Find the values of x if p + 1 = 0 and x2 + px - 6 = 0.
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Since, p + 1 = 0 it means p = -1.
Given,
x2+px−6=0⇒x2+(−1)x−6=0⇒x2−x−6=0⇒x2−3x+2x−6=0⇒x(x−3)+2(x−3)=0⇒(x+2)(x−3)=0 (Factorising left side) ⇒x+2=0 or x−3=0 (Zero-product rule) x=−2 or x=3x^2 + px - 6 = 0 \\[0.5em] \Rightarrow x^2 + (-1)x - 6 = 0 \\[0.5em] \Rightarrow x^2 - x - 6 = 0 \\[0.5em] \Rightarrow x^2 - 3x + 2x - 6 = 0 \\[0.5em] \Rightarrow x(x - 3) + 2(x - 3) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] x = -2 \text{ or } x = 3x2+px−6=0⇒x2+(−1)x−6=0⇒x2−x−6=0⇒x2−3x+2x−6=0⇒x(x−3)+2(x−3)=0⇒(x+2)(x−3)=0 (Factorising left side) ⇒x+2=0 or x−3=0 (Zero-product rule) x=−2 or x=3
Hence, the values of x are -2 , 3.
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Solve the following equation by factorisation:
x(x−7)=32\sqrt{x(x - 7)} = 3\sqrt{2}x(x−7)=32
Use the substitution y = 3x + 1 to solve for x :
5(3x + 1)2 + 6(3x + 1) - 8 = 0.
Find the values of x if p + 7 = 0, q - 12 = 0 and x2 + px + q = 0.
If x = p is a solution of the equation x(2x + 5) = 3, then find the values of p.
