Solve the following equation by factorisation:
a2x2 + (a2 + b2)x + b2 = 0, a ≠ 0.
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Given,
a2x2+(a2+b2)x+b2=0⇒a2x2+a2x+b2x+b2=0⇒a2x(x+1)+b2(x+1)=0⇒(a2x+b2)(x+1)=0 (Factorising left side) ⇒a2x+b2=0 or x+1=0 (Zero-product rule) ⇒a2x=−b2 or x=−1⇒x=−b2a2 or x=−1a^2x^2 + (a^2 + b^2)x + b^2 = 0 \\[0.5em] \Rightarrow a^2x^2 + a^2x + b^2x + b^2 = 0 \\[0.5em] \Rightarrow a^2x(x + 1) + b^2(x + 1) = 0 \\[0.5em] \Rightarrow (a^2x + b^2)(x + 1) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow a^2x + b^2 = 0 \text{ or } x + 1 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow a^2x = -b^2 \text{ or } x = -1 \\[0.5em] \Rightarrow x = -\dfrac{b^2}{a^2} \text{ or } x =-1 \\[0.5em]a2x2+(a2+b2)x+b2=0⇒a2x2+a2x+b2x+b2=0⇒a2x(x+1)+b2(x+1)=0⇒(a2x+b2)(x+1)=0 (Factorising left side) ⇒a2x+b2=0 or x+1=0 (Zero-product rule) ⇒a2x=−b2 or x=−1⇒x=−a2b2 or x=−1
Hence, the roots of given equation are −b2a2,−1.-\dfrac{b^2}{a^2}, -1.−a2b2,−1.
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a2x2 + 2ax + 1 = 0 , a ≠ 0.
x2 - (p + q)x + pq = 0
3x2+10x+73=0\sqrt{3}x^2 + 10x + 7\sqrt{3} = 03x2+10x+73=0
43x2+5x−23=04\sqrt{3}x^2 + 5x - 2\sqrt{3} = 043x2+5x−23=0