Question:
The length of the sides of a triangle are \(x + 1\), \(9 - x\) and \(5x - 3\). The number of values of x for which the triangle is isosceles is:
The length of the sides of a triangle are \(x + 1\), \(9 - x\) and \(5x - 3\). The number of values of x for which the triangle is isosceles is:
Show Hint
First equate each pair of sides to find candidate x-values, then check the triangle inequality on each candidate since not every algebraic solution gives real side lengths.
Updated On: Jul 13, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Set up the three cases where two sides are equal.
A triangle is isosceles when at least two of its three sides are equal in length. Here the three sides are \(x+1\), \(9-x\) and \(5x-3\). We get an isosceles triangle only when one of these three pairs matches, so there are three equations to check.
Step 2: Case 1, first side equals second side.
\[ x+1 = 9-x \]
\[ 2x = 8 \]
\[ x = 4 \]
At \(x=4\), the sides become \(5\), \(5\) and \(17\). For three lengths to form a real triangle, the sum of any two sides must be strictly greater than the third side. Here \(5+5=10\), which is less than \(17\), so no triangle can close up with these lengths. This case is rejected.
Step 3: Case 2, first side equals third side.
\[ x+1 = 5x-3 \]
\[ 4 = 4x \]
\[ x = 1 \]
At \(x=1\), the sides become \(2\), \(8\) and \(2\). Again check the triangle rule: \(2+2=4\), which is less than \(8\), so these three lengths also fail to form a triangle. This case is rejected too.
Step 4: Case 3, second side equals third side.
\[ 9-x = 5x-3 \]
\[ 12 = 6x \]
\[ x = 2 \]
At \(x=2\), the sides become \(3\), \(7\) and \(7\). Check the triangle rule for all three pairs: \(3+7=10>7\), and \(7+7=14>3\). Every pair satisfies the rule, so this is a genuine, valid isosceles triangle.
Final Answer:
Out of the three algebraic cases, only \(x=2\) gives side lengths that can actually close up into a triangle. The other two values, \(x=1\) and \(x=4\), solve the equal sides equation but fail the basic triangle rule, so they do not correspond to real triangles at all. There is exactly one value of \(x\) for which the figure is a genuine isosceles triangle. \[ \boxed{1} \]
A triangle is isosceles when at least two of its three sides are equal in length. Here the three sides are \(x+1\), \(9-x\) and \(5x-3\). We get an isosceles triangle only when one of these three pairs matches, so there are three equations to check.
Step 2: Case 1, first side equals second side.
\[ x+1 = 9-x \]
\[ 2x = 8 \]
\[ x = 4 \]
At \(x=4\), the sides become \(5\), \(5\) and \(17\). For three lengths to form a real triangle, the sum of any two sides must be strictly greater than the third side. Here \(5+5=10\), which is less than \(17\), so no triangle can close up with these lengths. This case is rejected.
Step 3: Case 2, first side equals third side.
\[ x+1 = 5x-3 \]
\[ 4 = 4x \]
\[ x = 1 \]
At \(x=1\), the sides become \(2\), \(8\) and \(2\). Again check the triangle rule: \(2+2=4\), which is less than \(8\), so these three lengths also fail to form a triangle. This case is rejected too.
Step 4: Case 3, second side equals third side.
\[ 9-x = 5x-3 \]
\[ 12 = 6x \]
\[ x = 2 \]
At \(x=2\), the sides become \(3\), \(7\) and \(7\). Check the triangle rule for all three pairs: \(3+7=10>7\), and \(7+7=14>3\). Every pair satisfies the rule, so this is a genuine, valid isosceles triangle.
Final Answer:
Out of the three algebraic cases, only \(x=2\) gives side lengths that can actually close up into a triangle. The other two values, \(x=1\) and \(x=4\), solve the equal sides equation but fail the basic triangle rule, so they do not correspond to real triangles at all. There is exactly one value of \(x\) for which the figure is a genuine isosceles triangle. \[ \boxed{1} \]
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