Question

if one of the roots of x square + kx - 6 is equal to zero is 3 then find the value of k also find the other route​

Answer 1

Step-by-step explanation:

one of the root=3

x²+kx-6=0

3²+3k-6=0

9-6+3k=0

3+3k=0

1+k=0

k=-1

x²+kx-6=0

x²-x-6=0

x²-3x+2x-6=0

x(x-3)+2(x-3)=0

(x-3)(x+2)=0

roots of the quadratic equation is -2 and 3

Answer 2

$\Huge {\sf{Answer :}}$

One root is 3 .

We have to find other root and value of k

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Equation = x² + kx - 6 = 0

Put x = 3 in equation,

(3)² + k(3) -6 = 0

⇒9 + 3k - 6 = 0

⇒3k + 3 = 0

⇒3k = -3

⇒k = -3/3

⇒k = -1

✪ Value of k is -1 ✪

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For another root put value of k in equation.

x² -1x - 6 = 0

Now split the middle term

⇒x² -3x + 2x - 6 = 0

⇒x(x - 3) + 2(x - 3) = 0

⇒(x - 3)(x + 2) = 0

So,

✪ Roots are x = 3 , -2

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