2+1)\) and product is \(\sqrt2\). \[ \sqrt2+1=\sqrt2+1 \] \[ \sqrt2 \times 1=\sqrt2 \] Splitting the middle term: \[ x^2-\sqrt2x-x+\sqrt2=0 \] Taking common factors: \[ x(x-\sqrt2)-1(x-\sqrt2)=0 \] \[ […]

Solve the Following Quadratic Equation by Factorization Question: \[ \sqrt{2}x^2-3x-2\sqrt{2}=0 \] Solution Given: \[ \sqrt{2}x^2-3x-2\sqrt{2}=0 \] Product of the coefficient of \(x^2\) and the constant term: \[ (\sqrt{2})(-2\sqrt{2})=-4 \] We split the middle term \(-3x\) as \(-4x+x\): \[ \sqrt{2}x^2-4x+x-2\sqrt{2}=0 \] \[ \sqrt{2}x(x-2\sqrt{2})+\frac{1}{\sqrt{2}}(x-2\sqrt{2})=0 \] \[ (x-2\sqrt{2})\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)=0 \] Multiplying the second factor by \(\sqrt{2}\): \[ (x-2\sqrt{2})(2x+1)=0 \]

Solve the Following Quadratic Equation by Factorization Question: \[ 4\sqrt{3}x^2+5x-2\sqrt{3}=0 \] Solution Given: \[ 4\sqrt{3}x^2+5x-2\sqrt{3}=0 \] Product of the coefficient of \(x^2\) and the constant term: \[ (4\sqrt{3})(-2\sqrt{3})=-24 \] We split the middle term \(5x\) as \(8x-3x\): \[ 4\sqrt{3}x^2+8x-3x-2\sqrt{3}=0 \] \[ 4x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)=0 \] \[ (\sqrt{3}x+2)(4x-\sqrt{3})=0 \] Therefore, \[ \sqrt{3}x+2=0 \quad \text{or} \quad 4x-\sqrt{3}=0 \] \[

Solve the Following Quadratic Equation by Factorization Question: \[ \sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=0 \] Solution Given: \[ \sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=0 \] Product of the coefficient of \(x^2\) and the constant term: \[ (\sqrt{3})(-2\sqrt{3})=-6 \] We split the middle term \(-2\sqrt{2}x\) as \(-3\sqrt{2}x+\sqrt{2}x\): \[ \sqrt{3}x^2-3\sqrt{2}x+\sqrt{2}x-2\sqrt{3}=0 \] Taking common factors: \[ x(\sqrt{3}x-3\sqrt{2}) +\sqrt{2}(\,x-\sqrt{6}\,) =0 \] \[ x(\sqrt{3}x-3\sqrt{2}) +\frac{\sqrt{2}}{\sqrt{3}}(\sqrt{3}x-3\sqrt{2}) =0 \] \[ (\sqrt{3}x-3\sqrt{2})

Solve the Following Quadratic Equation by Factorization Question: \[ 3\sqrt{5}x^2+25x-10\sqrt{5}=0 \] Solution Given: \[ 3\sqrt{5}x^2+25x-10\sqrt{5}=0 \] Product of the coefficient of \(x^2\) and the constant term: \[ (3\sqrt{5})(-10\sqrt{5})=-150 \] We split \(25x\) as \(30x-5x\): \[ 3\sqrt{5}x^2+30x-5x-10\sqrt{5}=0 \] \[ 3\sqrt{5}x(x+2\sqrt{5})-5(x+2\sqrt{5})=0 \] \[ (x+2\sqrt{5})(3\sqrt{5}x-5)=0 \] Therefore, \[ x+2\sqrt{5}=0 \quad \text{or} \quad 3\sqrt{5}x-5=0 \] \[ x=-2\sqrt{5} \] \[

Solve the Following Quadratic Equation by Factorization Question: \[ x^2-(\sqrt{3}+1)x+\sqrt{3}=0 \] Solution Given: \[ x^2-(\sqrt{3}+1)x+\sqrt{3}=0 \] We need two numbers whose product is \(\sqrt{3}\) and sum is \(\sqrt{3}+1\). \[ \sqrt{3}\times 1=\sqrt{3} \] \[ \sqrt{3}+1=\sqrt{3}+1 \] Splitting the middle term: \[ x^2-\sqrt{3}x-x+\sqrt{3}=0 \] Taking common factors: \[ x(x-\sqrt{3})-1(x-\sqrt{3})=0 \] \[ (x-\sqrt{3})(x-1)=0 \] Therefore, \[ x-\sqrt{3}=0 \quad

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x}, \qquad x\ne 0,-1,2 \] Solution Given: \[ \frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x} \] Multiplying both sides by \(10x(x+1)(x-2)\): \[ 20x(x-2)+15x(x+1) =46(x+1)(x-2) \] \[ 20x^2-40x+15x^2+15x =46(x^2-x-2) \] \[ 35x^2-25x =46x^2-46x-92 \] \[ 11x^2-21x-92=0 \] Factorizing: \[ 11x^2-44x+23x-92=0 \] \[ 11x(x-4)+23(x-4)=0 \] \[ (x-4)(11x+23)=0 \] Therefore, \[ x-4=0 \quad \text{or} \quad

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}, \qquad x\ne 1,-1,\frac14 \] Solution Given: \[ \frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1} \] Multiplying both sides by \((x+1)(x-1)(4x-1)\): \[ 3(x-1)(4x-1)+4(x+1)(4x-1) =29(x+1)(x-1) \] \[ 3(4x^2-5x+1)+4(4x^2+3x-1) =29(x^2-1) \] \[ 12x^2-15x+3+16x^2+12x-4 =29x^2-29 \] \[ 28x^2-3x-1 =29x^2-29 \] \[ x^2+3x-28=0 \] Factorizing: \[ x^2+7x-4x-28=0 \] \[ x(x+7)-4(x+7)=0 \] \[ (x+7)(x-4)=0 \] Therefore, \[

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}, \qquad x\ne -1,\frac{1}{3} \] Solution Given: \[ \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1} \] Multiplying both sides by \(2(x+1)(3x-1)\): \[ 6(3x-1)-(x+1)(3x-1)=4(x+1) \] \[ 18x-6-(3x^2+2x-1)=4x+4 \] \[ -3x^2+16x-5=4x+4 \] \[ -3x^2+12x-9=0 \] \[ 3x^2-12x+9=0 \] \[ x^2-4x+3=0 \] Factorizing: \[ x^2-3x-x+3=0 \] \[ x(x-3)-1(x-3)=0 \] \[ (x-3)(x-1)=0 \] Therefore, \[ x-3=0

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{5+x}{5-x}-\frac{5-x}{5+x}=\frac{15}{4}, \qquad x\ne 5,-5 \] Solution Given: \[ \frac{5+x}{5-x}-\frac{5-x}{5+x}=\frac{15}{4} \] Taking LCM on the left side: \[ \frac{(5+x)^2-(5-x)^2}{(5-x)(5+x)} =\frac{15}{4} \] Using the identity \(a^2-b^2=(a-b)(a+b)\): \[ \frac{\big[(5+x)-(5-x)\big]\big[(5+x)+(5-x)\big]} {25-x^2} =\frac{15}{4} \] \[ \frac{(2x)(10)}{25-x^2} =\frac{15}{4} \] \[ \frac{20x}{25-x^2} =\frac{15}{4} \] Cross-multiplying: \[ 80x=15(25-x^2) \] \[ 80x=375-15x^2 \] \[ 15x^2+80x-375=0