Question

find the zero of p(x) and verify the relationship of x^2-7x+12​

Answer 1

Given Equation

x² - 7x + 12 = 0

To Find

Zeroes and Verify relationship

Now Take

x² - 7x + 12 = 0

Factorise the equation

x² - 4x - 3x + 12 = 0

x(x - 4) - 3(x - 4) = 0

(x - 4)(x - 3) = 0

x - 4 = 0 and x -3 = 0

x = 4 and x = 3

We get

α = 4  and β = 3

Now we verify the relationship

Sum of zeroes

(α + β) = -b/a

Product of zeroes

(αβ) = c/a

Where

a = 1 , b = -7 and c = 12

Now Put the value

(α + β) = -b/a

( 3+ 4) = -(-7)/1

7 = 7

(αβ) = c/a

(4×3) = 12/1

12 = 12

Hence proved

Answer 2

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${\large{\pmb{\sf{\underline{Explanation...}}}}}$

Given that: We have provided with an equation as x²-7x+12

To find: The zero of p(x) [Polynomial] and verify the relationship of x²-7x+12

Full Solution:

~ Firstly let us find the zeroes of the polynomial x²-7x+12

★ Firstly let us factorise the given expression by using middle term splitting method:

➼ x²-7x+12

➼ x² - 4x - 3x + 12 = 0

➼ x(x-4) -3(x-4) = 0

Taking common terms together firstly then let's solve...

➼ (x-4) (x-3)

➼ x-4 = 0 or x-3 = 0

➼ x = 0+4 or x = 0 + 3

➼ x = 4 or x = 3

➼ α and β = 4 and 3 respectively

• Henceforth, we get the zeroes of the polynomial x²-7x+12

~ Now let's verify the relationship of the polynomial x²-7x+12

★ Firstly by using the formula let us compare:

➼ Sum of zeroes is given by -b/a

➼ Product of zeroes is given by c/a

➼ On comparing we get,

• a as 1
• b as -7
• c as 12

★ Now let's compare!

~ Firstly,

➼ α+β = -b/a

➼ 4+3 = -(-7)/1

➼ 4+3 = +7/1

➼ 7 = 7...LHS = RHS

~ Secondly,

➼ αβ = c/a

➼ 4(3) = 12/1

➼ 4 × 3 = 12/1

➼ 12 = 12...LHS = RHS

• Henceforth, we have verifed the relationship of the polynomial x²-7x+12.

${\large{\pmb{\sf{\underline{Additional \; Knowledge...}}}}}$

Some knowledge about Quadratic Equations -

★ Sum of zeros of any quadratic equation is given by ➝ α+β = -b/a

★ Product of zeros of any quadratic equation is given by ➝ αβ = c/a

★ Discriminant is given by b²-4ac

• Discriminant tell us about there are solution of a quadratic equation as no solution, one solution and two solutions.

★ A quadratic equation have 2 roots

★ ax² + bx + c = 0 is the general form of quadratic equation

★ D > 0 then roots are real and distinct.

★ D = 0 then roots are real and equal.

★ D < 0 then roots are imaginary.

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$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)^{2} \: = \: a^{2} + 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a-b)^{2} \: = a^{2} - 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)(a-b) \: = \: a^{2} - b^{2}}\end{gathered}$