Question

Find the roots of the quadratic equation
(4p - 1)² = 4p².
​

Answer 1

We are given with a quadratic equation, but before solving the equation, let's convert it to the standard form of a quadratic equation i.e ax² + bx + c = 0 first ;

${:\implies \quad \sf (4p-1)^{2}=4p^{2}}$

Using the identity (a - b)² = a² + b² - 2ab, we will be having

${:\implies \quad \sf (4p)^{2}+(1)^{2}-2\times 4p\times 1=4p^{2}}$

${:\implies \quad \sf 16p^{2}-8p-4p^{2}+1=0}$

${:\implies \quad \sf 12p^{2}-8p+1=0}$

So, now if we compare it with the standard form of quadratic equation, we will be having a = 12, b = -8 and c = 1, so now Discriminant (D) = b² - 4ac = (-8)² - 4 × 12 × 1 = 64 - 48 = 16, now as D > 0, so two real and distinct roots exist, so now by quadratic formula we know that ;

• ${\boxed{\bf{x=\dfrac{-b\pm \sqrt{D}}{2a}}}}$

So, now solving for x, we will be having ;

${:\implies \quad \sf x=\dfrac{-(-8)\pm \sqrt{16}}{2\times 12}}$

${:\implies \quad \sf x=\dfrac{8\pm 4}{24}}$

${:\implies \quad \sf x=\dfrac{8+4}{24}\:\:,\:\: \dfrac{8-4}{24}}$

${:\implies \quad \sf x=\dfrac{12}{24}\:\:,\:\: \dfrac{4}{24}}$

${:\implies \quad \boxed{\bf{x=\dfrac{1}{2}\:\:,\:\: \dfrac{1}{6}}}}$