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Question: How do you find the roots, real and imaginary of \[y=-2{{x}^{2}}-8x+16-{{\left( x-5 \right)}^{2}}\] ...

How do you find the roots, real and imaginary of y=−2x2−8x+16−(x−5)2y=-2{{x}^{2}}-8x+16-{{\left( x-5 \right)}^{2}} using the quadratic formula?

Explanation

Solution

This is the question of algebraic expression as it consists of variables, coefficients, constants, and mathematical operations such as addition, subtraction, multiplication and division. In the given question of an expression, you just need to simplify the expression by using mathematical operations and evaluate further. The quadratic formula provides the solution for the quadratic equation:
ax2+bx+c=0a{{x}^{2}}+bx+c=0. In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows: Roots of the quadratic equation= −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.

Formula used:
The quadratic formula provides the solution for the quadratic equation:
ax2+bx+c=0a{{x}^{2}}+bx+c=0
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}

Complete step by step answer:
Given quadratic equation,
y=−2x2−8x+16−(x−5)2y=-2{{x}^{2}}-8x+16-{{\left( x-5 \right)}^{2}}
Simplifying the above quadratic equation we get,
As we know that, (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}
y=−2x2−8x+16−(x2−10x+25)y=-2{{x}^{2}}-8x+16-\left( {{x}^{2}}-10x+25 \right)
y=−2x2−8x+16−x2+10x−25y=-2{{x}^{2}}-8x+16-{{x}^{2}}+10x-25
Combining the like terms, we get
y=−3x2+2x−9y=-3{{x}^{2}}+2x-9
Writing the above equation in a standard form, we get
−3x2+2x−9=0-3{{x}^{2}}+2x-9=0
The quadratic formula provides the solution for the quadratic equation:
ax2+bx+c=0a{{x}^{2}}+bx+c=0
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Determine the quadratic equation’s coefficients a, b and c:
The coefficient of the given quadratic equation −3x2+2x−9=0-3{{x}^{2}}+2x-9=0 are,
a = -3
b = 2
c = -9
Plug these coefficient into the quadratic formula:
−b±b2−4ac2a=−2±22−(4×−3×−9)2×−3\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-2\pm \sqrt{{{2}^{2}}-\left( 4\times -3\times -9 \right)}}{2\times -3}
Solve exponents and square root, we get
⇒−2±4−(108)−6\Rightarrow \dfrac{-2\pm \sqrt{4-\left( 108 \right)}}{-6}
Performing any multiplication and division given in the formula,
⇒−2±−104−6\Rightarrow \dfrac{-2\pm \sqrt{-104}}{-6}
Hence, the value of discriminant i.e. d=b2−4ac=−104d=\sqrt{{{b}^{2}}-4ac}=\sqrt{-104} is negative
Therefore there will be no real roots.
The roots of the quadratic equation will be imaginary.
Now, solving further
⇒−2±−104−6\Rightarrow \dfrac{-2\pm \sqrt{-104}}{-6}
We got two values, i.e.
⇒−2+−104−6 and⇒−2−−104−6\Rightarrow \dfrac{-2+\sqrt{-104}}{-6}\ and\Rightarrow \dfrac{-2-\sqrt{-104}}{-6}
As we know that,
i=imaginary number=−1i=imaginary\ number=\sqrt{-1}
Thus,
The imaginary roots will be;
⇒x=−2+i104−6 and⇒−2−i104−6\Rightarrow x=\dfrac{-2+i\sqrt{104}}{-6}\ and\Rightarrow \dfrac{-2-i\sqrt{104}}{-6}
Hence, this is the required answer.

Note: To solve or evaluation these types of expression, we need to know about the:
Solving quadratic equations using the formula
Simplifying radicals
Find prime factors
The general form of quadratic equation is ax2+bx+c=0a{{x}^{2}}+bx+c=0, where a b and c are the numerical coefficients or constants, and the value of xx is unknown one fundamental rule is that the value of a, the first constant can never be zero.