Solve :
3x2−26x+2=03x^2 - 2\sqrt{6}x + 2 = 03x2−26x+2=0
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⇒3x2−26x+2=0⇒3x2−6x−6x+2=0⇒3x(3x−2)−2(3x−2)=0⇒(3x−2)(3x−2)=0⇒(3x−2)2=0⇒(3x−2)=0⇒3x=2⇒x=23⇒x=2×33×3⇒x=63.\Rightarrow 3x^2 - 2\sqrt{6}x + 2 = 0 \\[1em] \Rightarrow 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \\[1em] \Rightarrow \sqrt{3}x (\sqrt{3}x - \sqrt2) - \sqrt2(\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2) (\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2)^2 = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow \sqrt{3}x = \sqrt2 \\[1em] \Rightarrow x = \dfrac{\sqrt2}{\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{\sqrt2 \times \sqrt3}{\sqrt{3} \times \sqrt3} \\[1em] \Rightarrow x = \dfrac{\sqrt6}{3}.⇒3x2−26x+2=0⇒3x2−6x−6x+2=0⇒3x(3x−2)−2(3x−2)=0⇒(3x−2)(3x−2)=0⇒(3x−2)2=0⇒(3x−2)=0⇒3x=2⇒x=32⇒x=3×32×3⇒x=36.
Hence, x = 63\dfrac{\sqrt6}{3}36.
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2x+9+x=13\sqrt{2x + 9} + x = 132x+9+x=13
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If x2 - 3x + 2 = 0, values of x correct to one decimal place are :
2.0 and 1.0
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3.0 and 2.0
3.0 or 2.0
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5.00 or -1.00
5 or 1
5.0 and -1.0
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