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To find the value of [tex]\( x \)[/tex] for the right-angled triangle with sides [tex]\( (x+5) \)[/tex] cm, [tex]\( (x-3) \)[/tex] cm, and hypotenuse 9 cm, let's use the Pythagorean theorem. The theorem states:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are the lengths of the legs of the triangle, and [tex]\( c \)[/tex] is the length of the hypotenuse.
Here, [tex]\( a = x + 5 \)[/tex], [tex]\( b = x - 3 \)[/tex], and [tex]\( c = 9 \)[/tex]. Substituting these values into the Pythagorean theorem, we get:
[tex]\[ (x+5)^2 + (x-3)^2 = 9^2 \][/tex]
Now, let's expand and simplify this equation step-by-step.
1. Expand both squared terms:
[tex]\[ (x+5)^2 = x^2 + 10x + 25 \][/tex]
[tex]\[ (x-3)^2 = x^2 - 6x + 9 \][/tex]
2. Add these expanded expressions:
[tex]\[ x^2 + 10x + 25 + x^2 - 6x + 9 = 81 \][/tex]
[tex]\[ 2x^2 + 4x + 34 = 81 \][/tex]
3. Subtract 81 from both sides to set the equation to 0:
[tex]\[ 2x^2 + 4x + 34 - 81 = 0 \][/tex]
[tex]\[ 2x^2 + 4x - 47 = 0 \][/tex]
4. Divide the entire equation by 2 to simplify further:
[tex]\[ x^2 + 2x - 23.5 = 0 \][/tex]
Next, we need to solve this quadratic equation. The solutions to the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -23.5 \)[/tex]. Plugging these values into the quadratic formula:
[tex]\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-23.5)}}{2(1)} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 94}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{98}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 7\sqrt{2}}{2} \][/tex]
[tex]\[ x = -1 \pm \frac{7\sqrt{2}}{2} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x = -1 + \frac{7\sqrt{2}}{2} \][/tex]
[tex]\[ x = -1 - \frac{7\sqrt{2}}{2} \][/tex]
Evaluating these to three significant figures:
[tex]\[ x \approx 3.95 \][/tex]
[tex]\[ x \approx -5.95 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] rounded to three significant figures are:
[tex]\[ x \approx 3.95 \][/tex]
[tex]\[ x \approx -5.95 \][/tex]
Both values are valid solutions to the quadratic equation resulting from the given problem.
[tex]\[ a^2 + b^2 = c^2 \][/tex]
where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are the lengths of the legs of the triangle, and [tex]\( c \)[/tex] is the length of the hypotenuse.
Here, [tex]\( a = x + 5 \)[/tex], [tex]\( b = x - 3 \)[/tex], and [tex]\( c = 9 \)[/tex]. Substituting these values into the Pythagorean theorem, we get:
[tex]\[ (x+5)^2 + (x-3)^2 = 9^2 \][/tex]
Now, let's expand and simplify this equation step-by-step.
1. Expand both squared terms:
[tex]\[ (x+5)^2 = x^2 + 10x + 25 \][/tex]
[tex]\[ (x-3)^2 = x^2 - 6x + 9 \][/tex]
2. Add these expanded expressions:
[tex]\[ x^2 + 10x + 25 + x^2 - 6x + 9 = 81 \][/tex]
[tex]\[ 2x^2 + 4x + 34 = 81 \][/tex]
3. Subtract 81 from both sides to set the equation to 0:
[tex]\[ 2x^2 + 4x + 34 - 81 = 0 \][/tex]
[tex]\[ 2x^2 + 4x - 47 = 0 \][/tex]
4. Divide the entire equation by 2 to simplify further:
[tex]\[ x^2 + 2x - 23.5 = 0 \][/tex]
Next, we need to solve this quadratic equation. The solutions to the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -23.5 \)[/tex]. Plugging these values into the quadratic formula:
[tex]\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-23.5)}}{2(1)} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 94}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{98}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 7\sqrt{2}}{2} \][/tex]
[tex]\[ x = -1 \pm \frac{7\sqrt{2}}{2} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x = -1 + \frac{7\sqrt{2}}{2} \][/tex]
[tex]\[ x = -1 - \frac{7\sqrt{2}}{2} \][/tex]
Evaluating these to three significant figures:
[tex]\[ x \approx 3.95 \][/tex]
[tex]\[ x \approx -5.95 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] rounded to three significant figures are:
[tex]\[ x \approx 3.95 \][/tex]
[tex]\[ x \approx -5.95 \][/tex]
Both values are valid solutions to the quadratic equation resulting from the given problem.
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