Answer :
To solve for [tex]\(x\)[/tex], we need to determine which side is the hypotenuse in our right-angled triangle. The hypotenuse is always the longest side of the triangle. Given the sides [tex]\(x + 4\)[/tex] cm, [tex]\(x - 1\)[/tex] cm, and 6 cm, it is clear that 6 cm is the largest, so it must be the hypotenuse.
Using the Pythagorean theorem, which states:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
Assign [tex]\(a = x + 4\)[/tex], [tex]\(b = x - 1\)[/tex], and [tex]\(c = 6\)[/tex]. Now substitute these values into the Pythagorean theorem:
[tex]\[ (x + 4)^2 + (x - 1)^2 = 6^2 \][/tex]
Expand each term on the left-hand side:
[tex]\[ (x + 4)^2 = x^2 + 8x + 16 \][/tex]
[tex]\[ (x - 1)^2 = x^2 - 2x + 1 \][/tex]
Add these results together:
[tex]\[ x^2 + 8x + 16 + x^2 - 2x + 1 = 36 \][/tex]
Combine like terms:
[tex]\[ 2x^2 + 6x + 17 = 36 \][/tex]
Now, subtract 36 from both sides to set the equation to 0:
[tex]\[ 2x^2 + 6x + 17 - 36 = 0 \][/tex]
[tex]\[ 2x^2 + 6x - 19 = 0 \][/tex]
This is a quadratic equation, which can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\(a = 2\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -19\)[/tex], let's plug these values into the formula:
First, calculate the discriminant:
[tex]\[ b^2 - 4ac = 6^2 - 4(2)(-19) = 36 + 152 = 188 \][/tex]
Now apply the quadratic formula:
[tex]\[ x = \frac{-6 \pm \sqrt{188}}{2 \cdot 2} = \frac{-6 \pm \sqrt{188}}{4} \][/tex]
Simplify the square root term and the fraction:
[tex]\[ x = \frac{-6 \pm 2\sqrt{47}}{4} => x = \frac{-3 \pm \sqrt{47}}{2} \][/tex]
There are two solutions to this equation:
[tex]\[ x = \frac{-3 + \sqrt{47}}{2} \][/tex]
[tex]\[ x = \frac{-3 - \sqrt{47}}{2} \][/tex]
Since [tex]\(x\)[/tex] must be positive (as it represents a length), we only consider the positive solution:
[tex]\[ x = \frac{-3 + \sqrt{47}}{2} \][/tex]
To find the numerical value, we approximate:
[tex]\[ \sqrt{47} \approx 6.856 \][/tex]
So:
[tex]\[ x \approx \frac{-3 + 6.856}{2} \approx \frac{3.856}{2} \approx 1.928 \][/tex]
Therefore, to three significant figures, the value of [tex]\(x\)[/tex] is:
[tex]\[ x \approx 1.93 \][/tex]
Using the Pythagorean theorem, which states:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
Assign [tex]\(a = x + 4\)[/tex], [tex]\(b = x - 1\)[/tex], and [tex]\(c = 6\)[/tex]. Now substitute these values into the Pythagorean theorem:
[tex]\[ (x + 4)^2 + (x - 1)^2 = 6^2 \][/tex]
Expand each term on the left-hand side:
[tex]\[ (x + 4)^2 = x^2 + 8x + 16 \][/tex]
[tex]\[ (x - 1)^2 = x^2 - 2x + 1 \][/tex]
Add these results together:
[tex]\[ x^2 + 8x + 16 + x^2 - 2x + 1 = 36 \][/tex]
Combine like terms:
[tex]\[ 2x^2 + 6x + 17 = 36 \][/tex]
Now, subtract 36 from both sides to set the equation to 0:
[tex]\[ 2x^2 + 6x + 17 - 36 = 0 \][/tex]
[tex]\[ 2x^2 + 6x - 19 = 0 \][/tex]
This is a quadratic equation, which can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\(a = 2\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -19\)[/tex], let's plug these values into the formula:
First, calculate the discriminant:
[tex]\[ b^2 - 4ac = 6^2 - 4(2)(-19) = 36 + 152 = 188 \][/tex]
Now apply the quadratic formula:
[tex]\[ x = \frac{-6 \pm \sqrt{188}}{2 \cdot 2} = \frac{-6 \pm \sqrt{188}}{4} \][/tex]
Simplify the square root term and the fraction:
[tex]\[ x = \frac{-6 \pm 2\sqrt{47}}{4} => x = \frac{-3 \pm \sqrt{47}}{2} \][/tex]
There are two solutions to this equation:
[tex]\[ x = \frac{-3 + \sqrt{47}}{2} \][/tex]
[tex]\[ x = \frac{-3 - \sqrt{47}}{2} \][/tex]
Since [tex]\(x\)[/tex] must be positive (as it represents a length), we only consider the positive solution:
[tex]\[ x = \frac{-3 + \sqrt{47}}{2} \][/tex]
To find the numerical value, we approximate:
[tex]\[ \sqrt{47} \approx 6.856 \][/tex]
So:
[tex]\[ x \approx \frac{-3 + 6.856}{2} \approx \frac{3.856}{2} \approx 1.928 \][/tex]
Therefore, to three significant figures, the value of [tex]\(x\)[/tex] is:
[tex]\[ x \approx 1.93 \][/tex]
