62. If the roots of the quadratic polynomial [tex]\( p(x) = 5x^2 - 26x + 2m - 9 \)[/tex] are reciprocal to each other, then [tex]\( m = \)[/tex] ___



Answer :

Certainly! Let's solve this problem step by step.

We have the quadratic polynomial [tex]\( p(x) = 5x^2 - 26x + 2m - 9 \)[/tex].

Given that the roots of the polynomial are reciprocal of each other, let's denote the roots by [tex]\( r \)[/tex] and [tex]\( \frac{1}{r} \)[/tex].

### Step 1: Understand the relationship between the roots

For a quadratic polynomial [tex]\( ax^2 + bx + c = 0 \)[/tex], the sum of the roots [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] is given by:

[tex]\[ r_1 + r_2 = -\frac{b}{a} \][/tex]

And the product of the roots [tex]\( r_1 \cdot r_2 \)[/tex] is:

[tex]\[ r_1 \cdot r_2 = \frac{c}{a} \][/tex]

### Step 2: Write down the given quadratic polynomial

Here, [tex]\( p(x) = 5x^2 - 26x + 2m - 9 \)[/tex], so we identify:
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = -26 \)[/tex]
- [tex]\( c = 2m - 9 \)[/tex]

### Step 3: Find the product of the roots

Since the roots are reciprocals, the product of the roots [tex]\( r \cdot \frac{1}{r} \)[/tex] is 1.

Using the product of roots formula for quadratic equations:

[tex]\[ r_1 \cdot r_2 = \frac{c}{a} \][/tex]

Substitute [tex]\( c \)[/tex] and [tex]\( a \)[/tex]:

[tex]\[ r \cdot \frac{1}{r} = \frac{2m - 9}{5} \][/tex]

Since [tex]\( r \cdot \frac{1}{r} = 1 \)[/tex]:

[tex]\[ 1 = \frac{2m - 9}{5} \][/tex]

### Step 4: Solve for [tex]\( m \)[/tex]

To find [tex]\( m \)[/tex], we will solve the equation:

[tex]\[ 1 = \frac{2m - 9}{5} \][/tex]

Multiply both sides by 5:

[tex]\[ 5 = 2m - 9 \][/tex]

Add 9 to both sides:

[tex]\[ 5 + 9 = 2m \][/tex]

[tex]\[ 14 = 2m \][/tex]

Divide both sides by 2:

[tex]\[ m = 7 \][/tex]

So, the value of [tex]\( m \)[/tex] is [tex]\( \boxed{7} \)[/tex].

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