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Solve the following quadratic equations by factorization:
\(x^2+(a+\frac{1}{a})x+1 = 0\)
Solve the following quadratic equations by factorization:
\(x^2+(a+\frac{1}{a})x+1 = 0\)
Solve the following quadratic equations by factorization:
\(x^2-x-a(a+1)=0\)
Solve the following quadratic equations by factorization:
\(x^2-x-a(a+1)=0\)
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.
\(x^2-x-a(a+1)=0\)
⇒ x2 – a2 – x – a = 0
⇒ (x + a)(x – a) – 1(x + a) = 0
⇒ (x + a)(x – a – 1) = 0
⇒ x = -a, a + 1
Solve the following quadratic equations by factorization:
\(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\)
Solve the following quadratic equations by factorization:
\(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\)
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.
\(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\)
⇒ x2 –x -√3x + √3 = 0
⇒ x(x – 1) – √3(x – 1) = 0
⇒ (x – √3)(x – 1) = 0
⇒ x = √3, 1
Solve the following quadratic equations by factorization:
\(x^{2}+2ab=(2a+b)x\)
Solve the following quadratic equations by factorization:
\(x^{2}+2ab=(2a+b)x\)
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.
\(x^{2}+2ab=(2a+b)x\)
⇒ x2 – (2a + b)x + 2ab = 0
⇒ x2 – 2ax – bx + 2ab = 0
⇒ x(x – 2a) – b(x – 2a) = 0
⇒ (x – b)(x – 2a) = 0
⇒ x = b, 2a
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.
\(x^2+(a+\frac{1}{a})x+1=0\)
⇒ ax2 +a2x + x + a =0
⇒ ax(x + a) + 1(x + a) = 0
⇒ (ax + 1)(x + a) = 0
⇒ x = -a, -1/a