Question

Find the zeros of the quadratic polynomial and verify the relationship between the zeros and Co efficients t²-15

Answer 1

Given

• Quadratic Polynomial → t² - 15

To find

• Find the zeros of the quadratic polynomial and verify the relationship between the zeros and Co efficients t²-15

Solution

p(t) = t² - 15

Apply identity :

(a² - b²) = (a + b)(a - b)

→ (t)² - (√15)² = 0

→ (t + √15)(t - √15)

Either

→ t + √15 = 0

→ t = - √15

Or

→ t - √15 = 0

→ t = √15

Hence, -√15 and √15 are the zeros of given polynomial

Verification

★ Sum of zeros ★

→ -√15 + √15 = 0 = -b/a = -(coefficient of t)/(coefficient of t²)

★ Product of zeros ★

→ -√15 × √15 = -15 = c/a = (constant term)/(coefficient of t²)

Hence, verified

Answer 2

✯✯ QUESTION ✯✯

Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and Co efficients t²-15.

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✰✰ ANSWER ✰✰

$\longmapsto\tt{{t}^{2}-15=0}$

$\longmapsto\tt{{t}^{2}-\sqrt{{(15)}^{2}}=0}$

➥Using Identity : -

$\longmapsto\tt{{a}^{2}-{b}^{2}=(a-b)(a+b)}$

$\longmapsto\tt{(t-\sqrt{15)}\:(t+\sqrt{15)}}$

➥Now ,

$\longmapsto\tt{{t-\sqrt{15}=0}}$

$\longmapsto\tt\bold{t=\sqrt{15}}$

$\longmapsto\tt{{t+\sqrt{15}}=0}$

$\longmapsto\tt\bold{t=-\sqrt{15}}$

☛So , √15 and -√15 are the zeroes of polynomial t²-15...

➮Here : -

• a = 1
• b = 0
• c = -15

★Sum of Zeroes : -

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

$\longmapsto\tt{\sqrt{15}+(-\sqrt{15)}=\dfrac{0}{1}}$

$\longmapsto\tt{\sqrt{15}-\sqrt{-15}=0}$

$\longmapsto\tt\bold{0=0}$

$\orange\longmapsto\:\large\underline{\boxed{\bf\red{L.H.S}\blue{=}\pink{R.H.S}}}$

★Product of Zeroes : -

$\longmapsto\tt{\alpha\beta=\dfrac{c}{a}}$

$\longmapsto\tt{\sqrt{15}\times(-\sqrt{15)}=\dfrac{-15}{1}}$

$\longmapsto\tt\bold{-15=-15}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{L.H.S}\orange{=}\purple{R.H.S}}}$

✺ HENCE VERIFIED ✺