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For what values of K will the quadratic (4x+1)x2+(k+1)x+1=0 have equal roots. |
Answer» Given quadratic equation: (k+4)x2+(k+1)x+1=0. Since the given quadratic equation has equal roots, its discriminant should be zero. ∴ D = 0 ⇒ (k+1)2−4 × (k+4) × 1=0 ⇒ k2+2k+1−4k−16=0 ⇒ k2−2k−15=0 ⇒k2−5k+3k−15=0 ⇒(k−5)(k+3)=0 ⇒k−5=0 or k+3=0 ⇒k=5 or −3 Thus, the values of k are 5 and −3. For k = 5: (k+4)x2+(k+1) x+1=0 ⇒ 9x2+6x+1=0 ⇒ (3x)2+2(3x)+1=0 ⇒ (3x+1)2=0 ⇒ x=−1/3, −1/3 For k = −3: (k+4)x2+(k+1)x+1=0 ⇒ x2−2x+1=0 ⇒ (x−1)2=0 ⇒ x=1,1 Thus, the equal root of the given quadratic equation is either 1 or −13. |
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