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Answer 1

Question:

Find the value of k, such that the number given against each equation is one of its roots;

(i) x² + kx -21 = 0; 3

(ii) x² + kx -5 = 0; 1

(iii) x² +12x + k = 0; -5

Concept:

If one zero is given, then substitute the value of the given zero in the given quadratic equation and then find the value of k.

★ Here we go

• (i) x² + kx -21 = 0

f(x) = x² + kx -21 = 0

Substituting x as 3

f(3) =3² + 3k -21 = 0

➟ 9 + 3k -21 = 0

➟ 3k -12 = 0

3k = 12

k = 12 ÷ 3

k = 4

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• (ii) x² + kx -5 = 0

f(x) = x² + kx -5 = 0

Substituting x as 1

f(1) = 1² + k -5 = 0

➟1 +k -5 = 0

➟k -4 = 0

k = 4

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• (iii) x² +12x + k = 0

f(x) = x² +12x + k = 0

Substituting x as -5

f(-5) = (-5)² +12 x(-5) + k = 0

n25 -60 + k = 0

➟-35 + k = 0

k = 35

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Henceforth,

(i) k = 4

(ii) k = 4

(iii) k = 35

Thank Üh