Question

$\huge\sf\underline{\purple{❥}\pink{Q}\orange{U}\blue{E}\red{S}\green{T}\purple{I}\pink{O}\red{N}}$

$\sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$
Find the roots of the above quadratic equation.
And define quadratic polynomial.

Question is for ItzFrozenFlames
Let her answer.. ​​

Answer 1

Answer:

THEOREM ;

The tangent at any point of circle is perpendicular to the radius through the points of contact .

GIVEN

A circle with center O and AB is the tangent at a point P of a circle .

TO PROVE

~~~~~~~~OP ⟂ AB OP⊥AB

CONSTRUCTION

We take a point Q on AB other than P . Then Join OQ

Take a point R on OQ such that R lies on the circle .

~~~~~~~~~~~~~

~~~~~~~~~~~~~

PROOF ;

~~~~~~~~Let \:OQ \:interest \:circle \:at \:R LetOQinterestcircleatR

~~~~~~~~~~~~~ Now ;

{~~~~~~~~~~~~~⟹ OQ = OR + RQ} ⟹OQ=OR+RQ

{~~~~~~~~~~~~~⟹ OQ > OR} ⟹OQ>OR

{~~~~~~~~~~~~~⟹ OR < OQ} ⟹OR<OQ

{~~~~~~~~~~~~~⟹ OP < OQ} ⟹OP<OQ {∴ OR = OP (same \:radius)}∴OR=OP(sameradius)

{~~~~~~~~OP \:is \:the \:short \:line} OPistheshortline {(perpendicular)}(perpendicular)

~~~~~~~~~~~~\boxed{OP ⟂ AB}

OP⊥AB

Answer 2

$\huge\sf\underline{\purple{❥}\pink{Q}\orange{U}\blue{E}\red{S}\green{T}\purple{I}\pink{O}\red{N}}$

$\sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$

Find the roots of the above quadratic equation.

$\huge\sf\underline{\purple{❥}\pink{A}\orange {N}\blue{S}\red{W}\green{E}\purple{R}}$

$x = \frac{ - 5}{ \sqrt{2} } \: or \: x = \sqrt{ - 2}$

Step-by-step explanation:

$\sqrt{2 {x}^{2} } + 7x + 5 \sqrt{2} = 0 \\ \sqrt{2} {x}^{2} + 5x + 2x + 5 \sqrt{2} \\ x( \sqrt{2} x + 5) + \sqrt{2} ( \sqrt{2} x + 5) \\ ( \sqrt{2}x + 5)(x + \sqrt{2} ) \\ \\ roots \: of \: this \: equations \: are \\ the \: values \: for \: which \: : - \\ ( \sqrt{2}x + 5)(x + \sqrt{2} ) = 0 \\ ∴ \: \sqrt{2} x + 5 = 0 \: or \: x + \sqrt{2} = 0$

$so \: \: x = \frac{ - 5}{ \sqrt{2} } \: or \: x = \sqrt{ - 2}$

$\color{red}{About \:Quadratic \: polynomials\: :-}$

➯ A polynomial having degree 2 is called a quadratic polynomial.

➯ The form of quadratic polynomial is

$p(x) = a {x}^{2} + bx + c$

➯ Degree of the quadratic polynomial will be 2.

➯ Variable of the quadratic polynomial will be 1.

$\color{red}{For \:example :-}$

$p(x) = 2 {x}^{2} + 5x + 3 \\ 3 \:➝ \: constant \\ 2 \: and \: 5 \: ➝ \: coefficient \\ {}^{2} \: ➝ \: degree \\ x \:➝ \: variable \:$

➯ In a quadratic polynomial there are 2 zeros because it has degree 2.

$\color{red}{➯ The \: 2\: zeros \:are :-}$

$i) \: \: alpha( \alpha ) \: \\ ii) \: beta \: ( \beta )$

$\color{red}{For \:example :-}$

$p(x) = 2 {x}^{2} + 5 + 3 \\ = 2 {x}^{2} + 2x + 3x + 3 \\ = 2x(x + 1) + 3(x + 1) \\ = (x + 1)(2x + 3) \\ \\ As, \: \: p(x) = 0 \\ \\ ❥∴ \: (x + 1)(2x + 3) = 0 \\ x + 1 = 0 \\ x = - 1 \\ \\ ❥ \: 2x + 3 = 0 \\ 2x = - 3 \\ x = \frac{ - 3}{2} \\ \\ Here, \: \alpha = - 1 \\ \beta = \frac{ - 3}{2}$