If 23\sqrt{\dfrac{2}{3}}32 is a solution of equation 3x2 + mx + 2 = 0, find the value of m.
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Since, 23\sqrt{\dfrac{2}{3}}32 is a solution of equation 3x2 + mx + 2 = 0.
⇒3.(23)2+m(23)+2=0⇒3×23+2+m(23)=0⇒2+2+m(23)=0⇒4+m(23)=0⇒m(23)=−4⇒m=−4×32⇒m=−223⇒m=−26.\Rightarrow 3.\Big(\sqrt{\dfrac{2}{3}}\Big)^2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) + 2 = 0 \\[1em] \Rightarrow 3 \times \dfrac{2}{3} + 2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 2 + 2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 4 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow m\Big(\sqrt{\dfrac{2}{3}}\Big) = -4 \\[1em] \Rightarrow m = -4 \times \sqrt{\dfrac{3}{2}} \\[1em] \Rightarrow m = -2\sqrt{2}\sqrt{3} \\[1em] \Rightarrow m = -2\sqrt{6}.⇒3.(32)2+m(32)+2=0⇒3×32+2+m(32)=0⇒2+2+m(32)=0⇒4+m(32)=0⇒m(32)=−4⇒m=−4×23⇒m=−223⇒m=−26.
Hence, value of m is −26.-2\sqrt{6}.−26.
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The roots of the quadratic equation x2 - 6x - 7 = 0 are :
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1 and 7
-1 or 7
1 and -7
