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Find the value of k so that the quadratic equation has equal roots:(k -5)x*2 + 2(k-5)x + 2 = 0 |
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Answer» Step-by-step EXPLANATION: It is given that the expression ax 2 +bx+6=0 doesn't have distinct real roots. ⇒ The graph of the FUNCTION doesn't CROSS the x-axis. ⇒ The expression f(x)=ax 2 +bx+6 is ALWAYS greater than or equal to zero. Hence, f(x)≥0. ⇒ f(2)=a(2) 2 +b(2)+6≥0 ⇒ 4a+2b+6≥0 ⇒ 2a+b+3≥0 ⇒ 2a+b≥−3 my friend hope it will help u |
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