Answer :
Let's analyze the equations one by one to determine which one is in quadratic form.
1. Equation 1:
[tex]\[ 2(x + 5)^3 + 8x + 5 + 6 = 0 \][/tex]
Expanding and simplifying this equation, you can see that it is a polynomial of degree 3 in terms of [tex]\(x\)[/tex]. A quadratic equation must be of degree 2.
2. Equation 2:
[tex]\[ x^6 + 6x^4 + 8 = 0 \][/tex]
We can substitute [tex]\( y = x^2 \)[/tex]. Therefore, [tex]\(x^4\)[/tex] becomes [tex]\(y^2\)[/tex] and [tex]\(x^6\)[/tex] becomes [tex]\(y^3\)[/tex], resulting in:
[tex]\[ y^3 + 6y^2 + 8 = 0 \][/tex]
This is a cubic equation in [tex]\( y \)[/tex], rather than a quadratic equation.
3. Equation 3:
[tex]\[ 7x^6 + 36x^3 + 5 = 0 \][/tex]
We can substitute [tex]\( y = x^3 \)[/tex]. Therefore, [tex]\(x^6\)[/tex] becomes [tex]\(y^2\)[/tex], leading to:
[tex]\[ 7y^2 + 36y + 5 = 0 \][/tex]
This is a quadratic equation in [tex]\( y \)[/tex]. So, when we rewrite it in terms of [tex]\( y = x^3 \)[/tex], it takes on the form of a quadratic equation.
4. Equation 4:
[tex]\[ 4x^9 + 20x^3 + 25 = 0 \][/tex]
We can substitute [tex]\( z = x^3 \)[/tex]. Therefore, [tex]\(x^9\)[/tex] becomes [tex]\(z^3\)[/tex], leading to:
[tex]\[ 4z^3 + 20z + 25 = 0 \][/tex]
This is a cubic equation in [tex]\( z \)[/tex], rather than a quadratic equation.
Based on the above analysis, the equation quadratic in form is:
[tex]\[ 7x^6 + 36x^3 + 5 = 0 \][/tex]
Thus, the answer is:
[tex]\[ 7x^6 + 36x^3 + 5 = 0 \][/tex]
1. Equation 1:
[tex]\[ 2(x + 5)^3 + 8x + 5 + 6 = 0 \][/tex]
Expanding and simplifying this equation, you can see that it is a polynomial of degree 3 in terms of [tex]\(x\)[/tex]. A quadratic equation must be of degree 2.
2. Equation 2:
[tex]\[ x^6 + 6x^4 + 8 = 0 \][/tex]
We can substitute [tex]\( y = x^2 \)[/tex]. Therefore, [tex]\(x^4\)[/tex] becomes [tex]\(y^2\)[/tex] and [tex]\(x^6\)[/tex] becomes [tex]\(y^3\)[/tex], resulting in:
[tex]\[ y^3 + 6y^2 + 8 = 0 \][/tex]
This is a cubic equation in [tex]\( y \)[/tex], rather than a quadratic equation.
3. Equation 3:
[tex]\[ 7x^6 + 36x^3 + 5 = 0 \][/tex]
We can substitute [tex]\( y = x^3 \)[/tex]. Therefore, [tex]\(x^6\)[/tex] becomes [tex]\(y^2\)[/tex], leading to:
[tex]\[ 7y^2 + 36y + 5 = 0 \][/tex]
This is a quadratic equation in [tex]\( y \)[/tex]. So, when we rewrite it in terms of [tex]\( y = x^3 \)[/tex], it takes on the form of a quadratic equation.
4. Equation 4:
[tex]\[ 4x^9 + 20x^3 + 25 = 0 \][/tex]
We can substitute [tex]\( z = x^3 \)[/tex]. Therefore, [tex]\(x^9\)[/tex] becomes [tex]\(z^3\)[/tex], leading to:
[tex]\[ 4z^3 + 20z + 25 = 0 \][/tex]
This is a cubic equation in [tex]\( z \)[/tex], rather than a quadratic equation.
Based on the above analysis, the equation quadratic in form is:
[tex]\[ 7x^6 + 36x^3 + 5 = 0 \][/tex]
Thus, the answer is:
[tex]\[ 7x^6 + 36x^3 + 5 = 0 \][/tex]
