Solve the following equation by factorisation:
x(2x + 5) = 3
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Given,
x(2x+5)=3⇒2x2+5x=3⇒2x2+5x−3=0 (Writing as ax2+bx+c=0)⇒2x2+6x−x−3=0⇒2x(x+3)−1(x+3)=0⇒(2x−1)(x+3)=0 (Factorising left side) ⇒2x−1=0 or x+3=0 (Zero product rule) ⇒2x=1 or x=−3⇒x=12 or x=−3x(2x + 5) = 3 \\[0.5em] \Rightarrow 2x^2 + 5x = 3 \\[0.5em] \Rightarrow 2x^2 + 5x - 3 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[0.5em] \Rightarrow 2x^2 + 6x - x - 3 = 0 \\[0.5em] \Rightarrow 2x(x + 3) - 1(x + 3) = 0 \\[0.5em] \Rightarrow (2x - 1)(x + 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x - 1 = 0 \text{ or } x + 3 = 0 \text{ (Zero product rule) } \\[0.5em] \Rightarrow 2x = 1 \text{ or } x = -3 \\[0.5em] \Rightarrow x = \dfrac{1}{2} \text{ or } x = -3 \\[0.5em]x(2x+5)=3⇒2x2+5x=3⇒2x2+5x−3=0 (Writing as ax2+bx+c=0)⇒2x2+6x−x−3=0⇒2x(x+3)−1(x+3)=0⇒(2x−1)(x+3)=0 (Factorising left side) ⇒2x−1=0 or x+3=0 (Zero product rule) ⇒2x=1 or x=−3⇒x=21 or x=−3
Hence, the roots of given equation are 12\dfrac{1}{2}21, -3.
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If 23\dfrac{2}{3}32 and -3 are roots of the equation px2 + 7x + q = 0, find the values of p and q .
x2 - 3x - 10 = 0
3x2 - 5x - 12 = 0
21x2 - 8x - 4 = 0
