Solve the following equation by factorisation:
6p2 + 11p - 10 = 0
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Given,
6p2+11p−10=0⇒6p2+15p−4p−10=0⇒3p(2p+5)−2(2p+5)=0⇒(2p+5)(3p−2)=0 (Factorising left side) ⇒2p+5=0 or 3p−2=0 (Zero-product rule) ⇒2p=−5 or 3p=2⇒p=−52 or p=236p^2 + 11p - 10 = 0 \\[0.5em] \Rightarrow 6p^2 + 15p - 4p - 10 = 0 \\[0.5em] \Rightarrow 3p(2p + 5) - 2(2p + 5) = 0 \\[0.5em] \Rightarrow (2p + 5)(3p - 2) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2p + 5 = 0 \text{ or } 3p - 2 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow 2p = -5 \text{ or } 3p = 2 \\[0.5em] \Rightarrow p = -\dfrac{5}{2} \text{ or } p = \dfrac{2}{3}6p2+11p−10=0⇒6p2+15p−4p−10=0⇒3p(2p+5)−2(2p+5)=0⇒(2p+5)(3p−2)=0 (Factorising left side) ⇒2p+5=0 or 3p−2=0 (Zero-product rule) ⇒2p=−5 or 3p=2⇒p=−25 or p=32
Hence, the roots of given equation are −52-\dfrac{5}{2}−25, 23\dfrac{2}{3}32.
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3x2 = x + 4
x(6x - 1) = 35
23x2−13x=1\dfrac{2}{3}x^2 - \dfrac{1}{3}x = 132x2−31x=1
3(x - 2)2 = 147
