Question

Ruth 2x squar+7x+5 ruth 2 middle tarm​

Answer 1

Given:-

• $\sqrt{2}x^2 + 7x + 5\sqrt{2}$

To find:-

• Factorise through splitting the middle term.

Answer:-

Given that,

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

We observe that, (√2) * (5√2) = 10 and the middle term (coefficient of x) is 7.

For splitting the middle term, we need to find two numbers such that their product is 10 and sum is 7.

Through hit and trial, we find those two numbers to be 2 and 5 as 2 * 5 = 10, and 2 + 5 = 7.

So,

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

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

Taking $\sqrt{2}x$ as common for first and second term and $5$ as common for third and fourth term, we get,

$= \sqrt{2}x( x + \sqrt{2} ) + 5(x + \sqrt{2})$

$= (x + \sqrt{2}) ( \sqrt{2} x + 5)$

Hence the factorised form is,

$= \underline{ \underline{(x + \sqrt{2}) ( \sqrt{2} x + 5)}}$

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

Split the middle term :-

$\sf \:  \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\huge \orange{AηsωeR} ✍$

─━─━─━─━─━─━─━─━─━─━─━─━─

Concept of splitting middle term :-

• Factorization of Quadratic polynomials of the form ax² + bx + c.
• (i) In order to factorize ax² + bx + c we have to find numbers p and q such that p + q = b and pq = ac.
• (ii) After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:  \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2}$

$\bf \:❥︎ \: \underline{ Step :- 1.}$

$\sf \:  ⟼In \: order \: to \: factorize ,\sf \:  \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2}$

$\begin{gathered}\begin{gathered}\bf Here : \begin{cases} &\sf{a = \sqrt{2} } \\ &\sf{b = 7}\\ &\sf{c = 5 \sqrt{2} } \end{cases}\end{gathered}\end{gathered}$

$\sf \:  ❥︎ we \: find \: two \: numbers \: p \: and \: q \: such \: that$

$\sf \:  \red{p + q = b = 7}$

$\bf \: \purple{and}$

$\sf \:  \blue{pq =a × c = √(2) × 5 √(2) = 10}$

$\sf \:  Clearly, 2 + 5 = 7 \: and \: 2 × 5 = 10.$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:❥︎ \: \underline{ Step :- 2.}$

$\sf \:  Now, \: we \: split \: the \: middle \: term \: 7x \: as \: 2x + 5x$

$\bf \:so \: \sf \:  \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} \: can \: be \: rewritten \: as$

$\sf \:  = \sqrt{2} {x}^{2} + 2x + 5x + 5 \sqrt{2}$

$\sf \:  = \sqrt{2} {x}^{2} + ( \sqrt{2} \times \sqrt{2} ) x + 5x + 5 \sqrt{2}$

$\sf \:  = \sqrt{2} x(x + \sqrt{2} ) + 5(x + \sqrt{2)}$

$\sf \:  = (x + \sqrt{2} )( \sqrt{2} x + 5)$

─━─━─━─━─━─━─━─━─━─━─━─━─

${\boxed{\boxed{\sf\implies \:{ \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = (x + \sqrt{2} )( \sqrt{2} x + 5) }}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─