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2 years ago

The model represents x2 – 9x + 14.

Mathematics

2 answers:

Stels [109]2 years ago

6 0

Factorize the quadratic trinomial $x^2 - 9x + 14$ by the rule:

$ax^2+bx+c=a(x-x_1)(x-x_2), \text{ where } x_1,\ x_2 \text{ are its roots. }$

1. Find the roots:

$D=(-9)^2-4\cdot 14=81-56=25,\ \sqrt{D}=5,\\ \\ x_1=\dfrac{9-5}{2}=2,\ x_2=\dfrac{9+5}{2}=7.$

2. Factorize the polynomial:

$x^2 - 9x + 14=(x-2)(x-7).$

3. Only factor x-2 is given in options, then the correct choice is B.

mario62 [17]2 years ago

4 0

The factored form of x2 - 9x +14
(x-2)*(x-7)
so the answer is B.(x-2)

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Where , d is the common difference

Now , here we are given with ;

${\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}$

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

${: \implies \quad \sf T_{1}+(5-1)d=T_{5}}$

Putting the values and transposing 1st term to RHS , we have ;

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${: \implies \quad \sf d=-\dfrac{12}{4}}$

${: \implies \quad \sf d=-3}$

Now , as we got the common difference , so we can find out the missing terms now ;

${: \implies \quad \sf T_{2}= T_{1}+(2-1)d}$

${: \implies \quad \sf T_{2}= 8 +d}$

${: \implies \quad \sf T_{2}= 8-3}$

${: \implies \quad \bf \therefore \: T_{2}= 5}$

Now

${: \implies \quad \sf T_{3}= T_{1}+(3-1)d}$

${: \implies \quad \sf T_{3}= 8 +2d}$

${: \implies \quad \sf T_{3}= 8-6}$

${: \implies \quad \bf \therefore \: T_{3}= 2}$

Also ,

${: \implies \quad \sf T_{4}= T_{1}+(4-1)d}$

${: \implies \quad \sf T_{4}= 8 +3d}$

${: \implies \quad \sf T_{4}= 8-9}$

${: \implies \quad \bf \therefore \: T_{4}= -1}$

Now , The given table can be written as ;

${\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}$

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

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