Question

Find the value of k such that the polynomial x2 (k + 6)x + 2(2k 1) has sum of its zeros equal to half of their product.

Answer 1

Correct Question:-

Find the value of k such that the polynomial x² - (k + 6)x + 2(2k - 1) has sum of its zeros equal to half of their product.

Answer:-

k = 7

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Solution:-

❍ x² - (k + 6)x + 2(2k - 1)

If I want to expand the above equation then it will looks like as

x² - kx + 6x + 4k - 2

But wait! It is in the form ax² + bx + c.

Where, a = 1, b = -(k + 6) and c = 2(2k - 1)

We have to find the value of k.

We know that,

➭ Sum of zeros = -b/a

→ -[-(k + 6)]/1

→ (k + 6)/1

→ k + 6

➭ Product of zeros = c/a

→ 2(2k - 1)/1

→ 4k - 2

Also given that, Sum of it's zero is equal to half of it's product.

Sum of zeros = 1/2 × Product of zeros

→ k + 6 = 1/2(4k - 2)

→ k + 6 = 1/2 × 2(2k - 1)

→ k + 6 = 2k - 1

→ k - 2k = - 1 - 6

→ - k = -7

→ k = 7

•°• Value of k is 7.

Answer 2

${\large\bf{\mid{\overline{\underline{Correct\:Question:-}}}\mid}}$

Find the value of k such that the polynomial x²- (k + 6)x + 2(2k- 1) has sum of its zeros equal to half of their product.

$\Large\bold\star\underline{\underline\textbf{Concept\:used}}$

→ The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.

→ The product of the roots of a quadratic equation is equal to the constant term (the third term), divided by the leading coefficient.

or we can say that, for Equation ax² + bx + c = 0

→ sum of Roots = -b/a

→ Product of Roots = c/a .

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$\underline {\underline{\LARGE{{\bf{\green{S}}}{\mathfrak{o}}{\mathfrak{\orange{l}}}{\mathfrak{\red{u}}}{\mathfrak{\pink{t}}}{\mathfrak{\purple{i}}}{\mathfrak{\blue{o}}}{\mathfrak{\red{n}}}}}} : \:$

$\bf \: our \: equation \: x^{2} - (k + 6)x + 2(2k- 1) = 0 \:$

$\sf \: here \: a =1 \\ \sf \: b = - (k + 6) \\ \sf \: c =2(2k - 1) \\ \\ \tt \: putting \: values \: we \: get \\ \\ \green{\bf \: sum \: of \: roots = - \frac{b}{a} = (k + 6)} \\ \blue{\bf \: product \: of \: roots = \frac{c}{a} = 2(2k - 1)}$

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According to Question Now,

$\boxed{\tt \: sum \: of \: its \: zeros = \frac{1}{2} \: of \: their \: product} \\ \\ \large\bf \: so \\ \\ \red\longmapsto\:\rm \: k + 6 = \frac{1}{ \cancel2} \times \cancel2(2k - 1) \\ \\ \red\longmapsto\:\rm \: k + 6 = 2k - 1 \\ \\ \red\longmapsto\:\rm \: 2k - k = 6 + 1 \\ \\ \red\longmapsto\: \bold{\boxed{\large{\boxed{\orange{\small{\boxed{\large{\red{\bold{\:k = 7}}}}}}}}}}$

Hence, Value of K will be (7) .

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$\bf\red\bigstar\underline\textbf{ \blue{Extra}\:Brainly\: \pink{Knowledge}}\red\bigstar$

If A•x^2 + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

D = B^2 - 4•A•C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots.

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