Answer :
Sure! Let's solve the equation [tex]\(36x^4 + 9 = 45x^2\)[/tex] step by step to find the real solutions.
1. Rewrite the Equation:
Start by moving all terms to one side of the equation to set it to zero:
[tex]\[
36x^4 - 45x^2 + 9 = 0
\][/tex]
2. Substitution:
Notice that this is a quadratic in form if we let [tex]\(y = x^2\)[/tex]. Rewrite the equation using this substitution:
[tex]\[
36y^2 - 45y + 9 = 0
\][/tex]
3. Solve the Quadratic Equation:
Now, solve the quadratic equation [tex]\(36y^2 - 45y + 9 = 0\)[/tex]. You can use the quadratic formula:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 36\)[/tex], [tex]\(b = -45\)[/tex], and [tex]\(c = 9\)[/tex]. Substitute these values into the formula:
[tex]\[
y = \frac{-(-45) \pm \sqrt{(-45)^2 - 4 \times 36 \times 9}}{2 \times 36}
\][/tex]
[tex]\[
y = \frac{45 \pm \sqrt{2025 - 1296}}{72}
\][/tex]
[tex]\[
y = \frac{45 \pm \sqrt{729}}{72}
\][/tex]
[tex]\[
y = \frac{45 \pm 27}{72}
\][/tex]
So the values for [tex]\(y\)[/tex] are:
[tex]\[
y = \frac{45 + 27}{72} = 1 \quad \text{and} \quad y = \frac{45 - 27}{72} = \frac{1}{2}
\][/tex]
4. Revert Substitution:
Now revert the substitution [tex]\(y = x^2\)[/tex] back to find [tex]\(x\)[/tex].
- For [tex]\(y = 1\)[/tex]:
[tex]\[
x^2 = 1 \quad \Rightarrow \quad x = \pm 1
\][/tex]
- For [tex]\(y = \frac{1}{2}\)[/tex]:
[tex]\[
x^2 = \frac{1}{2} \quad \Rightarrow \quad x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} = \pm \frac{1}{2}
\][/tex]
5. List the Real Solutions:
The real solutions for the original equation are:
[tex]\[
x = -1, \; x = -\frac{1}{2}, \; x = \frac{1}{2}, \; x = 1
\][/tex]
These are the real solutions: [tex]\(-1, -\frac{1}{2}, \frac{1}{2}, \text{and } 1\)[/tex].
1. Rewrite the Equation:
Start by moving all terms to one side of the equation to set it to zero:
[tex]\[
36x^4 - 45x^2 + 9 = 0
\][/tex]
2. Substitution:
Notice that this is a quadratic in form if we let [tex]\(y = x^2\)[/tex]. Rewrite the equation using this substitution:
[tex]\[
36y^2 - 45y + 9 = 0
\][/tex]
3. Solve the Quadratic Equation:
Now, solve the quadratic equation [tex]\(36y^2 - 45y + 9 = 0\)[/tex]. You can use the quadratic formula:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 36\)[/tex], [tex]\(b = -45\)[/tex], and [tex]\(c = 9\)[/tex]. Substitute these values into the formula:
[tex]\[
y = \frac{-(-45) \pm \sqrt{(-45)^2 - 4 \times 36 \times 9}}{2 \times 36}
\][/tex]
[tex]\[
y = \frac{45 \pm \sqrt{2025 - 1296}}{72}
\][/tex]
[tex]\[
y = \frac{45 \pm \sqrt{729}}{72}
\][/tex]
[tex]\[
y = \frac{45 \pm 27}{72}
\][/tex]
So the values for [tex]\(y\)[/tex] are:
[tex]\[
y = \frac{45 + 27}{72} = 1 \quad \text{and} \quad y = \frac{45 - 27}{72} = \frac{1}{2}
\][/tex]
4. Revert Substitution:
Now revert the substitution [tex]\(y = x^2\)[/tex] back to find [tex]\(x\)[/tex].
- For [tex]\(y = 1\)[/tex]:
[tex]\[
x^2 = 1 \quad \Rightarrow \quad x = \pm 1
\][/tex]
- For [tex]\(y = \frac{1}{2}\)[/tex]:
[tex]\[
x^2 = \frac{1}{2} \quad \Rightarrow \quad x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} = \pm \frac{1}{2}
\][/tex]
5. List the Real Solutions:
The real solutions for the original equation are:
[tex]\[
x = -1, \; x = -\frac{1}{2}, \; x = \frac{1}{2}, \; x = 1
\][/tex]
These are the real solutions: [tex]\(-1, -\frac{1}{2}, \frac{1}{2}, \text{and } 1\)[/tex].
