Question

r^2-13r+36=0.
find the value of r?​

Answer 1

Required Answer:-

Given:

• r² - 13r + 36 = 0

To find:

• The values of r.

Solution:

We have,

➡ r² - 13r + 36 = 0

We have to split -13 into two parts such that sum of those parts is -13 and their product is 36.

We found that,

• -4 - 9 = -13
• (-4) × (-9) = 36

So,

➡ r² - 13r + 36 = 0

➡ r² - 4r - 9r + 36 = 0

➡ r(r - 4) - 9(r - 4) = 0

➡ (r - 4)(r - 9) = 0

By zero product rule,

➡ Either (r - 4) = 0 or (r - 9) = 0

So,

➡ r = 4,9

★ Hence, the values of r are 4 and 9.

Answer:

• The values of r are 4 and 9.

Verification:

Let us verify our result.

When r = 4,

r² - 13r + 36

= (4)² - 13 × 4 + 36

= 16 - 52 + 36

= 52 - 52

= 0 which is true.

Again, when r = 9

r² - 13r + 36

= (9)² - 13 × 9 + 36

= 81 + 36 - 117

= 117 - 117

= 0 which is also true.

Hence, 4 and 9 are the zeros of the equation. (Verified)

Answer 2

Answer:

First, we need to factor the quadratic expression by playing with multipliers for 36 (1x36, 2x18, 3x12, 4x9, 6x6) which add to 13:

(r−9)(r−4)=0

Now we can solve each term for

0 :

Solution 1)

r−9=0

r−9+9=0+9

r−0=9

r=9

Solution 2)

r−4=0

r−4+4=0+4

r−0=4

r=4