Question

If the squared difference of the zeros of the quadratic equation f(x)=x^2+px+45 is equal to 144, find the vaue of p .​

Answer 1

Answer:

p = 18 or p = (-18)

Step-by-step explanation:

We are given,

f(x) = x² + px + 45

ax² + bx + c = 0

where a = 1, b = p, c = 45

Now, according to the Question,

Squared difference of the zeroes of the Quadratic equation f(x) = 144

So,

Let the zeroes be 'a' and 'b'.

Thus, according to the Question,

(a - b)² = 144

Now,

We know that,

Sum of zeroes = -b/a

a + b = -p/1

(a + b) = -p ------- 1

Also,

Product of zeroes = c/a

a × b = 45/1

ab = 45 ------- 2

There is a formula to get (a - b)² from (a + b)²

Let me just prove it for you......

Now,

We know that,

(a - b)² = a² - 2ab + b² ------- 3

(a + b)² = a² + 2ab + b² -------- 4

Hence,

Subtracting eq.3 and eq.4

(a - b)² - (a + b)² = (a² - 2ab + b²) - (a² + 2ab + b²)

Opening the brackets,

(a - b)² - (a + b)² = a² - 2ab + b² - a² - 2ab - b²

(a - b)² - (a + b)² = -4ab

∴ (a - b)² = (a + b)² - 4ab

Thus,

If (a - b)² = 144

then, from above,

(a - b)² = (a + b)² - 4ab = 144

So,

(a + b)² - 4ab = 144

From eq.1 and eq.2 we get,

(-p)² - 4(45) = 144

p² - 180 = 144

p² = 144 + 180

p² = 324

p = ±√324

p = ± 18

Hence,

p = 18 or p = (-18)

Hope it helped and believing you understood it........All the best