Find the value of k for which x = 1 is root of the equation $x^{2} + k x + 3 = 0$ .

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Solution

$It is given that (x = 1) is a root of (x^{2} + k x + 3 = 0) . Therefore, (x = 1) must satisfy the equation . \Rightarrow {(1)}^{2} + k \times 1 + 3 = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 Hence, the required value of k is - 4 .$

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(i) Find the value of k for which x=1 is a root of the equation $x^{2} + k x + 3 = 0$ . Also, find the other root.
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k

for which the equation

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has both real roots,

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and

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Q. Find that non-zero value of k, for which the quadratic equation

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has equal roots. Hence find the roots of the equation.

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k x (x - 3) + 9 = 0

has real equal roots.

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Find the value of k for which x = 1 is root of the equation x2+kx+3=0.

It is given that (x=1) is a root of (x2 + kx + 3 = 0). Therefore, (x=1) must satisfy the equation.⇒ (1)2 + k × 1 + 3 = 0⇒ k + 4 = 0⇒ k = −4Hence, the required value of k is −4.

Find the value of k for which x = 1 is root of the equation $x^{2} + k x + 3 = 0$ .

$It is given that (x = 1) is a root of (x^{2} + k x + 3 = 0) . Therefore, (x = 1) must satisfy the equation . \Rightarrow {(1)}^{2} + k \times 1 + 3 = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 Hence, the required value of k is - 4 .$