Question

if one zeros of the quad . polynomial x^2-5x+k is -4 find k

Answer 1

Step-by-step explanation:

Given -

• One zeroes of the quadratic polynomial x² - 5x + k is -4

To Find -

• Value of k

Now,

• p(x) = x² - 5x + k = 0

Then,

» p(-4) = (-4)² - 5(-4) + k = 0

» 16 + 20 + k = 0

» k = -36

Hence,

The value of k is -36

Verification :-

x² - 5x - 36

By middle term splitt :-

» x² + 4x - 9x - 36

x(x + 4)-9(x + 4)

» (x + 4)(x - 9)

Zeroes are -

x + 4 = 0 and x - 9 = 0

• x = -4 and x = 9

Hence,

Other zero is 9

Answer 2

$\huge{\underline{\bf{\blue{Question:-}}}}$

if one zeros of the quad . polynomial x^2-5x+k is -4 find k.

━━━━━━━━━━━━━━━━━━━━━━━

$\large{\underline{\bf{\pink{Answer:-}}}}$

Value of k = -36.

$\large{\underline{\bf{\purple{Explanation:-}}}}$

$\large{\underline{\bf{\green{Given:-}}}}$

✰ A quadratic polynomial is given as:-

➜⠀⠀⠀⠀p(x) = x² - 5x + k

✰ One zero is given = -4

$\large{\underline{\bf{\green{To\:Find:-}}}}$

we need to find the value of k.

$\huge{\underline{\bf{\red{Solution:-}}}}$

Given zero = -4

so , x+ 4 is a factor of give polynomial.

x = -4

➩⠀⠀⠀⠀⠀p(x) = x² - 5x + k

➩⠀⠀⠀⠀⠀⠀⠀x = -4

putting value of x in given polynomial.

➩⠀⠀⠀⠀ (-4)² - (5 × -4) + k = 0

➩⠀⠀⠀⠀ 16 -(-20) + k = 0

➩⠀⠀⠀⠀16 + 20 + k = 0

➩⠀⠀⠀⠀ 36 + k = 0

➩⠀⠀⠀⠀ k = -36

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⠀⠀⠀⠀Varification :-

➜⠀⠀⠀⠀⠀p(x) = x² - 5x + k

putting value of k and x

➜⠀⠀⠀⠀-4² - 5×(-4) + (-36)

➜⠀⠀⠀⠀16 + 20 -36

➜⠀⠀⠀⠀36 -36

➜⠀⠀⠀⠀0

hence varified