Question

if the sum of roots of quadratic equation x^2 - (k + 6)x + 2(2k - 1) = 0 is equal to half of their product then find the value of k​

Answer 1

The given quadratic equation is :-

x² - (k + 6)x + 2(2k - 1) = 0

We know,

◘ Sum of zeros, S = -(Coefficient of x) / Coefficient of x²

→ S = -b/a

◘ Product of zeros, P = Constant term / Coefficient of x²

→ P = c/a

_____________________

Sum = -b/a

$\sf{\longrightarrow S=\dfrac{-[-(k+6)]}{1} }$

$\sf{\longrightarrow S=k+6 }$

_____________________

Product = c/a

$\sf{\longrightarrow P=\dfrac{2(2k-1)}{1} }$

$\sf{\longrightarrow P=2(2k-1) }$

_____________________

A/q,

→ S = P/2

Putting the values :-

$\sf{\longrightarrow k+6=\dfrac{2(2k-1)}{2} }$

$\sf{\longrightarrow k+6=2k-1}$

$\sf{\longrightarrow 6+1=2k-k}$

$\sf{\longrightarrow 7=k}$

$\bf{\longrightarrow k=7}$

Therefore, the value of k is 7.

Answer 2

$\underline{\red{Given:-}}$

• The sum of roots of quadratic equation x² - (k + 6)x + 2(2k - 1) = 0 is equal to half of their product .

$\underline{\blue{To\:Find:-}}$

• The value of k

$\underline{\pink{SOLUTION:-}}$

Formula :-

$✧ \sf \orange{Sum \:of \:zeros, S = -(Coefficient \:of \:x) / Coefficient \:of \:x²}$

→ $\boxed{S = -b/a}$

$✧ \sf \orange{Product \:of \:zeros,\: P = Constant\: term / Coefficient \:of \:x²}$

→$\boxed{ P = c/a}$

Now,

Sum = -b/a

$\sf{:\implies S=\dfrac{-[-(k+6)]}{1} }$

$\sf{:\implies S=k+6 }$

And,

Product = c/a

$\sf{➜ P=\dfrac{2(2k-1)}{1} }\\\\\\ \sf{➜ P=2(2k-1) }$

A/q,

→ S = P/2

Substituting the values :-

$\sf{:\implies k+6=\dfrac{2(2k-1)}{2} }\\\\\\</p><p> \sf{:\implies k+6=2k-1}$

$\sf{:\implies 6+1=2k-k}\\\\\\</p><p>\sf{:\implies 7=k}\\\\</p><p>\underline{\boxed{\pink{\bf{\longrightarrow k=7}}}}$

Therefore, the value of k is 7.

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