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Question

Question: How do you solve \({p^2} - 4p + 4 = 0\) using the quadratic formula?...

How do you solve p24p+4=0{p^2} - 4p + 4 = 0 using the quadratic formula?

Explanation

Solution

To solve this problem we should know about quadratic equations and quadratic formulas.
Quadratic equation: It is any equation that can be rearranged in standard form as ax2+bx+c=0a{x^2} + bx + c = 0 where xx is represent a unknown and a,bandca,b\,and\,c represent known number, where a0a \ne 0 if a=0a = 0 then it will change into a linear equation.
Quadratic formula: The formula that is used to calculate the solution of a quadratic equation.
Let quadratic equation is ax2+bx+c=0a{x^2} + bx + c = 0 then its solution will be,
x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step by step solution:
As given in question p24p+4=0{p^2} - 4p + 4 = 0 .
Compare it with ax2+bx+c=0a{x^2} + bx + c = 0 to calculate the value of a,bandca,b\,and\,c .
We get, a=1,b=4andc=4a = 1,b = - 4\,and\,c = 4
As we know, the quadratic formula is used to find the roots of a quadratic equation.
So, keeping value in the quadratic equation. We get,
x1,2=b±b24ac2a{x_{1,2}} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
x1,2=(4)±(4)24.1.42.1\Rightarrow {x_{1,2}} = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4.1.4} }}{{2.1}}
By further solving it. We get,
x1,2=4±16162=42=2\Rightarrow {x_{1,2}} = \dfrac{{4 \pm \sqrt {16 - 16} }}{2} = \dfrac{4}{2} = 2
From the above calculation. We get
The roots of the quadratic equation will be the same, that is x=2x = 2 .

Note: We can also do this quadratic equation by factorization by grouping which is also a suitable and easiest method. The quadratic equation is used in our daily life like applied physics, engineering and research and development. When we throw a ball its motion is parabolic whose path will be calculated by using a quadratic equation. So, so much machinery in defense is based on it.