Find the values of k for which the quadratic equation, $k x (x - 3) + 9 = 0$ has real equal roots.

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Solution

$k x (x - 3) + 9 = 0 \Rightarrow k x^{2} - 3 k x + 9 = 0 I f r o o t s a r e r e a l a n d e q u a l t h e n D = 0 b^{2} - 4 a c = 0 {(- 3 k)}^{2} - 4 \times k \times 9 = 0 9 k^{2} - 36 k = 0 9 k (k - 4) = 0 ∴ k = 0 o r k = 4$

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Find the values of k for which the quadratic equation, kx(x−3)+9=0 has real equal roots.

kx(x−3)+9=0⇒kx2−3kx+9=0IfrootsarerealandequalthenD=0b2−4ac=0(−3k)2−4×k×9=09k2−36k=09k(k−4)=0∴k=0ork=4

Find the values of k for which the quadratic equation, $k x (x - 3) + 9 = 0$ has real equal roots.

$k x (x - 3) + 9 = 0 \Rightarrow k x^{2} - 3 k x + 9 = 0 I f r o o t s a r e r e a l a n d e q u a l t h e n D = 0 b^{2} - 4 a c = 0 {(- 3 k)}^{2} - 4 \times k \times 9 = 0 9 k^{2} - 36 k = 0 9 k (k - 4) = 0 ∴ k = 0 o r k = 4$